Post on 28-Mar-2018
I
High Dimensional Euclidean Geometry
Wroted by Li Honglu
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Informed by the author of this book
June 7,2004
Brief Introduction
This book is composed of three parts:oblique axes transform,oblique axes draughting and
high-dimensional spacial analytic geometry.It has totally nine chapters:Relation and property
determined by singular linear transform;"Relation" method and preferred n-dimensional coordinate
system;Relation between figure and digit in preferred n-dimensional coordinate system;Figure
shape of preferred n-dimensional coordinate system;Figure making of preferred n-dimensional
coordinate system;Interleaving and distance between two linear figures;Included angle between two
linear figures and its linear solution;Jianshi solution of the included angle between two linear
figures;Application of High Dimensional Euclidean Geometry .This book is truly a monograph of
high-dimensional spacial analytic geometry,it is based on (but different from) euclidean space theory
in linear algebra.
This book has reference value for students and teachers of relative speciality in colleges and
universities and researchers working on operational research and graphic theory.
II
Catalog
High Dimensional Euclidean Geometry I
Brief Introduction ..................................................................................................................... I
Prefaces .....................................................................................................................................V
The First Chapter: Relation and property determined[1]
by singular linear transform .... 1
§1 Property and relations between image and image source of singular linear transform. 1
1.1 Concept of "relation"............................................................................................ 1
1.2 Relations between image and image source of singular linear transform ............ 2
1.3 Property of relations between image and image source under singular linear
transform .................................................................................................................... 2
§2 Operation rules of elements in relationσ ....................................................................... 3
2.1 Linear operation rules........................................................................................ 3
2.2 Transposition (of terms) rules ........................................................................... 3
2.3 Element exchange rules..................................................................................... 5
§3 Relations between coordinates of vectors under singular linear transform................... 7
Exercises .......................................................................................................................... 13
The Second Chapter: "Relation" method and preferred n-dimensional coordinate system
.................................................................................................................................................. 14
§1 "Relation" method and examples of preferred 4-dimensional and 5-dimensional
coordinate system............................................................................................................. 14
1.1 Examples of building preferred 4-dimensional coordinate system ................. 14
1.2 Examples of building preferred five-dimensional coordinate system ............. 16
§2 Oblique axes transform and preferred n-dimensional coordinate system ................... 17
2.1 Oblique axes transform and building of preferred n-dimensional system....... 17
2 . 1 . 1 Methods and steps of building preferred n-dimensional
system——oblique axes transform ..................................................................... 17
2.1.2 Structure of preferred n-dimensional system........................................ 19
2.1.3 Kinds of preferred n-dimensional system............................................. 20
§3 Property of preferred n-dimensional system ............................................................... 21
3.1 Relation between preferred n-dimensional system and oblique axes transform 21
3.2 The "preference" of preferred coordinates system .......................................... 22
§4 Punctiform figures in preferred n-dimensional system—generic point ...................... 23
4.1 Generic point、projection trace and opposite trace ........................................ 23
4.1.1 Concept of generic point ...................................................................... 23
4.1.2 Projection trace and opposite trace of generic point............................. 23
4.2 Property of generic points .......................................................................... 24
4.2.1 Shape of generic points ........................................................................ 24
4.2.2 The uniqueness of generic points relative to upright axis .................... 25
4.2.3 Uniqueness of projection trace or opposite trace.................................. 25
Exercises .......................................................................................................................... 26
The Third Chapter: Relation between figures and numbers in preferred n-dimensional
system ...................................................................................................................................... 28
§1 Translation trace of generic point—generic curved surface and generic curve........... 28
1.1 Concepts of generic curved surface and generic curve ................................... 28
III
1.2 Dimensions of generic curved surface and generic curve—dimension theorem29
1.3 Kinds of generic curved surface and generic curve......................................... 30
§2 Graphic rules of preferred n-dimensional system—three kind of graphic methods.... 30
2.1 Concepts of three graphic methods ................................................................. 30
2.2 Indirect graphic method and ordinary graphic method ................................... 31
2.2.1 Indirect graphic method........................................................................ 32
2.2.2 Common generic of points ................................................................... 33
2.2.3 Ordinary figured method...................................................................... 35
§3 Direct graphic method................................................................................................. 36
3.1 Single principal overlapping direction、oblique number and oblique coordinate
.................................................................................................................................. 36
3.2 Coordinate transform of points........................................................................ 37
3.3 The principle of direct graphic method ........................................................ 38
3.4 Drawing and distinguishing steps of direct graphic method .............................. 39
Exercises .......................................................................................................................... 42
The Fourth Chapter: Shape of figures in preferred n-dimensional system ...................... 44
§1 Shape of linear figures ................................................................................................ 44
1.1 Generic planes and their shape........................................................................ 44
1.2 Generic straight lines and their shape.............................................................. 46
1.3 Shape of many generic planes' intersection.................................................. 51
§2 Shape of non-linear figures ......................................................................................... 56
Exercises .......................................................................................................................... 58
The Fifth Chapter: Making of figures in preferred n-dimensional system....................... 59
§1 Cutting-trace method................................................................................................... 59
§2 Leading-axis method................................................................................................... 61
§3 Synthetical graphic method......................................................................................... 62
The Sixth Chapter: Interleaving and distance between two linear figures. ...................... 73
§1 Interleaving between two linear figures ...................................................................... 73
§2 Orthodromic space and normal space ......................................................................... 77
§3 Exterior sumof two linear figures................................................................................ 81
3.1 Concepts of exterior sumof two linear figures ................................................ 81
3.2 Property of outer sum...................................................................................... 82
3.3 Equation of outer sum ........................................................................................ 83
§4 Distance between two linear figures ........................................................................... 84
4.1 Distance between two parallel figures................................................................ 84
4.2 Distance between two mutually interleaving figures ...................................... 85
Exercises .......................................................................................................................... 86
The Seventh Chapter: Included angle questions and their linear solutions between two
linear figures ........................................................................................................................... 87
§1 Diversity of included angle questions bertween two linear figures of higher space ... 87
1.1 Nonuniqueness of included angle amounts ........................................................ 87
1.1.1 Two kind of projection methods................................................................. 87
1.1.2 Defination of included angle between two linear figures........................... 89
1.1.3 Common vector and uncommon vector——sameness of two linear figures'
IV
dimension ............................................................................................................ 92
1.2 Nonuniqueness of the solution of included angle questions............................... 94
§2 Linear solution of included angle questions between two linear figures .................... 95
2.1 Positive angle method ................................................................................... 95
2.2 Complementary angle method.......................................................................... 100
Exercises ........................................................................................................................ 104
The Eighth Chapter: Jianshi solution solution of included angle questions between
twolinear figures................................................................................................................... 105
§1 Orthogonal transform and principal axis questions................................................... 105
1.1 Orthogonal matrix and orthogonal transform................................................... 105
1.2 Concepts and property of exterior product among vectors—orthogonal-unitization
of vectors................................................................................................................ 106
1.3 Principal axis questions [14] ...........................................................................111
§2 Projective generic elliptic cylinder and projection of generic circle ..........................113
2.1 Projection of sufficient-order generic circle......................................................113
2.2 Projection of difficient-ordergeneric circle .......................................................115
§3 Jianshi solution solution of included angle questions between two linear figures .....118
3.1 Principle and steps of Jianshi solution solution.................................................118
3.2 Included angle questions between two planes...................................................119
3.3 Included angle questions between other linear figures..................................... 125
3.4 Other questions of Jianshi solution principle——exterior product method and
included angle ........................................................................................................ 129
Exercises ........................................................................................................................ 134
The Ninth Chapter: The application of High Dimensional Euclidean Geometry .......... 136
§1 Application of High Dimensional Euclidean Geometry in linear programming....... 136
1.1 Example one..................................................................................................... 136
1.2 Theory, steps and assumption of graphical method in preferred
n-dimensionalsystem.............................................................................................. 140
1.2.1 Theory ...................................................................................................... 140
1.2.2 Method、steps ......................................................................................... 144
1.3 An assumption.................................................................................................. 147
§2 Application of High Dimensional Euclidean Geometry in non-linear programming 150
2.1 Example 2......................................................................................................... 151
2.2 Methods and steps ............................................................................................ 153
2.2.1 Determination of search direction ............................................................ 153
2.2.2 Determination of search range ................................................................. 155
2.3 The case that objective generic curved surface is solid.................................... 158
References...................................................................................................................... 168
Postscript........................................................................................................................ 169
Solutions to part exercises.............................................................................................. 170
Indexes of special terms or symbols .............................................................................. 171
Appendix:solution process of part exercises ....................................................... 173
V
Prefaces
High Dimensional Euclidean Geometry is the author's accomplishment of seventeen years'
effort. During this time, he educated himself while he studied it ,his accomplishment was reflected in
this book.
This book is composed of three parts:"oblique axes transform",oblique axes draughting and
high-dimensional spacial and analytic geometry.
"Oblique axes transform"(the auther named it "relation method") is the theoretical basis of the
whole book. It formed another linear transform on the basis of the relation between image and image
source which is determined by singular linear transform.
Under singular linear transform, relevant concepts,properties and operation rules of relations
between image and image source are studied in depth in this book,it also overcame obstacles
appeared in "relation" method application.
"Oblique axes draughting" is the basic method in this book. On the basis of using oblique axes
transform to establish preferred n-dimensional coordinate system(an analogue figure of
n-dimensional spatial rectangular axis),this book advanced methods to study the relation between
graphics and its relevant algebraic equation in coordinate system. This includes three kind of graphic
methods(direct graphic method, indirect graphic method and general graphic method)which are the
core contents, it summarized all graphic rules of preferred n-dimensional coordinate system:graphic
principles and graphic methods of a random graph and recognition methods of objects be
graphed,etc.Besides, it also includes points having common generic theory、dimension theorem and
cutting-trace method、 leading-axis method、synthetical graphic method.All of this proceeded
beneficial study to preliminarily formed a more systematic, construction reasonable,convenient and
practical,novel graph theory system. The graphic methods are easily to learn and memerize, readers
who know engineering linear algebra can easily master it.
High dimensional spacial analytical geometry is the principal part of High Dimensional
Euclidean Geometry.Oblique axes draughting naturally extendes some relative concepts of
three-dimensional space to high space,then beginners can easily master it. But oblique axes transform
takes questions of point coincidence as a breakthrough, then large nembers of geometric problems
of high space can be readily solved.
VI
Readers who have read this book will find that although linear algebra has geometry concepts of
euclidean space,vector and coordinate etc,if you use linear algebra methods to deal with high
dimensional geometry questions,some other geometry concepts must be introduced. In this book,
the author introduced so-called orthodromic space,normal space ,orthodromic vector,normal vector,
common vector,uncommon vector when he discussed relations between figures denoted by system of
non-homogeneous linear equations and the figures denoted by its corresponding system of
homogeneous linear equations. Some above mentioned terms are defined by the author,some are
different from the popular,please pay attention to it when you read the book. To solve distances between linear figures, the author introduced concept of "exterior sum" and
discussed its property and equation. He extended some concepts and property about subspace in
linear algebra to generic linear figures. In solving included angles between linear figures, the author
put forward and solved two questions of included angle numbers and the nonuniqueness of its
solution. He used linear algebra theory to introduce Jianshi solution,and extended it to solving
included angles between two randon linear figures. He also extended the concept of "exterior
product"(another name"cross product" or "vector product")of vector of three-dimensional space to
hyperspace and put forward vector orthogonalization and another method to solve system of linear
equations. The author of this book used linear algebra methods to study graphic theory, his profitable
discussion may arouse interest and attention of some people concerned.
Shen Yidan
Beijing Institute of Technology
Hou Bingtao
Beijing Academy of Armoured Force Engenering
Pan Jianzhong
The Maths Graduate School of Chinese Academy of Sciences
July,1996
1
The First Chapter: Relation and property determined[1]
by
singular linear transform
Study of descriptive geometry and high-dimensional spacial analytic geometry often use linear
transform . This book introduced a "relation" method, viz. directly use relation between image and
image source in linear transform to solve some questions. Because elements of "relation" can
directly calculate under some certain conditions, the "relation" method has the advantage of
convenience and direct-view, then we can solve many questions not solved by ordinary linear
transform and make the method theoretical basis of this book. Under the staminal guidance and
personal participation of professor Hou Bingtao of Beijing Academy of Armoured Force
Engenering, the author studied property and operation rules of relations between image and image
source of singular linear transform in depth. Except theorem 3、6、10、11,the other seven theorems
and two inferences were strictly and scientifically proved by professor Hou himself. Of which,
theorem 4、5、8 and two inferences were newly put forward by professor Hou.
§§§§1 Property and relations between image and image source of
singular linear transform
1.1 Concept of "relation"
Many people are used to operate under equal relations.For unequal relations,many people are
used to " larger than" or "less than" relation.To top it all, people often try to change some unequal
relations into equal relations(viz. the relation between image and image source of singular
transform which will be discussed after),so it obliterates lively difference of different things and
virtually set obstacles to subsequent operation. This almost makes some study pause for a long
time.In fact, the relation between different things is complicated and omnifarious. For
example,"parallel"、"vertical"、"homogeneous" which are known very well by people.
Let's introduce a "relation" of more extensive meanings,some "expressions" of the
"relation" can directly add and multiplicate on some particular conditions.
In discrete mathematics[2]
, a1∈A1,a2∈A2,…,an∈An are regarded as a random set. Set
which is composed of all n-component ordered classes (a1,a2,…,an) is called Cartesian product of
A1,A2,…,An,it is denoted by A1×A2×…×An,viz.
A1×A2×…×An ={(a1,a2,…,an)|a i∈A i。i =1,2,…,n}.
A random subset A1×A2×…×An of Cartesian product
is called a n-component relation of A1,A2,…,An.
For example, A1,A2,…,An all can be regarded as sets of real number field R,random
variables (a1,a2,…,an) of linear space Vn of real number field R are apparently elements of
Cartesian product,therefore,Vn is a subset of Cartesian product A1×A2×…×An ,
it is called a n-component relation of A1,A2,…,An.
An important particular case is n=2, if an ordered pair set σis a subset of
A1×A2={(a 1,a 2)| a1∈A1,a2∈A2},
then σ is called a 2-component relation from A1 to A2 ,"relations" refered in this book belong
to this 2-component relation.
2
If σ is a 2-component relation from A1 to A2 , and
(a 1,a 2)∈σ (a 1∈A1, a 2∈A2) ,
then a 1 has relation σ with regard to a 2,marked as
a1 σ a2 .
Formulas used to express the relations between a1 and a2 are called "relational expressions",they are called "relations" for short.
1.2 Relations between image and image source of singular linear transform
Let's assume Vn is linear space of real number field R,σ is singular linear transform of
Vn,for random α∈Vn ,σ(α)=α′is called image of α,and α is called image source
of α. At the same time,σ(Vn)=Vn′is called image set,Vn is called image source set. Let's
assume σ is singular linear transform of linear space Vn ,α∈Vn,α′∈Vn′,if σ(α)=α′,
then nn VV ′⊂σ ,viz. σ is a subset of Vn×Vn′,so σ is called a 2-component relation from Vn
to Vn′,and (α,α′) ∈σ ,then α has relationσwith regard to α′,marked as α σ α′,
ifσ(α)≠α′,then ( ) , , σαα ∉′ marked as α α.
1.3 Property of relations between image and image source under singular linear
transform
Theorem 1:Under singular linear transform σ,relation σ is not self reflexive.
Proof:If ∀α∈Vn ,there is ασα,then σ is an identical transform,which is in
contradiction with the singularity of σ,so relation σ doesn't has reflexivity. ▌
(note:terminal signs“ ▌”shows that the proof is finished.The same to the next).
Theorem 2:Under singular linear transformσ,relationσis not symmetrical.
Proof:Assume A is the matrix of σ to fundus e1 ,e2 ,…,en ,then A2
is the matrix
ofσ2 .
Assume relationσhas symmetry,viz.∀α,β∈Vn, because σ(α)=β, andσ(β)=α,thenσ〔σ(α)〕=α
viz.σ2 is an identical transform,then A
2=I ,
I is a n-moment unit matrix,which is in contradiction with that the rank r of A is less than n ,then
relationσhasn't symmetry. ▌
Theorem 3:Relation σ is not transferable under singular linear transform σ.
Proof:Assume relationσhas transferability,viz.∀α,β,γ∈Vn ,
Because σ(α)= β, σ(β)=γ, σ(α)=γ,
Then σ[σ(α)]=σ(α ),
viz. σ(α)=α(σmay have generalized inverse[4]),
Therefore σ[σ(α)]=α,
but bothσandσ2 are not identical transform concluded by theorem 1 and theorem 2,so relationσhasn't transferability. ▌
3
§§§§2 Operation rules of elements in relationσσσσ
2....1 Linear operation rules
Theorem 4:Ifα1 σ α1′,α2 σ α2′, then for a random real number k1,k2,it has
k1α1+k2α2σ k1α1′+k2α2′.
Proof: Because σ(α1)=α1′,σ(α2)=α2′,
then σ(k1α1+k2α2)=k1σ(α1)+k2σ(α2)=k1α1′+k2α2′,
so k1α1+k2α2σ k1α1′+k2α2′
2....2 Transposition (of terms) rules
Definition 1:If α σ α ,then α is called a self reflexive element of relation σ.
Assume λis a character,A is a n×n matrix,then λI-A(I is a n-moment unit matrix) is
called characteristic matrix of A. Determinant of characteristic matrix is characteristic polynomial,roots of characteristic polynomial are called characteristic root of A.But when A is a n×r or r×n
matrix whose rank is r,l is a matrix whose elements of main diagonal are 1 and all the other
element is 0,both its columns and rows are the same with A .Its characteristic matrix can be
changed into diagonal matrix by elementary transform,elements of main diagonal include first
powerλ-λ0(or polynomial aboutλ,it can be decomposed into products of same or variedλ-λ0)
about λ ,all λ=λ0 (the same λ is calculated by the times it appeared) are characteristic
roots of A.
Assumeλ0 is a characteristic root of A,α={x1,x2,…,xn} is a nonzero vector,if , 2
1
02
1
=
nn x
x
x
x
x
x
AMM
λ
then α is called a characteristic vector of A,which belongs to characteristic rootλ0.
Theorem 5:Assume A is a matrix of singular linear transform σto fundus e1,e2,…,en in Vn,
then the necessary and sufficient conditions that any element α=x1e1+x2e2+…+xnen
of Vn is a nonzero self-reflexive element of relation σ is :transpose matrix A′of A has
characteristic rootλ0=1,and α is a characteristic vector of A′,which belongs toλ0=1.
Proof: Because ( ) , 2
1
21 =
n
n
e
e
e
xxx MLα and σ(α)=α,
then ( ) ( ) ( ) ( ) ,2
1
21
2
1
21
2
1
21 ===
n
n
n
n
n
n
e
e
e
xxx
e
e
e
Axxx
e
e
e
xxx MLMLML σασ
viz. (x1 x2 …xn)A=(x1 x2 … xn),
4
it is =′ nn x
x
x
x
x
x
A MM 2
1
2
1
(necessity after transposition);
Whereas,because (x1 x2 …xn)A=(x1 x2 … xn),
And α=x1e1+x2e2+…+xnen
Then ( ) ( ) ( ) ( ) ===
n
n
n
n
n
n
e
e
e
xxx
e
e
e
Axxx
e
e
e
xxx MLMLML 2
1
21
2
1
21
2
1
21 σασ ,
Viz. σ(α)=α (sufficiency). ▌
All linear combinations of self reflexive elements are reflexive elements................................................................ concluded from
theorem 4.
Theorem 6:When the matrix of singular linear transform σ is
=
−−− rrnrnrn
r
ccc
ccc
M
,2,1,
11211
1
1
1
L
LLLL
L
O
,
then among characteristic vectors α=x1e1+x2e2+…+xnen
belonging toλ0=1 of transposed matrix M , xr+1 = xr+2 =…= xn =0,
viz. α=x1e1+x2e2+…+xr er .
Proof:Because (x1 x2 …xn)A=(x1 x2 … xn)M=(x1 x2 …xn),
it is
=
+++
+++
+++
=
=
′
+
−+
−+
−+
−
−
−
n
r
rnrrnrrr
nrnr
nrnr
nrrnr
rn
rn
n
x
x
x
x
x
xcxcx
xcxcx
xcxcx
x
x
x
cc
cc
cc
x
x
x
M
M
M
LLLLL
LLLLLLL
LLLLL
L
LLLLLL
L
L
M
L
LLLO
L
L
M
1
2
1
,11
2,1122
1,1111
2
1
,1
2,12
1,11
2
1
0 0
0 0
1
1
1 after transposition,so xr+1 = xr+2 =…= xn =0,
viz. α=x1e1+x2e2+…+xr er ▌
By theorem 6,when the matrix of singular linear transform σ is M,then any row vectors of
M is self reflexive element of σ.
Because transposed matrix of A may not have characteristic rootλ0 = 1,even it has
characteristic rootλ0 =1,it isn't convenient to calculate. So marix M but not A is usually directly
5
used in actual practice. Or A is used,but A=M·N on some certain conditions,so it must be
changed into M by M=A·N -1
(refer to it in example 2 of §3 of the chapter).
Theorem 7:Assume ασ α1′+α2′,
then the necessary and sufficient conditions of α-α1′σ α2′
is :α1′is self reflexive element. Proof:Because σ(α-α1′)=σ(α)-σ(α1′)=α1′+α2′-σ(α1′),
then,the necessary and sufficient conditions of σ(α-α1′)=α2′,σ(α1′)=α1′
viz. α1′is self reflexive element. ▌
Inference:Assume α1+α2 σ β,the necessary and sufficient conditions of α2 σ β-α1
isα1 is self-reflexive element. ▌
2....3 Element exchange rules
Definition 2: If α σ β,and βσα,then α andβ are symmetrical elements of
relation σ. Theorem 8:Assume A is a matrix to fudus e1,e2,…,en of singular linear transform σ in
Vn ,then the necessary and sufficient conditions that any α and β of Vn is symmetrical
element is :the transposed matrix(A2)′of A
2 has characteristic root λ0 =1,both α and β are characteristic vectors belonging toλ0 =1 of (A
2)′.
Proof:Necessity
Because ( ) , 2
1
21 =
n
n
e
e
e
xxx MLα
Then ( ) ( ) =
n
n
e
e
e
Axxx ML 2
1
21ασ ,
Assume β=k1e1 +k2e2 +…+knen
Then ( ) ( )
=
n
n
e
e
e
AkkkL
L2
1
21βσ ,
From σ(α)=β,we can get (x1 x2 … xn)A=(k1 k2 … kn) (1)
From σ(β)=α,we can get (k1 k2 … kn)A=(x1 x2 … xn) (2)
Substitute formula (1) into formula (2), we can get (x1 x2 … xn)A2=(x1 x2 … xn),
tanspose it, then we get ( )
=
′
nn x
x
x
x
x
x
AMM
2
1
2
1
2. 〔If substitute formula (2) into formula (1),we can prove that β is also a characteristic vector
6
belonging toλ0 =1 of(A2)′〕.
Now let's prove the sufficiency , because (x1 x2 … xn)A2 = (x1 x2 … xn) (3)
And σ(α)=β,
viz. ( ) ,Axxx
n
n =
e
e
eML 2
1
21 β
then when we suppose β=k1e1 +k2e2 +…+knen ,
it has (x1 x2 … xn)A=(k1 k2 … kn) (4) 〔the same with formula (1)〕,substitute formula (4) into formula 3,it can get
(k1 k2 … kn)A=(x1 x2 … xn)
so ( ) ( )
=
n
n
n
n
x
x
x
xxx
e
e
e
AkkkM
LM
L2
1
21
2
1
21 ,
viz. σ(β)=α,viz.α,β are symmetrical elements. ▌
From theorem 8 ,theorem 4 and theorem 5,we know all linear combinations of symmetrical
elements are symmetrical elements,and self reflexive is a particular case of symmetry.
Theorem 9:At the double ends of relation σ,symmetrical elements can exchange with each
other. Proof:Assume α1 +α2 σ β1 +β2 ,
andα1 and β2 are symmetrical elements,because α2 σβ2 ,α1 σβ1,
and because β2 σα2 ,α1 +β2 σ β1 +α2 ▌
Inference:Assumeα0 andβ0 are symmetrical elements,
(1) if α +α0 σ β,then α σ β-β0 ;
(2) if α σ β +β0,then α-α0 σ β.
Theorem 10:When the matrix of relation σ is M (refered in theorem 6),σ doesn't
contain unequal symmetrical elements.
Proof:Assume α and β are characteristic vectors belonging to λ0 =1 of (M2)′,but α≠β. Regard M as n×n matrix (all the elements of the following column are zero),then
M2=M ,viz. matrix M is idempotent. From theorem 8, it has (x1 x2 … xn)(M
2)′=(x1 x2 … xn)
and (x1 x2 … xn)M′=(k1 k2 … kn),
but (x1 x2 … xn)(M2)′=(x1 x2 … xn)M′,
So it has (x1 x2 … xn) =(k1 k2 … kn),
viz.α=β ,which is in contradiction with the hypothesis α≠β. So from theorem 5,it only has ασα and βσβ,
but α β and β α. ▌
At last ,let's point out by the way,because relation σ is not transferable under singular linear
transform,continuously transmit relations such as α1σα2σ…σαm (m>2)should not be used.
7
§§§§3 Relations between coordinates of vectors under singular linear
transform
In theorem 6,we refered to a matrix whose elements of main diagonal of anterior r rows are
one and the other elements are zero,for the convenience to describe it, we call this n×r and
column nonsingular matrix M matrix. Sometimes,M is written into partitioned matrixs of upper and
lower parts.
=
C
IM ,
of which,I is r-order unit matrix,C is (n-r)×r (r<n) matrix. Now,we choose a fundus e1 ,e2 ,…,en in linear space Vn ,choose a fundus e1′ ,e2′ ,…,er′(r<n) of image set σ(Vn ) of Vn ,then the image of fundus e1 ,e2 ,…,en can be
represented not only by its own linearity but also by the linearity of fundus e1′ ,e2′ ,…,er′,viz. when assume
( )( )
( )
+++=
+++=
+++=
nnnnnn
nn
nn
aaa
aaa
aaa
eeee
eeee
eeee
L
LLLL
L
L
2211
22221212
12121111
σ
σ
σ
( )( )
( )
′
′
′
=
=
=
=
′++′+′=
′++′+′=
′++′+′=
rnn
nrnn
r
r
nnnn
n
n
rnrnnn
rr
rr
BA
bbb
bbb
bbb
B
aaa
aaa
aaa
A
bbb
bbb
bbb
e
e
e
e
e
e
e
e
e
eeee
eeee
eeee
MMM
L
LLLL
L
L
L
LLLL
L
L
L
LLLL
L
L
2
1
2
1
2
1
21
22221
11211
21
22221
11211
2211
22221212
12121111
hasit
, assume and
and
σ
σ
σ
σ
(A is n×n singular matrix,B is n×r and column nonsingular matrix) or can be represented as
′
′
′
=
rnrnn
r
r
nnnnn
n
n
n bbb
bbb
bbb
aaa
aaa
aaa
e
e
e
e
e
e
e
e
e
M
L
LLLL
L
L
M
L
LLLL
L
L
M
2
1
21
22221
11211
2
1
21
22221
11211
2
1
σ
Apparently, when assume
8
=
′
′
′
nr
N
e
e
e
e
e
e
MM
2
1
2
1
(N is r×n and row nonsingular matrix),it has A=B·N.
•= ′
′
′
= ′
′
′
= ′
′
′
=
nr
nr
nr
e
e
e
NB
e
e
e
B
e
e
e
N
e
e
e
e
e
e
A
e
e
e
B
MMMMMM
2
1
2
1
2
1
2
1
2
1
2
1
then
because
hasit condition thefrom Proof :
viz. A=B·N. ▌
′
′
′
=
nn e
e
e
e
e
e
MM
2
1
2
1
all when ly,Particular σ
(er+1′,er+2′,…,en′can be linearly represented by fundus e1′,e2′,…,er′),B is a M
′
′
′
=
′
′
′
=
=
′
′
′
==
rr
nn
nr
e
e
e
e
e
e
e
e
e
e
e
e
e
e
e
N
e
e
e
C
IMB
MM
MM
MM
2
1
2
1
2
1
2
1
2
1
2
1
hasit
be so,
, because Proof
. viz.matrix,
σ
σ
:
9
Viz
=
nr e
e
e
N
e
e
e
MM
2
1
2
1
σ
Now,N becomes a submatrix composed of the front r rows of matrix A,viz.
=
rnrr
n
n
aaa
aaa
aaa
N
L
LLLL
L
L
21
22221
11211 ,
and in the submatrix of A which is composed of the back n-r rows of matrix A
=
+++
nnnn
nrrr
aaa
aaa
N
L
LLLL
L
21
,12,11,1
0 ,
random rows can be linearly expressed by rows of N ,so,the submatrix of A which is composed of
the back n-r rows of A can be regarded as a product of (n-r)×r matrix
=
−−− rrnrnrn
r
r
ccc
ccc
ccc
C
,2,1,
22221
11211
L
LLLL
L
L
and N: N0 = C·N ,viz. A can be changed into partitioned matrix of upper and lower parts,and
further be changed into the product of matrixes
NMNC
I
NC
NI
NC
NA ⋅=
=
⋅
⋅=
⋅
= ,
and because A=B·N ,
( ) . matrixunit order - is then rIC
IMB == ▌
= +++
+++
nrnn
rrrr
rrrr
bbb
bbb
bbb
C
L
LLLL
L
L
21
,22,21,2
,12,11,1
Now ,
viz. C is the submatrix of B which is composed of the back n-r rows of B.
Theorem 11:Under singular linear transform σ,the relation between coordinates of vector α=x1e1+x2e2+…+xnen
and its image σ(α)=y1e1′+y2e2′+…+yrer′
10
under σ can be expressed as (y1 y2 … yr)=(x1 x2 … xn)B.
Particularly,when σ(e i)=e i′
( ) , ,,, fundusby drepresentelinearly becan ,,, . ,,2,1 2121 rnrr eeeeeeni ′′′′′′= ++ LLL
( ) ( ) ( )
( )
( )
( ) ( ) ( ) ( ) . then
and
because Proof
6
or
5 then
2
1
21
2
1
21
2
1
21
2
1
2
1
2
1
2
1
21
2
1
2
1
,2211
2,22211222
1,22111111
2121
•= ′
′
′
=== ′
′
′
++++=
++++=
++++=
=
−++
−++
−++
n
n
n
n
n
n
rnnn
n
nr
nrrnrrrrrr
nrnrr
nrnrr
nr
e
e
e
NBxxx
e
e
e
Axxx
e
e
e
xxx
e
e
e
B
e
e
e
A
e
e
e
e
e
e
xxx
e
e
e
N
e
e
e
xcxcxcxy
xcxcxcxy
xcxcxcxy
Mxxxyyy
MLMLMLMMMML
MM L LLLLLL LLLL
==,,=
,:σασ
σα
( ) ( ) ( )
( ) ( )
( ) ( )( )
( ) ( ) .
then, because , whenly,particular
. viz.
then
because and
2121
2121
2
1
21
2
1
21
2
1
21
1
21
Mxxxyyy
MBee
Bxxxyyy
e
e
e
NBxxx
e
e
e
Nyyy
e
e
e
Nyyy
e
e
e
yyy
nr
ii
nr
n
n
n
r
n
r
r
r
LL
LL
ML
ML
ML
ML
=
=′=
=
•=
=
′
′
′
=
σ
ασ
,
,
〔viz. formula (5)〕.Transpose formula(5)and write it into coordinate expressions,then we get
formula(6), this is the formula of the relation between coordinates of vector and its imageσ(α)
under σ when σ(ei )=ei′. ▌
Under nonsingular linear transform,the relation between vector and its coordinates can be
acquired by substituting transform into the vector.But under singular linear transform,it can only be
11
acquired by matrix operation〔just like the derivation of formula (6)〕.Now by "relation" method,the formula of the relation between vector and its coordinates can be acquired by substituting
transform into the vector just like nonsingular linear transform. But for the beginners who don't
skillfully master the matrix operation technique,this is a gospel without question.
Example 1: Under the relation
21321
21
321
213
212
211
or
into changed is
vector
5
3
5
3
32
43
eYeXZeYeXe
eYeX
ZeYeXe
eee
eee
eee
′+′++
′+′
++
−−
−−
−−
σ
σ
σ
σ
Please find out the coordinate transform formula of the vector.
Solution: Substitute the above transforms into vector Xe1+Ye2+Ze3,
( ) ( )
−−−=′
−−−=′
−−−+ −−−
+−+−+−++
ZYXY
ZYXX
eZYXeZYX
eeZeeYeeXZeYeXe
5
334
5
323
then sign, equal theof ends double thecompare
5
334
5
323
into expressionrelation theof endright thearrange
, 5
3
5
33243 then
21
212121321 σ
But it's not the form of formula (6). If you want to acquire the result of formula (6),the
matrix of σ must be changed into matrix M.
So,matrix operation can't be absolutely divorced from.
Example 2: Change the matrix of the example 1 into the formula of M,and find out the
relation between coordinates of the vector under new transform. Solution because when σ(e1)=e1′, σ(e2)=e2′, σ(e3)=e3′
(e3′can be linearly represented by fundus e1′,e2′) A=M·N ,then M=A·N -1,
−
−=
=−−
−−=
−−
−−=
−
32
43 so
, 132
43 ,
32
43 because and
1N
NN
So
5
3
5
3
1
1
32
43
5
3
5
332
43
−
=
−
−•
−−
−−
−−
=M
and then relation σ can be expressed as
12
′−′
′
′
213
22
11
5
3
5
3
eee
ee
ee
σ
σ
σ
Assume under the latter relation,vector Xe1+Y e2+Ze3
is changed into X "e1′+Y"e2′,
or Xe1+Y e2+Ze3 σ X "e1′+Y"e2′,
substitute the latter transform into vector Xe1+Y e2+Ze3
. 5
3
5
3 getsit 2121321
′−′+′+′++ eeZeYeXZeYeXe σ
arrange and transpose the double ends of the equal sign ,it gets
,
5
35
3
then
, 5
3
5
3 so,
, 5
3
5
3 viz.
, 5
3
5
3
2121
21321
21
−=′′
+=′′
′
−+′
+=′′′+′′′
′
−+′
+++
′
−+′
+
ZYY
ZXX
eZYeZXeYeX
eZYeZXZeYeXe
eZYeZX
σ
this is completely the same as the form of formula (6).
In actual application,we usually directly wrote the matrix of transform into the form of
M(but not the form of A),which can ensure that each relation of the transform has its reflexive
element ,and then it can has strong calculation function. Example 3: Under the relation
, or
, into changed is
vector
3
2
3
15
4
5
2
214321
21
4321
214
213
22
11
eYeXTeZeYeXe
eYeX
TeZeYeXe
eee
eee
ee
ee
′+′+++
′+′
+++
−−
−−
σ
σ
σ
σ
σ
Please find out the coordinate transform formula of the vector. Solution substitute the above
relations into vector Xe1+Y e2+Ze3+Te4 ,
, 3
2
3
1
5
4
5
2 getsit 2121214321
−+
−−+++++ eeTeeZYeXeTeZeYeXe σ
13
, 3
2
5
4
3
1
5
2 so
, 3
2
5
4
3
1
5
2 viz
, 3
2
5
4
3
1
5
2 into endright thearrange
2121
214321
21
eTZYeTZXeYeX
eTZYeTZXTeZeYeXe
eTZYeTZX
−−+
−−=′+′
−−+
−−+++
−−+
−−
σ
compare the double ends of equal sign,and then it gets
,
3
2
5
43
1
5
2
−−=′
−−=′
TZYY
TZXX
this is also the same as the result of formula (6).
Exercises
1.1 Please find out the change of the vector 3e1+5e2-4e3 under the transform
+−
+−
−
213
212
211
4
2
53
eee
eee
eee
σ
σ
σ
1.2 Change the matrix of the relation of the above exercise into M ,and find out the change of
the vector under new transform.
1.3 Find out the change of vector 4e1-5e2+3e3 under the transform
+− 213
22
11
23 eee
ee
ee
σ
σ
σ
1.4 Find out the change of vector 5e1-4e2+6e3 under the transform
− 213
22
11
59
eee
ee
ee
σ
σ
σ
1.5 Find out the change of vector 12e1-7e2+8e3 under the transform
+− 213
22
11
2
1
2
2
eee
ee
ee
σ
σ
σ
14
The Second Chapter: "Relation" method and preferred
n-dimensional coordinate system
In the first chapter,we have discussed the concept and proerty of "relation" method,and
studied its operation rules in depth. In this chapter,we will begin to discuss approaches of using
"relation" method in practice,viz. how to use "relation" method to build preferred n-dimensional
coordinate system. In order to help people master the method step by step,we will begin with the
process of building preferred 4-dimensional coordinate system.
§§§§1 "Relation" method and examples of preferred 4-dimensional
and 5-dimensional coordinate system
1....1 Examples of building preferred 4-dimensional coordinate system
For example assume e1 ,e2 ,e3 are unit vectors perpendicular to each other,they are fundi
of 3-dimensional euclidean space V3 ,σ1 is a singular linear transform of V3 .Regard V3 as
rectangular axis Oxyz ,fundi e1 ,e2 ,e3 are separately regarded as unit vectors of x axis, y axis
and z axis,and suppose under σ1, e1 ,e2 ,e3 and their images it has the lower relations
( )1
5
4
5
3
213
22
11
−− eee
ee
ee
σ
σ
σ
According to the above relations we can figure out an analogue image(hereinafter referred to
as model,refer to it in figure 1) of Oxyz . In the figure,directviewing figure of unit vector e3 of
vertical axis z and the vector - 3e1 - 4e2 of horizontal
coordinate plane xoy overlapped.
For the convenience,people always hope to describe things
in 3-dimensional or even higher dimensional space by using a
2-dimensional plane,it is difficult to avoid figures overlap. As
this is the case ,why still get it around but not face up to it?
Facing up to questions is solving questions. So,a vector between
two random points overlapped in figure 1 is called (of figure 1)
overlapping direction.
For example,in figure 1, point (0,0,5) of z axis and point
(-3,-4,0) of xoy plane overlapped, this shows that
overlapping direction is a vector between the two points.So,the figure 1
overlapping direction of the model can be denoted as
S=3e1 +4e2 +5e3 .
Arrange and transpose the third relation expression of the above formula (1),it has 3e1 +4e2 +5e3 σ1 0
or S σ1 0
This shows that in figure 1,all straight lines running parallel to overlapping direction S are
punctiform.But 0 S
15
this is because e3 isn't self reflexive element,it can't be moved to the right end of relation
expression. So,S and zero aren't symmetrical,it shows that the image of point can only be
punctiform.
Under the enlightenment of stereoscopic projection[5] ,we think transform like formula
(1)should appeared in pairs,viz. there should be relation σ2
+ 213
22
11
5
4
5
3eee
ee
ee
σ
σ
σ
(2) From formula (2),another model (figure 2) of Oxyz can be
acquired. Z axis of figure1 and figure 2 are symmetrical relative to
origin.Because the two models represent the same 3-dimensional
rectangular coordinate system Oxyz,then we call them a couple of
preferred 3-dimensional planes. Respectively draw out a vertical
and upright ray through the origin of the preferred 3-dimensional
plane as the fourth coordinate axis t , and this forms a model of a
couple of 4-dimensional rectangular coordinate systems Oxyzt ,it
is called a couple of preferred 4-dimensional coordinate
systems(referred to as 4-dimensional system or specific 4-dimensional figure 2
figure 3 figure 4
system,refer to it in figure 3 and figure 4).
In order to distinguish it expediently,we call the 4-dimensional system of figure 3 host system
and call the 4-dimensional system of figure 4 guest system. The corresponding relations can be
expressed as
( )3
5
4
5
3
414
2113
212
111
−−
ee
eee
ee
ee
σ
σ
σ
σ
( )4
5
4
5
3
424
2123
222
121
+
ee
eee
ee
ee
σ
σ
σ
σ
16
From figure 3 and figure 4,it is observed that preferred four-dimensional system is composed
of cross axis x , vertical axis y,upright axis t and axis z with inclining direction (be called oblique
axis),which are all through the origin. Of which,the direction and unit length of oblique axis z is
symmetrical relative to the origin in host system and guest system.
In generalized case,when building preferred four-dimensional system we only write out
formula (3) and formula (4)〔viz. not from formula (1) and formula (2) to formula (3) and formula
(4). So,for figure 1 and figure 2,we usually regard them as preferred four-dimensional system of
deleting upright axis t .
1....2 Examples of building preferred five-dimensional coordinate system
Based on the same method,we can also work out a couple of models of five-dimensional
rectangular coordinate systems Ox1x2x3x4x5 . For example,in Oxyzt of figure 3 and figure 4 ,order
x=x1, y=x2,z=x3, t=x4 ,
and choose a vector 3e1+4e2+6e4 of a three-dimensional subsystem Oxyt (here marked as
Ox1x2x4 ) in Oxyzt as overlapping direction. Because 3e1+4e2+6e4 σ1 0 ,
then it has ( ).436
12114 eee +−σ
but oblique axis is symmetrical relative to the origin in guest system,so it has
( ).436
12124 eee +σ
Separately substitute the two relations into formula (1) and formula (2) and make a diagram
according to it,then a couple of preferred four-dimensional planes(just call it for the moment,refer
to it in figure 5) are acquired.
figure 5
Respectively draw out a vertical and upright ray through the origin of the preferred
four-dimensional plane as upright axis x5,then a couple of preferred five-dimensional coordinate
systems are acquired(refer to it in figure 6). The corresponding relations are
( )5
6
4
6
35
4
5
3
515
2114
2113
212
111
−−
−−
ee
eee
eee
ee
ee
σ
σ
σ
σ
σ
17
( )6
6
4
6
35
4
5
3
525
2124
2123
222
121
+
+
ee
eee
eee
ee
ee
σ
σ
σ
σ
σ
figure 6
§§§§2 Oblique axes transform and preferred n-dimensional coordinate
system
According to the above ideas and methods of building preferred four-dimensional system and
five-dimensional system,we will discuss some questions of building preferred n-dimensional
system.
2....1 Oblique axes transform and building of preferred n-dimensional system
2....1....1 Methods and steps of building preferred n-dimensional system————————oblique axes
transform
Preferred n-dimensional system actually is a kind of analogue image ( referred to as model)of
imaginary n-dimensional rectangular coordinate system Ox1x2…xn .Owing to the enlightment of
stereoscopic projection,this kind of models always appear in pairs. For the convenience of
calculation,the transform adopted are always symmetrical.
The detailed practice is :first break up Ox1x2…xn into n-2 three-dimensional subspaces
Ox1x2x3,Ox1x2x4,…,Ox1x2xn –1 ,Ox1x2xn ,and make every subspace has common level coordinate
plane x1ox2 . Second,choose vectors
S1=a11e1+a12e2+a3e3 ,
S2=a21e1+a22e2+a4e4 , … … … …,
Sn----3333=an--3,1e1+an-3, 2 e2+an-1en----1 ,
from the front n-3 subspaces.〔a3 ,a4 ,… , an-1 of every formula are nonzero numbers,ai1,ai2
are i numbers when group( i=1 ,2 ,… ,n-3 ) aren't zero at the same time;e1 , e2 ,…,en
are respectly unit vectors of x1 , x2 , …, xn 〕as overlapping direction. Then,under σ1 it has
S1 σ1 0 , S2 σ1 0, … ,Sn-3 σ1 0.
Because the n-2 subspaces have x1ox2 plane aggregately,and the xn axis of the subspace
Ox1x2xn is regarded as upright axis,e1 , e2 ,…,en are determined to be self reflexive elements,
18
so the two front terms of Si (i=1 ,2 ,… ,n-3) can be moved to the right end of the relation.
Then it has ( ) ( ); 3,2,1 1
2211
2
12 −=+−+
+ nieaeaa
e ii
i
i Lσ
but underσ2 , signs of the right end of the relations are opposite,viz.
( ) ( ); 3,2,1 1
2211
2
22 −=++
+ nieaeaa
e ii
i
i Lσ
write it into complete relation:
( )
( )
( )
( )
( )
( )
( )
( )8
1
1
1
7
1
1
1
2
22,311,3
3
21
222121
4
24
212111
3
23
222
121
1
22,311,3
3
11
222121
4
14
212111
3
13
212
111
+
+
+
+−
+−
+−
−−
−
−
−−
−
−
nn
nn
n
n
nn
nn
n
n
ee
eaeaa
e
eaeaa
e
eaeaa
e
ee
ee
ee
eaeaa
e
eaeaa
e
eaeaa
e
ee
ee
σ
σ
σ
σ
σ
σ
σ
σ
σ
σ
σ
σ
LLLLLL
LLLLLL
〔a3 , a4 , … , an-1 of the two formulas are nonzero numbers,ai1 , ai2 are numbers of i group
(when i =1 , 2 , … , n-3 ) aren't zero at the same time〕. Make a diagram according to
formula (7) and formula (8),a couple of analogue images( refer to it in figure7,in the figure, ai1=
a 1 ,ai2= a 2 ;a1 , a2 , … , an-1>0 ) of n-dimensional rectangular coordinate system Ox1x2…xn
can be acquired ,we call the couple of analogue images a couple of preferred n-dimensional
coordinate system(referred to as n-dimensional system or specific n-dimensional system). Of
which, the n-dimensional system generated from formula (7) is called host system, the
n-dimensional system generated from formula (8) is called guest system.Accordingly,formula is
called positive transform or main transform,formula (8) is called subsidiary transform or assistant
transform.Preferred n-dimensional systems always appear in pairs,usually host system is on the top
or on the left,guest system is below or on the right. Sometimes,it appears alone. When it appears
alone, it is always host system if there is no special version .
19
figure 7
In formula (7) and formula (8),the first ,the second and the n-th couple of relations determine
the existence of rectangular coordinate system Ox1x2xn.If people assume Ox1x2xn is existent through
common practice,at the same time we think using letter as sign and using figure as sign have the ............................................same significance................ ,so respectively substitute........................ “ □ ” and “ ○ ” for “σ1” and “σ2”,
then formula (7) and formula (8) can be respectively written into
( )
( )10
9
22,311,311
22212144
21211133
22,311,311
22212144
21211133
+
+
+
−−
−−
−−
−−−−
−−−−
eaeaea
eaeaea
eaeaea
eaeaea
eaeaea
eaeaea
nnnn
nnnn
○○○□□□
LLLL
LLLL
for short. Here,we may as well call “ □ ” of the two formulas sign “image relation sign”,and call
“ ○ ” sign of“shadow relation”(of which,“α□β ”can be read “the image of α is β ”,“γ ○ δ ”can be read “the shadow of γ is δ” ).
The introduction of sign “ □ ” and “○” enhances visualization and visualizability of
the transform,and it also lessens much descriptive trouble in the following application.
Figures of axis Ox1x2…xn-1 in preferred n-dimensional system are always oblique.So,we call
them oblique axes,and accordingly,formula (7) and formula (8)〔or formula (9) and formula (10)〕can also be called oblique axes transform.................... .
2....1....2 Structure of preferred n-dimensional system
Preferred n-dimensional system is composed of cross axis x1 ,vertical axis x2 ,oblique axis
Ox3x4…xn-1 and upright axis xn which are all through the origin(under particular circumstances,
x1, x2 can be selected for upright axis or oblique axis,and xn can be select for cross axis,vertical axis
or oblique axis,etc. But without special version,x1 , x2 are still served as cross axis and vertical
axis,xn is still served as upright axis),we can also consider that it is composed of models of a
couple of n-1-dimensional rectangular coordinate systems and upright axes. But n-1-dimensional
model represents a n-1-dimensional subspace of Vn ,so we call them a couple of preferred
n-1-dimensional planes for the moment,they are formed of overlapping of the models of n-3
three-dimensional subspace Ox1x2x3,Ox1x2x4,…,Ox1x2xn-1,of Vn . They have common plane
x1ox2 ,but figures of each oblique axis sometimes are overlapped , it is difficult to differentiate and
not convenient to be identified.So separate the overlapping oblique axes when we make a diagram.
They are listed parallel to a straight line(the straight line should be perpendicular to every oblique
axis in essence)which is through the origin,the straight line represents the origin of every oblique
axis.
20
For upright axis,because its actual position is through the origin and perpendicularily goes
away from the plane,it is difficult to be drawn on figures.According to the draughting of oblique
axis,we draw it in a random direction through the origin.But in order to differentiate it from
oblique axes,figure direction and unit length of upright axis is identical in host system and guest
system,but direction of the same oblique axis is opposite in host system and host system.
Through choosing reasonable direction and unit length,the upright axis of specific system can
not only offer calculation convenience but also lead some elements (such as point,line,plane,etc)of
geometric figure drawn in preferred n-dimensional system to suitable position,thereby it make
the whole geometric figure occupancy small chart size and has satisfactory and goodly shape.
Speaking from this meaning ,the upright axis of preferred n-dimensional can be called leading
axis.
2....1....3 Kinds of preferred n-dimensional system
According to the permutation pattern of oblique axis,preferred n-dimensional system can be
divided into convergent oblique pattern and emanative oblique pattern.In formula (7) and formula
(8)〔or formula (9) and formula (10)〕, when λ1a11 =λ2a21 = … =λn-3 an-3,1 = a1, λ1a12 =λ2a22 = … =λn-3 an-3,2 =a2
(λ1,λ2,…,λn-3 are nonzero numbers), directions of every oblique axis are parallel.
So their figures are overlapped,we think this kind of preferred n-dimensional system is
convergent oblique pattern(but when concretely making a diagram ,we should avoid figures
overlapping of oblique axes.So we should differentiate them one from another and parallelly
arrange them on a straight line which is through the origin, use the straight line as their origin for
the moment). Figure 8 is a couple of emanative oblique pattern examples when n is even
number(accordingly,its transform is
( ) ( )
−−
−−
+
−−
−− 221111
2211
221144
221133
11
eee
eee
eee
eee
aaa
aaa
aaa
aaa
nn
ii
ii □□□□
LLL
LLL
i=3 ,4 , … ,n-1,readers are expected to write out its subsidiary transform),and figure 9
convergent oblique pattern examples when n is odd number(the corresponding transform is
( ) ( )
−−
−−
+
−−
−−
−−
++
221111
22
1
11
1
221155
221144
221133
11
2
2
eee
eee
eee
eee
eee
aaa
aaa
aaa
aaa
aaa
nn
ii
ii □□□□□
LLL
LLL
i=5 ,6 ,… ,n-1 , the readers are expected to write out its subsidiary transform).
21
figure 8
If it dissatisfies the above terms,viz. in formula (7) and formula (8)〔or formula (9),formula
(10)〕,if and only if λ1,λ2,…,λn-3 are zero at the same time, λ1a11 =λ2a21 = … =λn-3 an-3,1 , λ1a12 =λ2a22 = … =λn-3 an-3,2
are founded,figure directions of every oblique axis are emanative,we call this kind of preferred
n-dimensional system emanative oblique pattern.Figure 10 is a couple of preferred six-dimensional
systems of emanative oblique pattern.
figure 9
§§§§3 Property of preferred n-dimensional system
3.1 Relation between preferred n-dimensional system and oblique axes transform
The above discussion has shown:given a transform ,we can immediately make out the
corresponding preferred n-dimensional system. Vice versa,given a preferred n-dimensional
system,can we immediately make out the corresponding oblique axes transform? The answer is
affirmative.
For example,in the host system of figure10,points
(0,0,2,0,0,0),(0,0,0,3,0,0),(0,0,0,0,5,0)
figure 10
22
of every oblique axis respectively coincide with points
(-2,-1,0,0,0,0),(1,-1,0,0,0,0),(-3,-1,0,0,0,0)
of x1ox2 plane,it shows the overlapping directions of Ox1x2x3 ,Ox1x2x4 ,Ox1x2x5 respectively is
2e1 + e2 + 2e3, -e1 + e2 + 3e4 , 3e1 + e2 + 5e5 .
Because they are all punctiform,viz. under singular linear transform σ1,their images are
all zero,viz. 2e1 + e2 + 2e3 □ 0, -e1 + e2 + 3e4 □ 0, 3e1 + e2 + 5e5 □ 0.
From figure 10,the cross axis、the vertical axis and the upright axis are respectively x1 , x2 , x6 ,so,e1 , e2 are self reflexive elements. Transpose the above three relations then the abbreviated
formula of positive transform is acquired.
−−
−
−−
215
214
213
35
3
22
eee
eee
eee □□□
Because the oblique axes directions in guest system are opposite to those of host system,
then the assistant transform is
+
+−
+
215
214
213
35
3
22
eee
eee
eee ○○○
In fact,according to that vectors 2e3 , 3e4 , 5e5 of oblique axes x3,x4,x5 in host system
respectively coincide with vectors -2e1-e2 , e1- e2 , -3e1-e2
of plane x1ox2 ,we can immediately determine it has relation σ1 between them, so the
corresponding oblique axes transform can be directly written out.
3....2 The "preference" of preferred coordinates system
Defination 1:The overlapping direction of host system in preferred n-dimensional system
can also be called principal overlapping direction of preferred n-dimensional system,referred to
as principal overlapping direction (rampart:it is the rampart that be referred as ramparting walls,building by piling bricks can also be called ramparting walls in Beijing dialect. Here,“rampart”can
be extended as overlap or coincidence).
Defination 2 : All linear combinations of principal overlapping direction form a
n-3-dimensional subspace ,it is called principal overlapping space .
Here,we can catch a glimpse of the advantage of abbreviated formula (9) and abbreviated
formula (10) .
Order S1=a11e1+a12e2+a3e3 ,
S2=a21e1+a22e2+a4e4 , … … … …,
Sn----3333=an--3,1e1+an - 3, 2 e2+an-1en----1 ,
From the linear independence among e3, e4,…, en-1 ,we can extrapolate that S1,S2,…,Sn----3333 are linearly independent.Transpose formula(9),it has
S1 □ 0, S2 □ 0, …, Sn----3333 □ 0 ,
so ( ) ( )11 numbers are 03
1
i
n
i
ii kSk □∑−=
As will be readily seen,S1,S2,…,Sn----3333 are respectively overlapping directions of n-3
23
three-dimensional models of preferred n-1 dimensional plane in host system these overlapping
directions are called principal overlapping directions of preferred n-dimensional system. All their
linear combinations filled a n-3-dimensional subspace,viz. principal overlapping space.When we
take all numerical values of ki , formula (11) represents that principal overlapping space
accumulates into punctiform. So,the "preference" of preferred n-dimensional system is :it suits a
known﹑preferred principal overlapping space and make it accumulate into punctiform.
In our realistic﹑ three-dimensional world ,people can't make out unified models of
n-dimensional space,but only make out "preferred" n-dimensional system satisfying different
specific conditions according to different needs.
§§§§4 Punctiform figures in preferred n-dimensional system————generic
point
4....1 Generic point、、、、projection trace and opposite trace
4....1....1 Concept of generic point
In descriptive geometry,use a bundle of parallel light to irradiate a geometrical body,then the
geometrical body leaves shadow on the image plane. Figures on the image plane(also referred to as
projection plane) and completely coincide with the shadow are called projection of the geometrical
body. The direction of light travelling is called projection direction,this kind of projection is
called parallel projection[6]
.
Preferred n-1-dimensional plane is equivalent to be formed of parallel projection,so,random
punctiform figures of n-1-dimensional plane in host system are equivalent to principal overlapping
space. And because preferred n-dimensional system is composed of n-1-dimensional plane and
upright axis,we can imagine the whole n-dimensional space is formed of the translational trace of
preferred n-1-dimensional plane along the direction of the upright axis. So,random punctiform
figures in preferred n-dimensional system are equivalent to punctiform figures of preferred
n-1-dimensional plane. Or,random punctiform figures of preferred n-dimensional system(referred
to as host system)can represent figures parallel or congruent with principal overlapping space.
Defination 3: Random punctiform figures in preferred n-dimensional syetem are called
generic points(the meaning is “generally referred to as "all punctiform figures"):
(1)When a punctiform figure represents a figure equivalent to (or parallel and congruent )
principal overlapping space,it actually represents the dimension is n-3(the directviewing dimension
is zero),it is called sufficient-rank generic point,or directly call it n-3-rank generic point,the
following generic points referred to are of this kind if without special version;
(2)when it is ordered only to represent a p-dimensional figure(P<n-3) ,we call it deficient
-rank generic point or directly call it p-rank generic point;
(3)For a point of common meaning,because it represents the dimension P=0,sometimes it is
also called zero-rank generic point (but it use the original name under most circumstances.Since
then, "points " has no "generic" before them are referred to be this kind),it is regarded as a
particular case of generic point.
4....1....2 Projection trace and opposite trace of generic point
Parallel projection can be divided into two kinds:
1.The included angle between projection direction and image plane is less than 90°,it is
called oblique projection[6] ;
24
2.The projection direction is perpendicular to image plane, it is called orthographic
projection[6]
.
The following projections referred to are all orthographic projections if without special
version.
The intersection point of random geometrical figure (referred to as their original shape,but not
their projection) and the projection plane is called trace of the geometrical figure[6]
.
Defination 4: For the projection A′of a generic point A on preferred n-1-dimensional
plane, its trace on horizontal projection plane x1ox2 is called projection trace of generic point,marked as Ja′.
Defination 5: The point of generic point A which corresponds with its projection trace is
called opposite projection trace of A,it is called opposite trace for short,marked as Ja.
The concept of opposite trace can be explained as:translate n-1-dimensional plane along
upright axis and make the projection A′of A coincide with A,then the projection trace Ja′
become opposite trace Ja′.So,a random generic point on preferred n-1-dimensional plane ,its
projection trace and opposite trace are the same point (the point can be called "trace" or "trace
point' of the generic point).
From defination 4 and defination 5,we can extrapolate:
1)If a generic point has projection trace, then it must have opposite trace. Except
sufficient-rank generic points,it isn't all generic points have projection trace and opposite trace.
Such as some (not all) difficient-rank generic points ;
2)Projection trace or opposite trace of generic points are points like these:their coordinates
relative to oblique axes x3, x4,…, xn-1 are simultaneously zero (of which,the coordinate of
projection trace relative to upright axis is zero too;but the coordinate of opposite trace relative to
upright axis is nonzero uner generalized case).
Among all points represented by a random generic point in preferred n-dimensional system,only the positions of projection trace and opposite trace are easy to be found out, but if the the
positions of projection trace and opposite trace are determined,the position of generic point is
determined naturally.
People are used to study in three-dimensional space,but preferred n-dimensional system
often choose subsystem Ox1x2…xn to simulate spatial three-dimensional rectangular coordinate
system,projection trace or opposite trace nicely give the directviewing position of the generic point
in three-dimensional coordinate system. If you master concerned knowledge of projection trace or
opposite trace,then you catch hold of the trace of generic points.
4....2 Property of generic points
4....2....1 Shape of generic points
Theorem 1:Random points are in the shape of punctiform in host system,and they are in the
shape of line(in convergent oblique pattern) or plane (in emanative oblique pattern)in guest system .
Proof:Because generic points are parallel or equivalent to principal overlapping space,from
formula (11), we can immediately prove that generic points are in the shape of punctiform in host
system.
Simultaneously add ai1e1+ai2e2
(i=1 , 2 , … , n-3) to the two ends of formula(10),it has
S1 ○ 2(a11e1+a12e2 ),
S1 ○ 2(a11e1+a12e2 ),
25
… … … … …,
Sn-3 ○ 2(an-3,1e1+an-3,2e2 ),
So ( ) ( )12 2 2211
3
1
3
1
eaeakSk ii
n
i
i
n
i
ii +∑∑ −
=
−
=
○
(ki are numbers),in convergent oblique pattern λ1a1j =λ2a2j =…=λn-3an-3,j (j=1 ,2)
( )
( ) ( ) ( )13 22 so
21321 so
3
1
3
1
22112211∑ ∑−
=
−
=
+=+
=−==
n
i
n
i i
iiii
i
j
ij
eaeak
eaeak
jnia
a
λ
λ,;,,,L
When we take all numbers of ki ,the right end of formula (13) represents that all linear
combinations of a1e1+a2e2 fill a straight line. viz. generic points are in the shape of line in host
system of convergent oblique pattern.
In emanative oblique pattern,it always has two vectors which are linearly indeqendent among a11e1+a12e2 ,
A21e1+a22e2 , … … ,
an-3,1e1+an-3,2e2 ,
so,when take all numbers of ki ,all their linear combinations fill a two-dimensional plane, viz.
generic points are in the shape of plane in host system of convergent oblique pattern. ▌
4....2....2 The uniqueness of generic points relative to upright axis
Theorem 2:Coordinates of a random generic point relative to upright axis are unique .
Proof:First suppose coordinates of a generic point relative to upright axis are not unique,viz. besides xn =xn1 ,
coordinates of a generic point relative to upright axis still have
xn =xn2,
So, the generic point includes the vector (xn-1xn2 )en .
And because figures represented by generic points are parallel or equivalent to principal
overlapping space,and principal overlapping space is all linear combinations of principal
overlapping direction,then (xn-1xn2 )en .
is a linear combination of principal overlapping direction,from formua (11) we know
(xn-1xn2 )en□ 0 .
but from formula (7),formula (8),en is reflective element,it should has
(xn-1xn2 )en□ (xn-1xn2 )en
and viz. xn1 = xn2 . ▌
4....2....3 Uniqueness of projection trace or opposite trace
Theorem 3:Projection trace of a random generic point is unique,the opposite trace is
unique ,too.
Proof:If we can prove that the trace of a random generic point of n-1-dimensional plane is
unique,the theorem can be immediately proved.still use disproof :
Suppose a random generic point A of n-1-dimensional plane has at least two traces on x1ox2
26
,
:
) 0,,0,,( and
) 0,,0,,( plane
2
20100
2
20100
876
L
876
L
−
−
n
n
bbB
aaA
and α1=b10-a10 , α2=b20-a20 ,
then principal overlapping space must include the vector
221100 eeBA αα += ,
From formula(11) we know α1e1+α2e2 □ 0 ,
but e1,e2 are all reflective elements,it should has α1e1+α2e2 □ α1e1+α2e2 ,
viz. α1e1+α2e2=0 ,
so, α1=α2 = 0 ,
viz. a10 = b10 , a20 = b20 .
then A0,B0 are the same pont. ▌
Equations of generic point (referred to in theorem 3 of the four chapter) can be discussed after
solving the graphic rules of preferred n-dimensional system.
Exercises
2.1 What's principal overlapping direction? what's principal overlapping space? what's the
"preference' of preferred n-dimensional system?
2.2 what's generic point ? what kinds can generic point be divided into?why?
2.3 What's the projection trace and opposite trace of generic point ?what's its principal
character(note:its coordinates relative to some coordinate axes)?
2.4 Why say“the projection trace and opposite trace of a random generic point on preferred
n-1-dimensional plane are the same point”?
2.5 Please give out three characters of generic point .
2.6 According to the given transformation
−
−
+−
+−
525
2124
2123
222
121
515
2114
2113
212
111
3
2
6
53
1
3
2
3
2
6
53
1
3
2
ee
eee
eee
ee
ee
ee
eee
eee
ee
ee
σ
σ
σ
σ
σ
σ
σ
σ
σ
σ
make out a couple of preferred five-dimensional systems.
2.7 According to formula (9)、formula(10),change the transformation formula of the above
exercise(exercise 2.6) into abbreviated formula.
2.8 Suppose the given transform formulas are
27
+
+
+
−−
−−
−−
215
214
213
215
214
213
643
964
325
643
964
325
eee
eee
eee
eee
eee
eee ○○○□□□
then please make out a couple of preferred six-dimensional systems.
2.9 Assume the given transform formulas are
−−
−
+−
+
+
+−
−
−−
216
215
214
213
216
215
214
213
435
65 3
734
35 5
43 5
653
73 4
355
eee
eee
eee
eee
eee
eee
eee
eee ○○○○□□□□
then please make out a couple of preferred seven-dimensional systems.
2.10 Among the above preferred n-dimensional systems of exercise 2.6、exercise 2.7、exercise
2.8 and exercise 2.9,which are emanative oblique pattern? which are convergent oblique pattern?
2.11 Please write out the oblique axes transform formulas of the following preferred
six-dimensional system:
2.12 Please write out the corresponding principal transform and subsidiary transform formula
according to the following host system of preferred seven-dimensional system,and make out the
guest system of the preferred seven-dimensional system.
exercise 2.12 exercise 2.13
2.13 Please write out the corresponding principal transform and subsidiary transform formula
according to the following host system of preferred nine-dimensional system,and make out the
guest system of the preferred nine-dimensional system.
28
The Third Chapter: Relation between figures and numbers
in preferred n-dimensional system
In the front two chapters we have discussed concept of oblique axes transform ,preferred
n-dimensional system、generic point and their property. On the basis,this chapter will discuss
relation between figures and their corresponding algebraic equations in preferred n-dimensional
system. This inclues two aspects of contents:(1) Given figures of a preferred n-dimensional
system,we can immediately make out the corresponding algebraic equations;(2) Given a algebraic
equation,we can also immediately make out the corresponding figure of preferred n-dimensional
system.
§§§§1 Translation trace of generic point————generic curved surface and
generic curve
1....1 Concepts of generic curved surface and generic
curve
A
ssume during the course of a generic point of n-dimensional
system accompanied with n-1dimensional plane continuously
transposing along the upright axis,A is also continuously
transposing on the n-1-dimensional plane,then the trace of A is
in the shape of curve(figure 11).we call the curve-shape figure
generic curve .When A sufficient-rank,the curve-shape figure is
also called sufficient-rank generic curve ; When A is
difficient-rank ,the generic curve is also called difficient-rank figure11
generic curve.
And suppose there is a generic curve on n-1-dimensional
plane,it is regarded as the trace of generic point A. When
n-1-dimensional plane is continuously transposing along upright
axis , the generic curve is also transposing on the
n-1-dimensional plane. and its shape either continuously
changes or has no change (viz. the movement of the generic
curve on the n-1-dimensional planemay not be transposition),then the trace of the generic curve must be in the shape of
curved surface(figure12),the figure is apparently regarded as the
trace of A . So,we call the figure which is in the shape of curved
surface generic curved surface. When A is sufficient-rank,the
generic curved surface is also called sufficient-rank generic
curved surface;When A is difficient-rank ,the generic curved figure12
surface is also called difficient-rank generic curved surface.
Generic curved surface and generic curve are usually referred to as hypersurface and
hypercurve. Here,we call them generic curved surface and generic curve,first, in order to
29
accommodate to the concept of trace of generic point in preferred n-dimensional system;second,it has Chinese feature in language. It is believed to be more easily accepted.
A particular case of generic curved surface and generic curve is curved surface and curve of
ordinary meaning ,they are regarded as the trace of zero-order generic point,so sometimes they
are called zero-order generic curved surface and zero-zero generic curve. Since then,all curved
surface and curve without "generic" before them are to be this kind.
1....2 Dimensions of generic curved surface and generic curve————dimension theorem
Dimensions of generic curved surface or generic curve can be divided into actual dimension
(marked as W ) and directviewing dimension(marked as Z ). Here it is mainly referred to be actual
dimension.
For generic curved surface and generic curve (include generic point) determined by system of
linear equations,they can be solved by linear algebra theory. Such as the actual dimension of
figures determined by system of homogeneous linear equations
( )∑=
==n
i
iij jxa1
2 1 0 L,,
is W=n-r,of which,r is the order of coefficients matrix of the equation set. But for the figures
determined by system of non linear equations,this method is inconvenient . Here,we introduce
another method.
Figures of preferred n-dimensional system can all be regarded as the trace of generic point.
From the analysis of the second chapter we know,the actual dimensions W of generic point is equal
to its orders (marked as J ). And the directviewing dimension of generic point is zero,so,the
dimensions of generic point can be expressed as W=Z+J=0+J=J ;
but generic curve is the trace of generic point transposing along certain paths,on the moving
paths ,it has directviewing dimension Z=1. So,the actual dimension of generic curve is
W=Z+J=1+J,
viz. its actual dimension is one dimension more than its above dimension of generic point; Generic
curved surface can also be regarded as the trace of generic curve transposing along certain paths,apparently,the moving path has one directviewing dimension. So,the directviewing dimension of
generic curved surface is Z=2,one dimension more than that of generic curve,two dimensions
more than that of generic point,viz. the actual dimension of generic curved surface
W=Z+J=2+J.
In addition,there are some curved surfaces of directviewing dimension Z=3 and somen
curves of directviewing dimension Z=2,they are all transposition traces of difficient-order generic
point and separately be called difficient-order generic solid and difficient-order generic curved
surface. The dimension of difficient-order generic curved surface has suited W=Z+J ;Difficient-ordergeneric solid is the transposition trace of difficient-order generic curvd surface,it is
three dimensions more than the directviewing dimensions of above difficient-order generic point,viz. W=Z+J=3+J.
Generalize the above description,we have the following dimension theorem:
Theorem 1:The relation between the actual dimension W of random figures in preferred
n-dimensional system and their directviewing dimension Z and order J is
W=Z+J. ▌
30
1....3 Kinds of generic curved surface and generic curve
Generic curved surface and generic curve can be divided into the sufficient-order and the
difficient-order,they can also be divided into the linear and the nonlinear according to equation
property. Of which,the generic curved surface and generic curve of linear equation are respectively
called generic plane and generic straight line,they can be regarded as a particular case of generic
curved surface and generic curve. Preferred n-1-dimensional plane can be regarded as a particular
case of generic plane,it is called generic coordinate plan.....................e ..
Generic plane、generic straight line and generic point are generally called linear fiqures,and
generic curved surface and generic curve of nonlinear equation are generally called non-linear
fiqures.
Equations of generic curved surface and generic curve will be discussed wehn solving the
figured rules of preferred n-dimensional system.
§§§§2 Graphic rules of preferred n-dimensional system————three kind of
graphic methods
2....1 Concepts of three graphic methods
Graphic rules of preferred n-dimensional system is how to identify the figured principle of
figures,graphic methods and the objects be figured,these contents form the basic principle and
methods of oblique axes draughting and high-dimensional spacial and analytic geometry.
Because figures of preferred n-dimensional system are regarded as the trace of generic point,and in the final analysis,generic point is regarded as the trace of points(moving along principal
overlapping direction and directions parallel to all linear combinations). In this meaning,we can
build the relation between the directviewing figures of generic curved surface or generic curve and
their corresponding algebraic equations:all points coordinates on the generic curved surface or
generic curve satisfy the equations of the generic curved surface or generic curve;All points
coordinates that aren't on the generic curved surface or generic curve don't satisfy their equations.
Contrary,all points that satisfy equations of generic curved surface or generic curve must be on the
generic curved surface or generic curve ;all points that don't satisfy equations of generic curved
surface or generic curve must not be on the generic curved surface or generic curve.
Judge a point to be on the generic curved surface(or generic curve) or not,it is not onoy
determined by direct-view(because for difficient-order generic curved surface and difficient-order
generic curve,some points are on the figures by direct-view,but they are not on them by the
analytic property of equations),but also be determined by figured rules(viz. three figured methods).
Defination 1: Coordinates of points be figured can be directly read from figures without
explanations ,this kind of figured method is called direct graphic method..
Defination 2: Coordinates of points be figured can't be directly read from figures, they need
explanations(include the analytic property of equations or proper auxiliary lines ) to give
coordinates of the point or point out some of their geometric properties,this kind of figured method
is called indirect graphic method..
Defination 3: Between the above two methods,method that using a sufficient-order generic
point and its translation trace to express all points belonging to them, it is called an ordinary figured
method.
Direct graphic method.s uses couples of preferred n-dimensional system,they represent the
31
coordinates of ordinary points(viz. zero-order generic point);both indirect graphic method.s and
ordinay figured methods use single n-dimensional system(referred to host system). Of which,indirect graphic method.s are not only used for ordinary points but also be used for difficient-order
and sufficient-order generic points,but ordinary figured methods are only used for sufficient-order
generic points and their translation
trace.
Before introducing three kind of
graphic methods,let's introduce
along-axis measuring method by a
visualized example[6],this is the
necessary basis of constituting figured
rules of preferred n-dimensional system——because the position of a random
point of n-dimensional euclidean space
in a random preferred n-dimensional
system is determined by this method.
Example 1: The preferred
five-dimensional system of figured 13 figure13
is generated by transform
−−
−−
214
213
233
232
eee
eee □□ ,
let's introduce the method of determining the position of point B(2,1,1,-3,2) in the system:
Accoring to coordinates of B relative to each axis
x1 = 2 , x2 = 1 , x3 = 1 , x4=-3 , x5 = 2
and take the following steps first:begin with the origin and make
11 2eOB = ;
secondly:begin with point B1 and make 221 eBB = ;
thirdly:begin with point B2 and make 332 eBB =
or 2132 51 ee.BB −−□ ;
fourthly:begin with point B3 and make 443 3eBB −=
or 2143 23 eeBB +□ ;
last(viz. fifthly),begin with point B4 and make 54 2e= BB .
then,the position of point B(2,1,1,-3,2) is determined (referred to in figure13).
2....2 Indirect graphic method and ordinary graphic method
Contents of direct graphic method is much,we will discuss it in §3 of this chapter. Here we
will first discuss indirect graphic method. and ordinary figured method.
32
2....2....1 Indirect graphic method
When coordinates of a point relative to each axis are known,first we find the proper position
of the point in preferred n-dimensional
system by along-axis measuring method,secondly label its coordinates relative to
each axis beside it,or point out it among
corresponding words or equations. Figure 14
is an example of removing auxiliary point
and auxiliary line of figure13 and label its
coordinates under point B to represent the
position of point B(2,1,1,-3,2) in the
preferred five-dimensional system.
We often take indirect graphic method
for points that can't be directly figured,sometimes we take indirect graphic method figure 14
for points that can be directly figured. The following are examples of building equations of generic
curved surface by indirect graphic method.
Example 2: figure 15 is a generic curved
surface whose centre is on the origin and the
radius is R of four-dimensional space,please find
out the equation.
Solution:The exercise give that the distance
between a random point of the generic curved
surface and the orgin is R,suppose the random point
is M( x1 , x2 , x3 , x4 ) ,
then it has |OM|=R,
viz. Rxxxx =+++ 2
4
2
3
2
2
2
1 ,
square the two ends of it,it has figure 15
22
4
3
3
2
2
2
1 Rxxxx =+++ .
this is the equation of the generic curved surfaceto be solved.
The generic curved surface is called generic sphere.
Example 3: It is known that a generic plane of preferred four-dimensional system
perpendicularily halve the line segment between point A(3,5,2,6) and point B(8,7,4,3)(refer to
it in figure16),please find out its equation.
Solution:from the exercise we know the generic curved surface to be solved is trace of points
that are equidistant from point A and point B,a random point on the generic plane
M(x1 , x2 , x3 , x4 )
has relation |AM|=|BM| ,viz.
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )2
4
2
3
2
2
2
1
2
4
2
3
2
2
2
1 34786253 −+−+−+−=−+−+−+− xxxxxxxx
Square the two ends of equal mark and simplify it ,it has
5x1+2x2+2x3-3x4-32=0,
33
This is the equation of generic plane to be solved.
As will be readily seen,the figures and methods used in "analytical geometry of three
dimensions" of High Mathematics [7],are mostly equivalent to the indirect graphic method. to be
referred to.
2....2....2 Common generic of points
The principle of ordinary graphic method is
based on points having common generic theory,and points having common generic theory
condense oblique axes transform to a kind of
simple sign and make the description more
shortcut and convenient.
Defination 4: Suppose the principal
overlapping direction of a preferred
n-dimensional system is
S1=r11e1+r12e2+r3e3,
S2=r21e1+r22e2+r3e3 , … … … , figure 16
Sn-3=rn-3,1e1+rn-3,2e2+rn-1en-1 ,
(in each formula,ri1 , ri2 aren't zero at the same time,ri+2 ≠0;i=1 ,2 ,… ,n-3 ),then its
coefficient matrix.
=
−−− 12,31,3
42221
31211
nnn rrr
rrr
rrr
RLLL
is called principal overlapping matrix.
Defination 5: In preferred n-dimensional system, if more than two points A1,A2,…,Am
(m≥2) belong to the same generic pointA (or explain that they have common generic A ),we call
these points common-generic,marked as
A1A2…Am or A~ A1A2…Am
Theorem 2:Suppose more than n-2 points A1A2…Am (m≥n-2),the vector of a random point
Ai(1≤i≤m) to other points is
{ } ( )jimiijAAaAA jnjjji ≠+−== ; ,,1,1,,1,,, 21 LLL ,
then it make the necessary and sufficient conditions of
A1A2…Am
is ,matrixes
=+++
−−−
mnmm
niii
niii
n
aaa
aaa
aaa
aaa
A
L
LLLL
L
L
LLLL
L
21
,12,11,1
,12,11,1
11211
composed of coordinates of these vectors can be changed into principal overlapping matrix R by
34
elementary transform.
Proof :For the convenience of description,we may suppose i=m and j=1 ,2 ,… ,m-1
of ji AA .If A can be changed into principal overlapping matrix R,then vectors 1AAm ,2AAm ,…, 1−mm AA are respectively a linear combination of principal overlapping direction S1,
S2,…,Sn-3 ,so when order
( )numbers are 0 because
3
1
22311311
22212144
21211133
i
n
i
i
,n,nnn
kSk
rrr
rrr
rrr
∑−=
−−−−
−−
−−
−−
□ □□□ eee
eee
eee
LLL ,
〔refer to it in formula (11) of the second chapter〕,so it also has
( )numbers are 0 3
1
i
n
i
jm kAA∑−=
□
then ,in the preferred n-dimensional system taking S1,S2,…,Sn-3 as principal overlapping
direction ., A1A2…Am
this is sufficiency.
If A can't be changed into principal overlapping matrix R,then despite a random group of S1,S2,…,Sn-3 are choosed as principal overlapping direction ,among A1,A2,…,Am there are always
such points,the vectors of them to other points can't be linearly represented by the group S1,S2,…,Sn-3 ,so,in preferred n-dimensional system taking a random group S1,S2,…,Sn-3 as
principal overlapping direction ,they all can't make
( )numbers are 0 3
1
i
n
i
jm kAA∑−=
□ ,
viz. they can't make A1A2…Am
this is necessity. ▌
From theorem 2 of the second chapter,the coordinates of generic points to upright axis are
uniquely determined,so the coordinates of A1,A2,…,Am (m≥n-2) to xn are the same. But when
the coordinates of A1,A2,…,Am (m≥n-2) to xn are not identical but identical to another axis
xk (1≤k≤n-1),we can take axis xk as upright axis and change xn into another axis(accordingly,the nonzero column suffix of the k column in principal overlapping matrix are all changed into n),we can still order A1A2…Am
In actual application,we usually first take the opposite trace of generic points. But for the
other points(altogether n-3), we take the vector terminal which takes the opposite trace as starting
point and parallel to every principal overlapping direction.
35
2....2....3 Ordinary figured method
Suppose there a group of points A1,A2,…,Am (m≥n-2) ,they satisfy the conditions of
theorem,find out the principal overlapping matrix R according to the coordinates of these points,then write out the corresponding oblique axis transformation,in order to make out preferred
n-dimensional system which orders
A~ A1A2…Am
(m≥n-2) ,in the n-dimensional system,first find out the projection trace of generic point A and
mark out by Ja′,then make out nnaa eaJJ □′
(point is marked as Ja A,Ja is the opposite trace of generic point A ) according to the coordinates
of these points to the coordinate xn = an of upright axis,rub out the connecting line between Ja′
and Ja (is marked as A ) then the ordinary figure to generic point is finished. The method can be
generalized into three steps:
(1) choose preferred n-dimensional system that can make
A~ A1A2…Am
(m≥n-2) ;
(2) determine the position of projection trace
Ja′of generic point A ;
(3) determine the opposite trace Ja viz. the
position of generic point A (so we mark it as A
not Ja ).
In ordinary figured method we should pay
attention to:while we figure a generic point,we
must mark out its position of projection trace. If
we don't mark out the position of projection trace,
then it indicates that the generic point is on figure 17
generic coordinate plane and it coincides with the trace point.
Example 4: Figure 17 shows a generic plane of preferred six-dimensional system,please find
out its equation.
Solution: The preferred six-dimensional system that the generic plane is in is generated by
transform
5213
4212
3211
115
214
213
5 3 3
5
is system theofdirection ramparting principal theindicatesit
335
5
eeeS
eeeS
eeeS
eee
eee
eee
+−−=
+−−=
+−−=
+
+
+□□□
So, random generic points on the generic plane are all correspondent with the vectorS1,S2,S3 .
In figure17,generic points C and D are on the generic plane and they concide wit their trace,then we can easily know the two points are
36
Jc (-4,3,0,0,0,0),Jd (-2,0,0,0,0,0),
but the projection trace of generic point A is
Ja′(-4,0,0,0,0,0),
so its oppposite trace is Ja (-4,0,0,0,0,6).
We suppose the generic plane is composed of the translation trace of generic point C along
. 3
1
32 suppos and
and
625
214
eeJJ
eeJJ
JJJJ
ac
dc
acdc
−=−=
−==
S
S
,
Respectively regard the coordinates of S1,S2,S3 ,S4,S5 as elements of the following
determinants from the second row to the sixth row,then the normal direction vector of the generic
plane { }1 5 3 1 2 3
200010
000032
010011
005033
000511
50
1
654321
,,,,,=
−
−
−
−
−
=
eeeeee
H ,
is acquired.Choose a known point Jd and a random point
X (x1 , x2 , x3 , x4 , x5 , x6 ) ,
and use dot-method formula [7]
{ } ( ){ } .xxxxxxXJ d ,,,,,,,,,, 021 5 3 1 2 3 654321 =−−⋅=⋅H
then we get the general equation of the generic plane
065323 654321 =++++++ xxxxxx .
§§§§3 Direct graphic method
3.... 1 Single principal overlapping direction、、、、 oblique number and oblique
coordinate
Defination 6:Suppose the principal overlapping direction of a preferred n-dimensional system
is ( )3 2 1 222211 −=++= ++ nieaeaea iiiii ,,, LS ,
of which, ai1 and ai2 aren't zero at the same time,ai+2 ≠0 , and at least two numbers of
each group of ai1 ,ai2 ,ai+2 are relatively prime ................... And suppose
{ }0 call then we
1321321
22,32212
11,32111
,,,,,
,
,
−−
−
−
=+++
=+++
=+++
nn
n
n
aaaa
aaaa
aaaa
LL
L
L
SSS
is single linear combination of principal overlapping direction,it is also called single principal
overlapping direction of the preferred n-dimensional system.
Defination 7 : From the third coordinate to the (n-1) -th coordinate a3 ,a4 ,…,an-1 of
single principal overlapping direction { }0 1321 ,,,,, −= naaaa LS
37
are called oblique axes transform number,they are called oblique number for short,we use a vector
S(xs) to denote them,viz. S(xs) ={0,0,a3,a4,…,an-1,0} .
Defination 8: The coordinates x3 , x4 , … ,xn-1 of a random point A(x1 , x2 ,… ,xn ) of
preferred n-dimensional to each oblique axis are called oblique coordinates of A,we also use a
vector A(xb) = {0,0,x3,x4,…,xn-1,0}
to denote it.
3....2 Coordinate transform of points
Supppose images of a point A(x1 , x2 ,… ,xn )
under singular linear transform σ1 and σ2 are
A01 ( x1′, x2′,0 ,… ,0 , xn′),A02 (x1", x2", 0 ,… ,0 ,xn") 〔transform formula like formula (9) and formula (10) in the second chapter〕,regard A as the
terminal point of vectorOA (O is the origin),then the transform of A coordinate can be regarded
as the transform of OA coordinate. Successively substitute formula(9) and formula(10) of the
second chapter into OA , then it has
,,○
□
OA
21
1
2,3
4
4
223
3
122
11
1
1,3
4
4
213
3
111
21
1
2,3
4
4
223
3
122
11
1
1,3
4
4
213
3
111
nnn
n
n
n
n
n
nnn
n
n
n
n
n
exexa
ax
a
ax
a
ax
exa
ax
a
ax
a
ax
exexa
ax
a
ax
a
ax
exa
ax
a
ax
a
axOA
+ +++++
+ ++++
+ −−−−+
+ −−−−
−
−
−
−
−
−
−
−
−
−
−
−
L LL L
but it also has the terms , ○□ , 0201 OAOAOAOA
.
viz.
2211
2211
nn
nn
exexexOA
exexexOA
′′+′′+′′
′+′+′○□ ,
( )
( )2
1 hasit so
1
1
2,3
4
4
223
3
1222
1
1
1,3
4
4
213
3
1111
1
1
2,3
4
4
223
3
1222
1
1
1,3
4
4
213
3
1111
=′′
++++=′′
++++=′′
=′
−−−−=′
−−−−=′
−
−
−
−
−
−
−
−
−
−
−
−
nn
n
n
n
n
n
n
nn
n
n
n
n
n
n
xx
xa
ax
a
ax
a
axx
xa
ax
a
ax
a
axx
xx
xa
ax
a
ax
a
axx
xa
ax
a
ax
a
axx
L
L
L
L
The above formula(1) and formul(2) can be derived from formula(7)、formula(8) of the second
38
chapter and formula(6) of the first chapter.
Suppose the relation between the oblique coordinates A(xb) and the oblique numbers S(xs) of
point A is A(xb) =λS(xs) ,
(it is read as “the oblique coordinates of A are equal to the oblique numbers of λS ”) ,viz.
λ====−
−
1
1
4
4
3
3
n
n
a
x
a
x
a
xL
(λ is a number),because a11 + a21 + … + an-3,1 = a1 ,
a12 + a22 + … + an-3,2 = a2 ,
then,the above two formulas are changed into
( )
( )4
3
222
111
222
111
=′′
+=′′
+=′′
=′
−=′
−=′
nn
nn
xx
axx
axx
xx
axx
axx
λ
λ
λ
λ
because A(xb) =λS(xs) ,
it has x3 =λa3 , x4 =λa4 , … , xn-1 =λan-1 ,
add up formula (3) and formula(4) ,it has
( )
( )
( )
( )5
1,,4,3
2
1
2
1
11
33
222
111
′′=′=
=
−==
=
′′+′=
′′+′=
−−
nnn
nn
ii
xxx
ax
niax
ax
xxx
xxx
λ
λ
λ
LL
L
LL
For direct graphic method, point coordinate transform under oblique axes transform has great
significance. In fact,formula(3) and formula(4) are basis of the figured methods when point A of
preferred n-dimensional satisfies A(xb) =λS(xs) ,and formula(5) is the basis of reading the
coordinates of the points be directly figured.
3....3 The principle of direct graphic method
Theorem 3:The necessary and sufficient conditions of a random point A of n-dimensional
space can be directly figured in a couple of preferred n-dimensional system is
A(xb) =λS(xs)
(λ is a number).
39
Proof:From formula (3),formula(4),formula(5),if coordinates of A satisfy
A(xb) =λS(xs) ,
then it can be directly figured in preferred n-dimensional system which takes S as single principal
overlapping direction,this is sufficiency.
If coordinates of A don't satisfy A(xb) =λS(xs) ,
What about the result? Please look at the following example:in figure 18,
S1 =3e1 + 2e2 +2e3 ,S2 =3e1 + 2e2 +3e4 ,
then S={6,4,2,3,0}, S(xs) ={0,0,2,3,0} ,
coordinates of point A(0.9,-0.6,0.5,0.75,6) satisfy
A(xb) = 1 4 S(xs),
Figure 18 indicates that it can be directly figured in the couple of preferred five-dimensional
system. but it is also observed that,point (0.9,-0.6,1,0,6) ,(0.9,-0.6,1.5,-0.75,6) and
point A belong to the same generic point,their position in host system and guest system concide
with A
figure 18
completely.
If we don't regulate the relevance between the oblique coordinates and oblique numbers of
points,it will arouse chaos,The direct diagrammatic representation of points become yeasty
talk.This is necessity. ▌
From theorem 3,the following two inferences can be immediately acquired.
Inference 1:The necessary and sufficient conditions that more than
two points A1,A2,…,Am (m≥2)
of n-dimensional space can be directly figured in the same couple of preferred n-dimensional
system is λ1A1(xb) =λ2A2(xb) = … =λmAm(xb) = S(xs)
(λ1,λ2,…,λm are numbers). ▌
Inference 2: More than two points of four-dimensional space can be directly figured in the
same couple of four-dimensional system . ▌
3.4 Drawing and distinguishing steps of direct graphic method
For a random point A(x10 , x20 , … ,xn0 )
of n-dimensional space,we can directly figure it by the following steps:
First,choose a group of nonzero oblique axes transform numbers according to the oblique
coordinates of A a3 ,a4 ,… ,an-1
and represent it by vector S(xs) ,in order to make
A(xb) =λS(xs) .
Second,determine a group of principal overlapping directions according to S(xs)
40
Si = ai1e1 + ai2e2 + ai+2ei+2
(i= 1 ,2 ,… ,n-3 ) notice:ai+2 ≠ 0 ,ai1 and ai2 aren't zero at the same time,and at least two
numbers of ai1 ,ai2 ,ai+2 are relatively prime ................... Then make out a couple of preferred n-dimensional
system according to the group of principal overlapping directions .
Third,respectively mark out the projection A′(x10 , x20 , … , xn-1,0 , 0 )
of A on generic coordinate plane by along-axis measuring method in host system and guest system,last respectively mark out the position of A in host system and guest system(if the projection A′of
A is not marked out,it indicates that A is on generic coordinate plane),then the direct figure of A is
to be finished.(figure18).
The distinguish (viz. the reading of
coordinates of A and other ppints)process is
finished in guest system. Take the guest system
of figure 18(figure19)as example:first find out
the projection trace position of generic point A
that A is on and mark it out by Ja′,connect
Ja′and A′,from the front two equalities of
formula(5) we know,the projection of A on
plane x1ox2 is at the midpoint A"of line
segment Ja′A′,
respectively figure out vertical lines to axes figure 19
x1 , x2 through point A",then its coordinates to axes x1 , x2 are x1 = x10 ,x2 = x20;and
( ) ( )xsxb SAAA λ==′′′ ,
then coordinates of A to each oblique axis are
x3 =λa3 = x30 ,
x4 =λa4 = x40 , … … … … ,
xn-1 =λan-1 = xn-1,0 ,
this is identical to formula (5). From directly measurement,it has nn exAA 0=′ ,
so the coordinates of A to upright axis are xn = xn0 (refer to it in figure 19).
figure 20
Besides,the figured objects of direct graphic method also includes such points as projection
traces or opposite traces:In A(xb) =λS(xs) when λ=0,the coordinates of A to each oblique axis
are all zero. So,in host system and guest system,their (include their projection on the generic
41
coordinate plane) positions to axis x1 , x2 are identical.
Figure 20 is an example of using theorem 3 and reference 1(in the figure,upright axis xn is
omitted).In the preferred n-dimensional system,point A1 and point A2 are directly figured at the
same time. But figure 21 gives the reading results of the coordinates of the two points:
A1 (a11 , a12 , a3 , a4 , … , an-1 , 0 ),
A2 (a21 , a22 , 2a3 , 2a4 , … , 2an-1 , 0 ).
The following is an example of building
generic curved surface equation using theorem 3
and reference2:
Example 5: In preferred four-dimensional
system of figure 22,there is a curved surface in the
shape of cylindroid (in host system,upright axis x4
is omitted),please find out its equation according to
the conditions given by the figure.
Solution:The cut trace of the cylindroid-shape
surface cut by generic coordinate plane is a piece of
elliptical continuum,after being read,the four
points of the elliptical continuum separately are: figure 21
( ) , 0,2
,2
,2
, 0 , , 0 , 0 , 0,2
,2
,2
, 0,0,2
,2
−−
−
−
RRRDRC
RRRB
RRA
figure 22
apparently,they aren't coplanar,and their distance to the origin is:
|OA|=|OB|=|OC|=|OD|=R ,
then we can extrapolate that the distance of a random point M (x1 , x2 , x3 , 0 ) of the elliptical
continuum to the origin is |OM|=R ,viz.
Rxxx =++ 2
3
2
2
2
1
square the two endas,it has 22
3
2
2
2
1 Rxxx =++ .
Now look at the elliptical continuum where x4 =x40 ,it is identically equal to the elliptical
continuum of generic coordinate surface,this shows that when the elliptical continuum with
generic coordinate surface translates along upright axis,its position and shape with respect of
upright axis are changeless,so,the equation of the cylindroid-shape surface has no relation with
42
x4.then,equation 22
3
2
2
2
1 Rxxx =++
without variable x4 is the equation of the cylindroid-shape surface,we call the cylindroid-shape
surface generic cylindrical surface.
The elliptical continuum of generic coordinate surface can be called generic circular curve or
generic circle for short,it is generic intersecting line of generic cylindrical surface and generic
coordinate surface,that is to say ,points of the generic circle is not only on the generic cylindrical
surface but also on the generic coordinate surface. then,a random point of the generic circle
simultaneously satisfies equations of generic cylindrical surface and generic coordinate surface.
So,equations of the generic circle are simultaneous system of equations of generic cylindrical
surface and generic coordinate surface
.04
22
3
2
2
2
1
=
=++
x
Rxxx
In fact,the generic circle is a sphere of three-dimensional space . Its centre is on the origin
and its radius is R.
From the above analysis, it is observed that a generic curved surface of preferred
n-dimensional system can be expressed by an equation
f ( x1 , x2 , … , xn ) = 0
among x1 , x2 , … , xn ,contrarily,an equation
f ( x1 , x2 , … , xn ) = 0
among x1 , x2 , … , xn can express a generic curved surface of n-dimensional system. and generic
curve serves as intersecting line of more than two generic curved surface
f1 ( x1 , x2 , … , xn ) = 0
f2 ( x1 , x2 , … , xn ) = 0 … … … … …,
fm ( x1 , x2 , … , xn ) = 0
(m≥2),its equations are simultaneous system of equations of esquations of these generic curved
surface
( )( )
( )
=
=
=
0
0
0
21
212
211
nm
n
n
x,,x,xf
x,,x,xf
x,,x,xf
L
LLL
L
L
(m≥2).
Exercises
3.1 How to determine the dimension of generic curved surface and generic curve?
3.2 What are linear figures? What are non-linear figures?
3.3 What are figured rules of preferred n-dimensional system?
3.4 What are three kind of graphic methods? Why there are neither more nor less than three
kind of graphic methods?
3.5 In the following preferred six-dimensional system,please mark out the position of point (3,5,2,7,-6,4) according to along-axes measuring method.
43
3.6 Suppose the principal ramparting direction of a preferred five-dimensional system is
2e1 +3e2 + 5e3 , 4e1 + 6e2 + 3e4 ,
the projection trace of a generic point is Ja′(2,-3,0,0,0) ,
its opposite trace is Ja (2,-3,0,0,4) ,
please picture A.
3.7 Suppose there are five points
A0 (2,-1,0,0,0,0,5), A1 (5,3,4,0,0,0,5),
A2 (5,3,0,3,0,0,5), A3 (5,3,0,0,5,0,5) , A4 (5,3,0,0,0,4,5), please make out a preferred seve n-dimensional system and make
A0A1A2A3A4 .
3.8 Please write out equations of the following generic plane according to the methods of
example 4.
3.9 Please picture point A (2,5,4,6,5,6)
by direct graphic method.
3.10 Please determine coordinates of point A according to conditions offerred by the
following figures.
3.11 Please judge which points of the following can be directly pictured in the same couple of
preferred n-dimensional systems,which can't be directly pictured:
A1 (3,2,5,3,3,5), A2 (5,4,7,1,5,6),A3 (2,5,3,6,4,3), A4 (4,-3,10,6,6,7), and picture the points that can be directly pictured in the same couple of n-dimensional system.
44
The Fourth Chapter: Shape of figures in preferred
n-dimensional system
Figures in preferred n-dimensional system can be divided into linear and non-linear according
to their property.Their shape is referred to whether they are in the shape of directviewing
plane,line or point or not in a preferred n-dimensional system. The following respectively narrate
them.
§§§§1 Shape of linear figures
1....1 Generic planes and their shape
When the equation of generic curved surface F
f (x1 ,x2 ,… ,xn ) =0
is a first-order equation about x1 ,x2 ,… ,xn ,F represents a generic plane(refer to it in example
3 and example 4 of the third chapter),it can be regarded as a particular case of generic curved
surface,and generic coordinate plane can be regarded as a particular case of generic plane,its
equation is always xn = 0 .
The same generic plane isn't always in the shape of plane in different preferred n-dimensional
system,different generic planes are not always in the shape of plane in the same preferred
n-dimensional system. In order to simplify it, we are always ready to change generic planes we
studied into planes as many as possible. Our desire can at least be satisfied for each separate
generic plane.
Example 1: Suppose there is a generic plane of six-dimensional space
F: x1 +3x2 +4x3 +2x4 +x5 +3x6 + 5 = 0
please change it into a figure which is in the shape of plane.
solution::::choose four coordinates on F, their x 6 are all zero at the same time
A0(-5,0,0,0,0,0),A1(1,2,-3,0,0,0),
A2(1,2,0,-6,0,0),A3(1,2,0,0,-12,0),
then it gets a group of vectors
. 1226
626
326
521303
421202
321101
eeeAA
eeeAA
eeeAA
−+==
−+==
−+==
S
S
S
,
,
order A~ A0A1A2A3
then it gets a preferred six-dimensional system which take S1 ,S2 ,S3 as principal overlapping
direction,A is a sufficient-order generic point of the six-dimensional system andA is on F. So,F
can be regarded as the trace of A,viz. the orders of F. J = 6-3 = 3 . From linear algebra theory,we
know the actual dimensions of F are W = 6-1 = 5 ,then according to dimension theorem of the
third chapter,the directviewing dimensions of F are
Z = W-J = 5-3 = 2 ,
viz. F is changed into a figure in the shape of plane. (refer to it in figure 23).
Example 2: Suppose an equation of a generic plane in five-dimensional space is
F:3x3 +2x4 -x5 –3 = 0
45
please change it into plane .
solution:Because coefficients of the front two terms are zero, points that whose coordinates
relative to x5 are identical and satisfy other conditions of theorem 2 in the third chapter are
ompletely
figure 23
exist. So,choose x5 asvertical axis,and change x2 into upright axis. So,choose three points on F
A0 (2,4,0,0,-3), A1 (5,4,2,0,3), A2 (5,4,0,3,3).
then it gets . 633 623 5412053110 eeeAAeeeAA ++=++= ,
order A~ A0A1A2 ,
viz. use transform
−−
−−
514
513
2
632
eee
eee □□
then we can change F into plane.(figure 24).
Normally,for a random generic plane whose
a1 and a2 aren't zero at the same time(or a1 and
a2 are zero at the same time,but a3 , a4 , … , an-1 are also zero at the same time)
F:a1 x1 + a2 x2 + … + an xn + a = 0
On it, we can all find n-2 points A0 ,A1 ,… ,An-3 figure 24
with coordinates xn are identical and make
302010 −nAAAAAA , , , L
linearly indeqendent and satisfy other conditions of theorem 2 in the third chapter,order
A~ A0A1A2…An-3 ,
then it gets a preferred n-dimensional system(sometimes, 302010 −nAAAAAA , , , L can
become its principal overlapping direction),A is a sufficient-order generic point of the
n-dimensional system and A is on F,viz. F is regarded as the trace of A .
46
So,the order of F is J = n – 3 ,its actual dimension is W = n –1 ,So from dimension
theorem W=Z+J ,then the directviewing dimension of F become
Z = W – J = ( n - 1) - ( n – 3 ) = 2 ,
viz.F become in the shape of plane.
But,like generic plane of example 2, coefficients a1 and a2 of the front two terms of their
equations are zero at the same time,but their coefficients from the third term to the n-1-th term are
not zero at the same time. If we still order x1 ,x2 ,x5 separately as cross axis ,vertical axis and
upright axis,because the points we choosed can't satisfy theorem 2 of the second chapter (the
coordinates of generic points relative to upright axis are uniquely determined),then they have no
common generic.〔for example,order x3 ,x4 ,…,xn-1 are zero at the same time,it has
n
na
ax −=
(when determine the opposite trace of generic points ,we should take values like this),if order one
of x3 ,x4 ,…,xn-1 isn't zero,the others are still zero at the same time (when determine the
terminal points of vectors parallel to every principal overlapping direction and take the opposite
trace of generic points as starting points,we should take values like this),but it has
n
na
ax −≠ 〕,
so it is impossible to change the generic plane into plane .But if we change one of x1、x2 into upright
axis,and change xn into cross axis(or vertical axis),then the generic plane can be changed into
plane.
It should be specialized that,when the coefficients of variable x3 ,x4 ,…,xn-1 of equations
of a generic plane are zero at the same time,no matter how take values to x3 ,x4 ,…,xn-1, they
all can make the equation be ture. So,the generic plane can still be changed into plane.
As the above description,we has:
Theorem 1: For a random generic plane of n-dimensional space
∑=
=+n
i
ii axaF1
0:
we can always change it into the shape of plane in a preferred n-dimensional system by proper
oblique axes transformation. ▌
When a generic plane isn't in the shape of plane in a preferred n-dimensional system,the
generic plane can be regarded as trace of difficient-order generic points,so it is also called
difficient-order (or n - 4-order ) generic solid.〔Proof:From linear algebra theories,the actual
dimension of generic plane is W = n – 1,and from dimension theorem W = Z + J ,and the
maximum directviewing dimension of a random figure Z≤3,so,when generic plane isn't in the
shape of plane,its order J = W - Z = ( n - 1 ) - 3 = n - 4〕.
1....2 Generic straight lines and their shape
In three-dimensional space,there is a concept that the intersection of two planes determines a
straight line,extend the concept to n-dimensional space,then it is the intersection of two generic
planes determines a generic straight line.
Suppose two generic planes
F1 :a11 x1 + a12 x2 + … + a1n xn + a1 = 0
F2 :a21 x1 + a22 x2 + … + a2n xn + a2 = 0
intersect on generic straight line L,then the equation of L is compatible system of equations
=++++
=++++
0
0
22222121
11212111
axaxaxa
axaxaxaL
nn
nn
L
L
:
47
( ) 2 , 1 0 or 1
==+∑=
jaxaLn
i
jiji:
A same generic straight line isn't always in the shape of line in different preferred
n-dimensional systems,different generic straight lines aren't always in the shape of line in a same
preferred n-diemensional system. But in order to simplify it,we are always ready to change generic
straight lines we studied into the shape of line. Our desire can at least be separately satisfied for
every single generic straight line.
Example 3: Suppose there is a generic straight line in six-dimensional space
, 021533
0145332
654321
654321
=++++++
=++++++
xxxxxx
xxxxxxL:
Please change it into the shape of line.
Solution:first find out four points on L
A0(-7,0,0,0,0,0), A1(-5,1,-7,0,0,0), , , 0 , 5
7 , 0 , 0 , 1 , 5 0 , 0 , 0 ,
3
7 , 1 , 5 32 −− −− AA
and order A0A1A2A3
then a preferred six-dimensional system is aquired. Because vector 302010 AAAAAA , , can be
regarded as the principal overlapping direction of the preferred six-dimensional system,and all of
its linear combinations are a sufficient-order generic point,so L is regarded as the trace of
sufficient-order generic points,viz. now its order is J = 6 – 3 = 3. From linear algebra theory,its
actual dimension should be W = 6 - 2 = 4,then from dimension theorem its directviewing
dimension is Z = 4 - 3 = 1,viz. L is changed into the shape of line(figure 25).
Example 4: Suppose there is a generic straight line in five-dimensional space
Figure25
,: 0223
074232
54321
54321
=++++
=+++++
xxxxx
xxxxxL
please change it into the shape of line.
Solution: First find out three points on L
A0(1,-3,0,0,0), A1(0,-7,7,0,0), A2(-4,-2,0,7,0),
48
and order A0A1A2
then a preferred five-dimensional system of emanative
oblique pattern,in the five-dimensional system,L is in the
shape of line(figure 26).
Example 5: Suppose
=+−+
=−−+
013
04322
421
321
xxx
xxxL:
is a plane of four-dimensional space,please change it into the
shape of line.
Solution:because the rank of the coefficients matrix of its figure 26
front two terms' coefficients is 1, a vector between two random points α11e1 +α12e2 +α3e3
can't makeα11 ,α12 zero at the same time andα3≠0. Then,change x4 into vertical axis and
change x2 into upright axis,choose point
( )( ), 2,2,0,5 and
1,0,0,2
1
0
A
A
on L, then a vector between the two points is acquired
3e1 + e4 +2e3 ,
Order A0A1 ,
viz. using the transform 2e3 □ -3e1 - e4 ,
can change L into the shape of line.(figure 27).
Let's analyze the front examples:in example 3,the rank of coefficients matrix
( )
=
× 153113
153132
2 njia
is r = 2 , of which the rank of coefficients
matrix ( it is a submatrix of ( a j i )2 × n ) of
variables x3 ,x4 ,x5
( )
=
−×+531
531
)3(22, nija
is r′= 1 . But in example 4, both the ranks of coefficients
matrix (aji)2×n and skew matrix (aj,i+2 )2×(n-3) are 2. So,the
generic straight lines of example 3 can be in the shape of line in figure 27
a convergent oblique pattern,but the generic straight lines of example 4 can only be in the shape of
line in a emanative oblique pattern. But,when we exchange variable x4 and x5 of the equation of
example 4 and newly arrange them,it will has
=++++
=+++++′
0223
074232
54321
54321
xxxxx
xxxxxL :
49
because the rank of the skew matrix becomes 1,then L′can be in the shape of line in a preferred
five-dimensional system of convergent oblique pattern.(figure 28).
Normally,for a random generic straight line of n-dimensional space
( )2 , 1 01
==+∑=
jaxaLn
i
jiji:
when the rank of coefficients matrix of its front columns is 2,we can find out n-2 points
A0 ,A1 ,A2 ,…,An–3 whose coordinates relative to xn are identical,Of which,A0 is the opposite
trace of a sufficient-order generic point on generic straight line L,A0 ,A1 ,A2 ,…,An–3 separately
is terminal point of every principal overlapping direction which takes A0 as starting point. The
following is our detailed proof.
First suppose xn =αn ,and suppose
2221
1211
aa
aaD =
(because the rank of coefficients matrix formed of the front two terms' coefficients of the
equation of L is 2,so D ≠ 0 ),and
, , 221
111
2
222
121
1ta
taD
at
atD ==
of which, t1 = - a1nαn - a1 , t2 = - a2nαn - a2 ,
when suppose x3 = x4 = … = xn-3 = 0 ,and D ≠ 0 ,according to Cram's rule,it has
,, 222
111
D
Dx
D
Dx ==== αα
Then point A0(α1 ,α2 ,0,…,0,αn )
is first choosen.
And suppose the other n–3 points are expressed by
Ak ( xk1 ,xk2 ,0 ,…,0 ,xk+2 ,0 ,…,0,αn )
( k=1 , 2 , … ,n -3 ) ,for the convenience ,we can suppose
( )
. which of
suppose and number realany can take , 0hen 0
21
22
122222222
111 1 12211
22121
21
22
+
+
++
++
+++
+
++
=
−=−−−=−−−=
−=−−−=−−−=
=≠
==
k,
k,
k
knnnnkk,k
nnnnkk,k
kk,k,
k,
kk
a
a
DtaaDkaaat
DtaaDaaat
xawa
a
Dx
λ
λααα
ααα
α
,
,
, ,
because we have already suppose D ≠ 0 , and when we suppose
, , 221
111
2
222
121
1
k
k
k
k
k
kta
taD
at
atD ==
according to Cram's rules,it has
50
,, 2
22
1
11D
Dx
D
Dx k
kk
k
kk ==== αα
viz. point A1(α11 ,α12 ,α3 ,0,…,0,αn),
A2(α21 ,α22 ,0,α4 ,0,…,0,αn), … … … … … … … ,
An-3(αn-3,1 ,αn-3,2 ,0,…,0,αn-1 , αn)
is also choosen.Order A0A1A2…An-3
a preferred n-dimensional system taking
302010 −nAAAAAA , , , L as principal overlapping
direction is acquired. Now,A0 ,A1 ,A2 ,…,An–3 have a
common sufficient-order generic point,and the generic point is
on L . So ,L can be regarded as trace of the generic point,viz.its
order is J = n- 3,from linear algebra theory,the actual
dimension of L is W = n – 2 ,then from dimension theorem
we know the directviewing dimension of L is Z = W – J = (n – 2 ) - (n – 3 ) = 1
viz. L become in the shape of line.
It is easy to know the rank of coefficients matrix figure 28
( )
=×
n
n
njiaaa
aaaa
22221
11211
2L
L
of L is r = 2 , when the rank of its skew matrix
( )
=−
−
−×1,22423
1,11413
)3(2n
n
njiaaa
aaaa
L
L
r′≤1(viz. r′=0 or r′= 1 ),it has λ1 =λ2 = … =λn-3 ,
then it has t12 = t22 = … = tn-3,2 .
so α11 =α21 = … =αn-3,1 , α12 =α22 = … =αn-3,2 ,
viz. the preferred n-dimensional system is convergent oblique pattern. When r′=2 ,only when λ1 =λ2 = … =λn-3 = 0
t12 = t22 = … = tn-3,2
is true,but because 02,1
2,2==
+
+
k
k
ka
aλ ,
it must has a23 = a24 = … = a2,n-1 = 0 ,
then ,the rank of skew matrix become r′=1 , it contradicts with the assumption r′=2. So the
n-dimensional system can only be emanative oblique pattern.
When the rank of the front two terms' coefficients matrix of L is 1,then the above point
A0 ,A1 ,A2 ,…,An–3 which simultaneously satisfy theorem 2 and theorem 3 of the second
51
chapter can't be simultaneously found out,so the generic straight line can't be directly changed into
the shape of line,now,we can choose another axis as cross axis (or vertical axis) instead of x1(or x2 ),and change x1 (or x2 ) into other axis,in order to change the rank of the two terms'
coefficients matrix of L corresponding to new cross axis and new vertical axis into 2,viz. change L
into the shape of line.
As the above description,we has:
Theorem 2::::For a random generic straight line of n-dimensional space
( )2 , 1 01
==+∑=
jaxaLn
i
jiji:
we can always make it in the shape of line in a preferred n-dimensional system by proper oblique
axes transform. Of which,when the rank of coefficients matrix (aji )2×n r=2,the rank of the front
two columns of submatrixs (ajk )2×2 is 2 ,when the rank of skew matrix (aj,i+2 )2×(n-3) r′≤1,the preferred n-diemensional system can be a convergent oblique pattern;But when r′= r = 2 ,the n-dimensional system can only be a emanative oblique pattern. ▌
When a generic straight line isn't in the shape of line but plane and solid in a certain preferred
n-dimensional system,it can also be regarded as trace of difficient-order generic point,so it can
also be called difficient-order generic plane or difficient-order generic solid.
1....3 Shape of many generic planes' intersection
Now let's discuss cases when more than three generic planes
0
0
0
11
221212
111111
=+++
=+++
=+++
mnmnmm
nn
nn
axaxaF
axaxaF
axaxaF
L
LLLLLL
L
L
:
:
:
( m≥3 ) intersect .Assume
( ) ;: 3 , , 2 , 1 01
≥==+∑=
mmjaxaMn
i
jiji L
is a figure determined by m generic planes' intersection,and assume the rank of coefficients matrix
of M ( aji )m×n , r = m ≤n ,and the system of equations is compatible,then M can be regarded as n –m -3-order generic solid (when n≥6,m≤n-3 )or n–m -2-order generic plane (n≥5,m≤n-2 )
or n-m-1-order generic straight line( n≥4, m≤n-1). Apparently,when m=n-2,M can be regarded
as a zero-order generic plane(viz. ordinary plane)or one-order generic straight line;When m=n-1,M can only be regarded as a straight line or an one-order generic point. When m=n ,M represents a
point. It is necessary to be pointed out, when m=3 ,sometimes M represents a sufficient-order
generic point.
Example 6: Assume a linear figure of five-dimensional space
,:
07631210
0434243
042244
54321
54321
54321
=−−++
=+−−++
=−−+−+
xxxxx
xxxxx
xxxxx
M
52
please change it into the shape of point.
Solution:Make elementary transformation for the augmented matrix of M
−−
−−−
−−
07631210
434243
422144
(the third row minus twice of the second row ,then minus the first row),it gets
,
410000
434243
422144
−−−
−−
viz. the equation of M becomes
,:
04
0434243
042244
5
54321
54321
=−
=+−−++
=−−+−+
x
xxxxx
xxxxx
M
Now,M can be regarded as intersection of a generic straight line
=+−−++
=−−+−+
0434243
042244
54321
54321
xxxxx
xxxxxL:
and a generic coordinate plane x5 = 4 .
In three-dimensional space,there is a concept that a plane and a straight line intersect to a
point. Extend the concept to higher space,then the intersection of a generic straight line and a
generic plane determines a sufficient-order generic point. Then,first change L into the shape of line
and then make it intersect with generic coordinate plane according to theorem 2, thus problems can
be solved.
In order to make people have deep impression on theorem 2,we will give a detailed solution
process of n-2 = 5-2 = 3 points A0 ,A1 ,A2 whose coordinates are all four with relation to x5
according to the proof methods of theorem 2.
. 483
124 16
48
412 then
8443 12442
443
44 Because
21
21
−====
=−×==+×=
==
DD
tt
D
, ,
, ,
so when order x3 = x4 = 0,it has
,, 1 4 22
11 −====
D
Dx
D
Dx
viz. point A0 (4,-1,0,0,4) is first choosen.
For the convenience,we can assume
( )
( ),
,
2 22
4
1 41
4
2,1
24
2,1
23
=====
=−=−
===
+
+
+
+
ka
Dxx
ka
Dxx
k
k
k
k
53
in turn,assume tk1 = t1 - D, tk2 = t2 –λk D
simultaneously, (k=1,2), of which λ1 =λ2 = - 2
(the rank of skew matrix is 1),so tk1 = 8 , tk2 = 16 ,
when assume 221
111
2
222
121
1
k
k
k
k
k
kta
taD
at
atD == ,
,,
,,
104
40 8
4
32 so
40163
84 32
416
48 hasit
22212111
22122111
===−=−
==
===−===
xxxx
DDDD
viz. point
A1 (-8,10,-4,0,4),A2 (-8,10, 0, 2,4),
are also choosen. Now
,21112
,41112
42102
32101
eeeAA
eeeAA
−−=
+−=
order M~ A0A1A2
then a preferred five-dimensional system taking
2010 , AAAA
as principal ramparting direction is acquired (because λ1 =λ2 = - 2 ,then the preferred
five-dimensional system is convergent oblique pattern),in the preferred five-dimensional system,M is in the shape of point (figure 29).
Theorem 3: Assume the rank of coefficients matrix (aji )3×n of consistent system of linear
equations ( )3 , 2 , 1 01
==+∑=
jaxan
i
jiji
is r =3;The rank of its front two columns of submatrixes
(ajk )3× 2 is r′= 2, the rank of front n-1 columns of
submatrixes ( ajl )3×(n-1) is r"= 2;and the rank of skew
matrix (from the third column to the n-1-th column of
submatrix) ( aj,i+2 )3×(n-3) is r = 2 or r≤1,then under proper
oblique axes transform,M represents a sufficient-order
generic point. Of which ,when r≤ 1,M can suit a
convergent oblique pattern,when r = 2, M only suits
emanative oblique pattern. figure 29
Proof : From theorem1 and theorem 2 of the second chapter,the coordinates of generic point
with relation to upright axis are uniquely determined,its opposite trace is also unique. So,the
solution of equation M with relation to xn should be unique,at the same time,when assume the
solutions of M with relation to x3,x4,…,xn-1 are all zero,its solutions with relation to x1,x2 should be unique.
Because r″= 2,then there must be numbersλ1 ,λ2,λ3 which aren't zero at the same
54
time ,they make ∑∑=
−
=
=3
1
1
1
0j
n
i
ijij xaλ
But from the equation of M,it apparently has
∑∑= =
=+3
1 1
0)(j
n
i
jijij axaλ
then ( ) ∑∑∑∑=
−
== =
=−+3
1
1
1
3
1 1
0j
n
i
ijij
j
n
i
jijij xaaxa λλ
viz. (λ1a1n +λ2a2n +λ3 a3n )xn= -λ1a1 +λ2a2 +λ3 a3
and because r =3,so λ1a1n +λ2a2n +λ3 a3n≠0 ,
then , 332211
332211 αλλλ
λλλ=
++
++−=
nnn
naaa
aaax
viz. the solution of M with relation to xn is unique.
In fact,because r = 3 ,r″= 2,then in the equations of three generic planes of M,at least
equations of one generic plane can be directly changed into xn =α by augmented matrix operation
of M,then the remainder two equations are equations of a generic straight line,then according to
the methods of theorem 2,we can find out n - 2 points
A0(α1,α2,0,…,0,α),
A1(α11,α12,α3,0,…,0,α), … … … … …,
Aj(αj1,αj2,0,…,0,αj+2,0,…,0,α), … … … … …,
An-3(αn-3,1,αn-3,2,0,…,0,αn-1,α),
on the generic straight line ,their coordinates with relation to xn are allα.When order
A1A2…An-3 ,
a preferred n-dimensional system taking
302010 , −nAAAAAA ,,L
as principal overlapping direction is acquired,because all linear combinations among
302010 , −nAAAAAA ,,L
is a sufficient-order generic point,and A0 ,A1 ,A2 ,…,An–3 are on M,both dimensions of
the sufficient-order generic point and dimensions of M are n-3, so M is a sufficient-order generic
point of the n-dimensional system,and A0 is its opposite trace.
Now ,three intersectant generic planes can be regarded as trace of M,viz. they are all in the
shape of plane. And from theorem 2 ,that three generic planes pairwise intersect determines three
generic straight lines. Because r = 3,then the rank of coefficients matrix of equations of a random
generic straight line is 2,assume the rank of its skew matrix is ( ) 3 , 2 , 1 ,=′′′ jr j when
1 ≤′′′r , it has 1 ≤′′′jr ,so the n-dimensional system M suits can be a convergent oblique pattern;When 2 =′′′r ,at least the rank of skew matrix of one generic straight line's equation is
55
2 =′′′jr ,from theorem 2,M only suits a emanative oblique pattern. ▌
Theorem 4 :under proper oblique axes transform,assume equations determined by m (m>3)
generic planes' intersection are consistent system of equations ( ) 3 . 21 01
>==+∑=
mmjaxaMn
i
jiji ,, , : L
and assume the rank of coefficients matrix of M is r,at least the intersection of three generic planes
satisfies theorem 3,then,
(1) when r = m,only three generic planes can be changed into the shape of plane;
(2) When 3<r<m,at most m-r+3 generic planes can be changed into the shape of plane;
(3) When r =3,sometimes, the m generic planes can all be changed into the shape of plane.
Proof:assume intersectant generic planes satisfying theorem 3 is at least the following
0
0
0three
331313
221212
11 1111
=+++
=+++
=+++
axaxaF
axaxaF
axaxaF
nn
nn
nn
L
L
L
:
:
::
then under proper oblique axes transform
(1) when r =
m,according to linear algebra theory,the dimension of the intersection of a
random other generic plane and the three generic planes is n- 4,apparently,the intersection can
only be a difficient-order generic point,other generic planes can only be regarded as trace of the
difficient-order generic point,viz. their orders are all J = n - 4,but from linear algebra theory,their actual dimensions are all W =n - 1,then according to dimension theorem,their directviewing
dimensions are all Z = W – J =3, viz. they are all difficient-order generic solid. So,only the three
generic planes can be changed into the shape of plane.
(2) when 3<r<m,assume the rank of coefficients matrix of the front r generic planes '
intersection ( ) 21 01
rkaxan
i
kiki ,, , L==+∑=
is r,then from question (1),only three generic planes can be changed into the shape of plane,and
assume they are separately F1,F2,F3,oter r -3 generic planes F4,F5,…,Fr can't be changed
into the shape of plane. Besides,there are still m-r generic planes Fr +1, Fr +2 ,…, Fm ,their normal
direction vectors are all a combination of normal direction vectors of front r generic planes. Assume
the normal direction vector of a certain generic plane such as Fr +1 and the normal direction vector
of F4,F5,…,Fr are linearly independent,viz. the normal direction vector of Fr +1 can only be
regarded as a combination of normal direction vectors of F1,F2,F3 ,and because the rank of
coefficients matrix of system of equations
( )
=+
==+
∑∑
=++
=n
i
riir
n
i
ii
axa
axa
1
1,1
1
0
3 , 2 , 1 0 τττ
is 3 ,the rank of front n -1 columns of submatrixes is 2 ,the rank of skew matrix is r"= 2 or
r"=1.So,the intersection of the four generic planes also satisfies the conditions of theorem 3,then
they intersect a same generic point,viz. Fr +1 can also be changed into the shape of plane. By the
56
same token,if the intersection of an other random generic plane Fr +s (1<s<m-r) and F1,F2,F3
satisfies theorem 3,it can also be changed into the shape of plane.
When the intersection of all the intersections of m-r generic planes and F1,F2,F3
( )
( )
−==+
==+
∑∑
=++
=n
i
riir
n
i
ii
rmaxa
axa
1
,
1
,,2,1 0
3 , 2 , 1 0
Lδ
τ
δδ
ττ
.
(3) When r =3,if change all of r of question (2) into 3,then it can be proved. ▌
§§§§2 Shape of non-linear figures
In preferred n-dimensional system,directviewing figures of non-linear generic curved surfaces
are not all in the shape of curved surface with zero thickness, they are always in the shape of
directviewing solid. So do directviewing figures of generic curves (non-linear),they are not all in
the shape of curve.
For example,in example 2 of the third chapter,no matter what oblique axes transformation we
use, we can't change generic sphere
into the shape of directviewing sphere,it is
always in the shape of global solid. In fact,they are traces of difficient-order generic
points,so they can be called difficient-order
generic sphere. Other generic curved surfaces
also have sphere cases. For this kind of generic
curved surfaces,we can call it solid (actually it
isn't solid) generic curved surface(the above
difficient-order generic spheres can all be
called solid generic sphere) , in order to
distinguish from another kind of generic
curved surface which is hollow generic surface
from direct view,such as
2
323121
2
4
2
3
2
2
2
1 4222224233 Rxxxxxxxxxx =−−−+++ . figure 30
Let's look at figure 30,the cut trace of the hollow generic sphere cut by generic coordinate
surface is a elliptic hollow generic curve,because the principal overlapping direction of the
four-dimensional system is 321 2 eee ++ ,so a random point of the generic curve represents a
straight line which is parallel to the principal overlapping direction.
In fact,the generic curve is a cylindrical surface of three-dimensional space,its directrix is a
circle and its generating line is parallel to vectore 321 2 eee ++ . And because the hollow
generic sphere can be regarded to be composed of a series of cut traces aquired by generic
coordinate surface translating along up and down directions of upright axis,from the figure it is
observed that, every cut trace is a elliptic generic curve,and every generic curve represents a
cylindrical surface of three-dimensional space which is parallel to the generating lines. The hollow
2
2
4
2
3
2
2
2
1 Rxxxx =+++
57
generic sphere can become "hollow",it is because of the existence of these generating lines.
Under the nonsingular linear transform
=′
+=′
+−−=′
++−=′
2
1
2
12
1
2
1
2
12
1
2
1
2
1
34
413
4212
4211
xx
xxx
xxxx
xxxx
equation 2
323121
2
4
2
3
2
2
2
1 4222224233 Rxxxxxxxxxx =′′−′′−′′−′+′+′+′
can be changed into 2
2
4
2
3
2
2
2
1 Rxxxx =+++ ,
this is just the generic cylindrical surface of example 5 in the third chapter.And under the inverse
transform
′
′+′+′=
=
′−′=
′+′−′−=
4
3214
3
212
3211
2
1
2
1
2
1
2
1
2
12
1
2
1
2
1
x
xxxx
x
xxx
xxxx ,
the latter can be changed into the former.So,some(not all) solid generic curved surface can be
changed into hollow generic curved surface,but all hollow generic curved surface can be changed
into solid ones.
The cases of generic curve are such in essence(for example,the above generic directrix of
figure 30 is a hollow generic ellipse;The cut trace on generic coordinate surface of example 2 and
example 5 in the third chapter is solid generic circle). Since then if generic curved surface and
generic curve referred to are hollow if without special versions,otherwise they are referred to as
solid generic curved surface and solid generic curve.
Whether it is solid or not, both the determinant of generic curved surface and generic curve
are acceptable for dimension theorem. For example,because the directviewing dimension of the
above solid generic sphere is Z=3,its order is J=0,then its actual dimension is
W = Z + J = 3 + 0 = 3,
and the directviewing dimension of the above hollow generic sphere is Z = 2,its order is J = 1,then
its actual dimension is W = 2 + 1 = 3,it is obvious that their actual dimensions are identical. For
the convenience,sometimes for generic curved surfaces whose actual dimension are W = n -1
and generic curves whose actual dimensions are W = n -2 ,whether they are solid or not,we all
separately call them sufficient-order generic curved surfaces and sufficient-order generic curves.
58
Exercises
4.1 Please find out a preferred four-dimensional system and make generic plane
3x1 + 2x2 - 4x3 + 5x4 + 3 = 0
in the shape of plane in the system.
4.2 Find out a six-diemensional system and make generic plane
2x1 -3 x2 + 5 x3 + 5 x4 - 4 x5 - 2 x6 – 5 = 0
in the shape of plane in the system.
4.3 Find out a preferred six-dimensional system and make generic plane
04432 6543 =+−+− xxxx
in the shape of plane in the system.
4.4 The following generic straight lines suit convergent-oblique pattern or emanative-oblique
pattern preferred five-dimensional system?Why? Find out a suited preferred five-dimensional and
make generic straight line
=−++−−
=+−−−+
0446543
0232425
54321
54321
xxxxx
xxxxx
in the shape of line in the system.
4.5 Find out a suited six-dimensional system and make generic straight line
=+−+−−+
=+++−−+
01364643
02432354
654321
654321
xxxxxx
xxxxxx
in the shape of line.
4.6 Find out a preferred five-dimensional system and make
=−+−++
=+−+++
0242322
0334433
54321
54321
xxxxx
xxxxx
in the shape of straight line in the system.
4.7 Find out a preferred six-dimensional system which makes
=+++−++
=+−+−++
=+++−++
04332
0236422
02432
654321
654321
654321
xxxxxx
xxxxxx
xxxxxx
in the shape of point(Can this system be a convergent-oblique pattern?If it can be a
convergent-oblique pattern then it should make six-dimensional system be found a
convergent-oblique pattern).
59
The Fifth Chapter: Making of figures in preferred
n-dimensional system
Figures of preferred n-dimensional system can be made out by three kind of graphic methods
according to the analyticity of their equations and different needs.
In the third and fourth chapter,we have discussed questions of graphic rules (viz. three kind of
pictographies) and the shape of figures,and elementarily know some methods of making figures.
Here,we will discuss more about methods of making figures in preferred n-dimensional system. It
has three aspects of contents for concreteness:(1) cutting-trace method;(2) leading-axis method;
(3) synthetical pictography.
§§§§1 Cutting trace method
In three-dimensional space,a surface of revolution can be made by a curve rotating round a
coordinate axis.But in n-dimensional system,most punctiform figures are nonzero-order generic
points.Under the rotation,once generic point leaves its original position ,directions of most
nonzero vectors belonging to the generic point will change,which may make generic point lose the
possibiltiy of being punctiform. Thus,it will lose equivalence relation with the generic point on
its original position.So,in a general way, attempting to use rotation method to make figures is
impracticable in preferred n-dimensional system.
But,we can image preferred n-dimensional system to be composed of the translation trace of
generic coordinate surface along upright axis. And the projection of generic coordinate surface with
relation to upright axis is zero,so,coordinates of random figures of generic coordinate surface with
relation to upright axis are uniquely determined. That is to say,we can get a series of cut traces of
some figures by translating generic coordinate surface along upright axis and then make out these
figures.
Example 1: Assume there is a generic circular conical surface in four-dimensional space
2
4
2
3
2
2
2
1 xxxx =++ ,
please make out its figure.
Solution:Substitute x4 = x40 into the equation of generic circular conical surface,it gets
2
40
2
3
2
2
2
1 xxxx =++ ,
and make it simultaneous with x4 = x40,then the cut trace
obtained is a generic circle
=
=++
404
2
40
2
3
2
2
2
1
xx
xxxx
(when x40 = 0,the cut trace is a generic circle retracting into a
point),then the generic circle is equivalent to a sphere of figure 31
three-dimensional space whose centre is on the origin and its radius is |x40| (figure 31). Choose two
points on the sphere
−− 40
4040 ,0,2
,2
xxx
A
60
and B ( 0,0,x40,x40 )
and order AB
then a preferred four-dimensional system is acquired,at the same time ,cut trace of the generic
circular conical surface at x4 = x40 is acquired.
When x4 changes from zero to positive and negative direction continuously,it will get the
trace of generic circular conical surface on a series of generic planes x4 = x4i
=
=++
i
i
xx
xxxx
44
2
4
2
3
2
2
2
1
(i =1 , 2 , …),then the continuously
varying cut traces form the directviewing
figure of the generic circular conical
surface(figure 32).
In the equations of the generic circular
conical surface,order x4 = Ct of which,C=3×105(km/s),t is time(unit:s=1 sec.),then
it has
222
3
2
2
2
1 tCxxx =++
this is the equation of light cone[9],we call it “Einstein light cone”.When x4≤0,every
point of axis x4 represents a black hole
translating to the origin figure 32
along axis Ct,and every generic circle represents that because of the great attractive effect of the
black hole near the black hole,light ray retracts to the centre of sphere in the shape of spherical
wave at the speed of three hundred thousand km per second. When x4 = 0,spherical wave retracts to
a point,it represents light ray has been inhaled into the black hole,there may be catastrophe at the
time,it may change into the white hole or other shiny
celestial body.Then,when x4≥0,every point of axis
x4 represents that the white hole or new celestial
body translates along the positive direction of axis
Ct with time going on,now each generic circle
represents that the spherical wave generated by the
white hole or new celestial body quickly expands at
the speed of enlarging the radius three hundred
thousand km per second.
Figures of other generic curved surface can be
made by this method. In fact ,for a curved surface,making out at least four or five its cut traces and
connecting them by smoothed curve is ok.
Both generic sphere figure 33
22
4
2
3
2
2
2
1 Rxxxx =+++
(refer to it in figure 15 of example 2 of the third chapter)of the third chapter and generic cylindrical
61
surface 22
3
2
2
2
1 Rxxx =++
(refer to it in figure 22 of example 5 of the third chapter) adopted this making method.
Figure 33 is a generic paraboloid of four-dimensional space
4
2
3
2
2
2
1 2xxxx =++
Figures in a preferred four-dimensional system are also made by the above methods.
§§§§2 Leading-axis method
Example 2: Please find out the directviewing figures of generic plane F
F:2x1 – 5x2 + 9x3 + 6x4 – 2x5 – 3x6 – 2x7 – 3 = 0
of seve n-dimensional space.
Solution:when assume the six posterior variables are all zero at the same time ,then it gets a
point of x1 (1.5,0,0,0,0,0,0),
Considering it is advisable to put the figures being about to be found out of F in first quartile
of plane x1ox2,so,it is better to put the oblique axes of preferred seve n-dimensional in the second
or the fourth quartile. Then we can find out the following solution vectors
6214
5213
4212
3211
624
924
324
224
eeeOA
eeeOA
eeeOA
eeeOA
−+−=
−+−=
++−=
++−=
,
,
,
in the leadingout group
F′:2x1 – 5x2 + 9x3 + 6x4 – 2x5 – 3x6 – 2x7 = 0
(O is the origin),order OA1A2A3A4
it gets the generic coordinate plane of preferred seve n-dimensional system which changed F′
into the shape of plane (because F′is the leadingout group of F,then figures of F are changed into
the shape of plane at the same time).Point (1.5,0,0,0,0,0,0) of axis x1 on the generic coordinate
plane is the trace of generic point A of F,find out the trace (4,1,0,0,0,0,0) of another point
of F on plane x1ox2 and mark out B ,connect AB ,then the directviewing figure of generic
straight line
=
=−−−++−
0
03326952
7
654321
0x
xxxxxxL :
(viz.the cut trace of F cut by generic coordinate plane) determined by intersection of generic plane
F and generic coordinate plane is finished.
Order x7 = 6,substitute it into the equation of F and make it simultaneous with the equation,then the cut trace
=
=−−−++−
6
015326952
7
654321
1x
xxxxxxL:
of F cut by generic plane x7 = 6 is acquired. In L1,because generic plane
F11:2x1 -5x2 + 9x3+6x4-2x5-3x6-15=0
doesn't include variable x7,then it is parallel to axis x7 and vertical to generic coordinate plane,so
62
it is called projective generic plane of L1. Make the equation of F11 simultaneous with axis x7=0,then the projection
=
=−−−++−′
0
015326952
7
654321
1x
xxxxxxL:
of L1 on generic coordinate plane is acquired. Choose two projection traces
Jc′(2.5,-2,0,0,0,0,0) and Jd′(5,-1,0,0,0,0,0)
of L1 on L1′,choose proper direction and unit length and find out the directviewing figures of x7,lead Jc′、Jd′to generic point C、D and make
(Jc,Jd are separately opposite traces of C and D ),connect AC,BD,CD,then the directviewing
figure of generic plane is finished(figure 34).
figure 34
It is observed from the example,under many cases we don't be anxious to find out the figure
of upright axis xn,but first find out the cut trace of the figure be made ,it is cut by generic
coordinate plane xn = 0 ;Then find out one or two projection trace of the figure's cut trace on plane
x1ox2 , the cut trace is cut by generic plane xn =xn0(xn0 ≠0);Then in order to lead the
projection trace to the blank of the figure and and make the figures have pleasing shape and occupy
smaller chart size, choose proper direction and unit length and make out the figure of upright axis
xn. Under this meaning,upright axis xn can also be called leading axis,and the draughting can be
called leading-axis method accordingly.
§§§§3 Synthetical graphic method
Synthetical graphic method is to synthetize three kind of graphic methods. Using more than
two kinds of graphic methods for a figure of n-dimensional space belongs to synthetical graphic
method.
In example 4 of the fourth chapter,the projection trace of generic point B on generic straight
line ,: 0223
074232
54321
54321
=++++
=+++++
xxxxx
xxxxxL
is far away from B ,if we solely use ordinary graphic method,it must make the chart size too
large,and then bring unnecessary trouble. So,we simultaneously used ordinary graphic method and
76eJJJJ ddcc =′=′
63
indirect method for L,separately mark out the trace of point A and point B ,the coordinates of
opposite trace(from the coordinates of Ja we know generic point A is on the generic coordinate
plane,refer to it in figure 26 of the fourth chapter). Let's look at the following example.
Example 3: assume in six-dimensional space
=+++−+−
=++−+−−
=++−+−−
0242 22
048443244
012 422
654321
654321
654321
xxxxxx
xxxxxx
xxxxxx
M:
is regarded as difficient-order generic plane,find out its figure according to ordinary graphic
method and direct graphic method.
Solution:from linear algebra theory,we know the actual dimension of M is W=3,because
Z=2,then from dimension theorem we know the order of the difficient-order generic plane to be
found out is J=1,viz. it is regarded as the trace of one-order generic point. And one-order generic
point represents a straight line,so we can assume four one-order generic vertex
of the one-order generic plane are separately lines:
=−
=+
=+++−+−
=++−+−−
=++−+−−
=−
=+
=+++−+−
=++−+−−
=++−+−−
=
=+
=+++−+−
=++−+−−
=++−+−−
=
=+
=+++−+−
=++−+−−
=++−+−−
0 2
06
0242 22
048443244
012 422
0 2
016
0242 22
048443244
012 422
0
02
0242 22
048443244
012 422
0
012
0242 22
048443244
012 422
6
1
654321
654321
654321
4
6
1
654321
654321
654321
3
6
1
654321
654321
654321
2
6
1
654321
654321
654321
1
x
x
xxxxxx
xxxxxx
xxxxxx
A
x
x
xxxxxx
xxxxxx
xxxxxx
A
x
x
xxxxxx
xxxxxx
xxxxxx
A
x
x
xxxxxx
xxxxxx
xxxxxx
A
:
:
:
:
Regard variable coefficients of the four straight lines' equations as elements of the following
determinants,then,it gets direction vectors of the four straight lines
64
5432
654321
56152
100000
000001
211122
443244
142211
eeee
eeeee
−++=−−
−−−
−−−
=
e
S
Now consder compressing M along vector S (viz. is regarded as single principal overlapping
direction of preferred six-dimensional system),and make the directviewing shape of M of host
system be in accord with that in the first generic plane of M
F1:x1 -x2 -2x3 +2x4 - 4x5 + x6 +12=0.
Therefor,find out points (-12,0,0,0,0,0), (4,-14,15,0,0,0), (-19,5,0,6,0,0) and (-21,11,0,0,-5,0) on F1 and make them have common generic,viz.
make use of transform
+−−
+−
−
−−
−
+−
215
214
213
215
214
213
1195
576
141615
1195
576
141615
eee
eee
eee
eee
eee
eee
○○○□□□
then a couple of preferred six-dimensional systems of
emanative-oblique pattern satisfying the two requests
are acquired. figure 35
Find out the directviewing figure A1A2A3A4 of generic plane F1 according to ordinary graphic
rules in host system,the figure is in accord with the
figure of M in direct view,then in fact it is regarded as
figure of difficient-order generic plane M (figure
35).In the figure,among the four difficient-order
generic vertexes , the traces of A1,A2 are separately
midpoint of M
A11 (-12,0,0,0,0,0) and A21 (-2,10,0,0,0,0),
the opposite traces of A3 and A4 are separately points
of M
A31 (-16,-2,0,0,0,2)and A41 (-6,8,0,0,0,2),
Separately mark the four points(trace or opposite
trace)into guest system,and then separately choose a figure 36
point on four straight lines denoted by the four difficient-order generic vertexes which are parallel
to each other
A12(-12,2,15,6,-5,0);
A22(-2,12,15,6,-5,0);
A32(-16,0,15,6,-5,2);
A42(-6,10,15,6,-5,2).
65
vertexes are also marked into guest system.
Because four three-dimensional subsystems
Ox1x2x3、Ox1x2x4、Ox1x2x5、Ox1x2x6
of the six-dimensional system we imagined are all right handed systems,then plane A12A22A32A42
should be on top of plane A11A21A31A41 ,when connect the eight vertexes in order to form
hexahedron to be soved,the edge line near the observer should be drawn in real line,the other three
edge lines behind the hexahedron should be drawn in broken lines(figure 36).
Example 4: Diagrammatize the following nonlinear programming questions by ordinary
graphic method and indirect graphic method:
Objective function: t =5x1+6x2+4x3+3x4-2x5
constraint condition:
≥
≤++++
≤++++
≤+++
≤++++
≤++++
0,,,,
3322234
3963 25
21 4 43
334 52
44536 2
54321
54321
54321
4321
54321
54321
xxxxx
xxxxx
xxxxx
xxxx
xxxxx
xxxxx
In order to introduce some graphic methods of points when the direction of oblique axis is
opposite,in this question, the coefficient of x5 is turned into negative number which is contrary to
routine.
Solution::::choose the following three points in figures (it is a generic plane,it is also called
objective generic plane) denoted by objective function
(4,-2,-2,0,0,0),(6,-3,0,-4,0,0),(4,-2,0,0,4,0),
use transform
+−
−
−
215
214
213
244
364
242
eee
eee
eee □□□
then generic coordinate plane of a six-dimensional system is acquired(refer to it in figure 37).
First regard every equation of constraint condition as equality,that is to say,regard them as
equations of generic plane(be called constraint generic plane).
The cut traces acquired by intersection of every constraint generic plane and every
two-dimensional coordinate plane is called track lines in every two-dimensional coordinate palne.
For example,the trace line of generic plane (is called the third constraint generic plane,by
analogy,other generic plane can be called the first,the second and the fourth,the fifth constraint
generic palne)denoted by the third equation of constraint condition on plane x1ox2 is trough point A
(7,0,0,0,0,0) and point B (0,5.25,0,0,0,0),a triangluar area is formed between AB and the original
point O(refer to it in figure 37),traces(include their intersection points ) of other constraint generic
planes on plane x1ox2 are all on the outside of the area △OAB(to lessen confusion of figures,other trace lines are omitted).
As will be readily seen,the triangle and its inner points satisfy all the constraint conditions,most points on the outside of the triangluar area don't satisfy any constraint condition,only few
66
figure 37
points can satisfy the constraint condition. So,we call the triangle and its inner area feasible
region,the area outside the triangle unfeasible region.
For preferred n-dimensional system of emanative-oblique pattern,it isn't difficult to find out
trace lines and of every constraint generic plane on every two-dimensional coordinate plane and
intersection points of these trace lines,But it isn't easy to find them out in n-dimensional system of
convergent-oblique pattern.
For example,on plane x1ox4,we easily found out the trace line of the third constraint generic
plane,the trace line is through point A and point C (0,0,0,5.25,0,0).Because traces of other generic
planes are all on the outside of feasible region,then they are omitted.
But for plane x1ox3,there is some difficult. In fact,the figure of axis x3 is also through the
origin,in order to distinguish them, just arrange them parallelly on a straight line which is through
the origin. So,we regard the figure of axis x4 as axis x3 for the moment: first make out the trace
point D(0,0,6.6,0,0,0) of the second constraint generic plane on axis x3,then make DD′to x4
through D ,and make the line segment throughing the origin and denoting endpoints of x3、x4
parallel and intersect axis x4 at D′. Originally,the position of D′is the position of point on axis
x4,but now it reverses the position of the host and the guest,it become a point of axis x3. The other
end of the trace line is point (33,0,0,0,0,0) of x1.
The trace line of the third constraint generic plane intersects axis x1 at point A,the other end is
point (0,0,21,0,0,0) of axis x3. The two trace lines intersect at point 0,0,0,7
39,0,
7
36L (figure
37).
On plane x2ox3,the second and the third constraint generic planes separately intersect axis x2 at
point B and point (0,16.5,0,0,0,0),and they intersect axis x3 at point D and point (0,0,21,0,0,0) ,and the two trace lines intersect at point G(0,4,5,0,0,0);On plane x2ox4,the trace line of the third
constraint generic plane separately intersect axis x2 and axis x4 at point B and point C;
On plane x2ox5,the trace line of the second constraint generic plane intersect axis x2 and axis
x5 at point (0,16.5,0,0,0,0), and point (0,0,0,0,8.25,0),The trace of the third constraint generic
plane is through point B of axis x2 and parallel to axis x5,the two trace lines intersect at point E
(0,5.25,0,0,5.625,0).
67
Trace lines of every constraint generic planes on plane x3ox5 and plane x4ox5 and the
draughting of intersection point are the most difficult.
Originally,coordinates of points on positive direction of axis x5 will lessen values of objective
function,but considering the demand in futural nonlinear programming,it's necessary to introduce
graphic methods under this case.
Here,we have two means of resolution :One is to make a plane rectangular coordinate
system additionally,take x3 as cross axis,take x5 as vertical axis,order x1=x2=x4=0 in equations
of every constraint generic plane,then find out their trace lines and intersections in the plane
rectangular coordinate system;The other is borrowing-axis method,viz. take x1 or x2 as axis x3 or
axis x5 for the moment.
The anterior method is familiar to people,not only people who learn analytical geometry but
also those who learn descriptive geometry,they are all familiar with this method. So,let's
emphasize on borrowing-axis method:borrow the positive direction of axis x1 as axis x5,according to the solution of trace point D,axis x4 is again borrowed as axis x3.
figure 38
Put scales of axis x5 on top of axis x1,mark out trace points of every constraint generic plane
according to the new scales,connect every corresponding trace point and gain trace lines of every
constraint generic plane on plane x3ox5. Trace lines of the second and the fourth constraint generic
plane on plane x3ox5 separately intersect axis x3 at point D (0,0,6.6,0,0,0) and point
Q(0,0,39,0,0,0),they intersect x5 at point T(0,0,0,0,8.25,0) and point U(0,0,0,0,6.5,0),the two trace
lines DT and QU intersect at point
0,
13
81,0,
13
21,0,0R
(figure 38). Because the distances of some points are too small to see ,we magnify contents of the
circle in figure 38,for the convenience of calculating and drafting ,change axis x1 as axis x5,change scales of axis x3 into the same as those of axis x5(refer to it in figure 39a).RS and RV are
line segments which represent coordinates of R with relation to axis x3 and axis x5. Traces of other
68
generic planes and intersections are all on the outside of
feasible region enclosed by quadrilateral ODRU,so they
are omitted.
Figure 39b introduced resetting method of these trace
points and intersections: because the trace points of
intersections solved in the question are near from each
other,it is difficult to describe the principle clearly ,then
choose a random point N as example to describe it. In the
figure, WN and PN separately represent coordinates of
point N with relation to axis x3 and axis x5. The resetting
process is equivalent to make the borrowed axis x1 revolve
aroundthe origin,during the revolve ,point P arrives at
point P′ along arc ⌒PP ′,point N arrives at N′along
arc ⌒NN ′ . The position of point N′is the last correct
position of N. figure 39
On plane x3ox4,we again use borrowing-axis method(borrow axis x1 as axis x3,the process is
omitted),trace lines of the first constraint generic plane separately intersect axis x3 and axis x4 at
point
0,0,
3
440,0,0,point and 0,0,0,
3
22,0,0 ,Trace lines of the second constraint generic
plane separately intersect the two axes at point D (0,0,6.6,0,0,0) and point (0,0,0,33,0,0),Traces of
the third constraint generic plane separately intersect the two axes at point (0,0,21,0,0,0) and
point C (0,0,0,5.25,0,0).
The three trace lines DH、HI、CI intersect at point H、I . Of which,coordinates of point H are
;, , , , , 009
22
9
5500 54321 ====== txxxxx
coordinates of point I are
.0021
82
21
11300 54321 ====== txxxxx , , , , ,
Other trace lines and intersections are on the outside of feasible region enclosed by pentagon
ODHIC,so they are omitted.
On plane x4ox5,trace lines of the third constraint generic plane is straight lines which is though
point C of axis x4 and parallel to axis x5;Trace lines of the fourth and the fifth constraint generic
plane are separately connecting lines among point
(0,0,0,13,0,0), point (0,0,0,6.6,0,0)
of axis x4 and point (0,0,0,0,6.5,0), point (0,0,0,0,16.5,0)
of axis x5.The three trace lines intersect at point
J (0,0,0,5.25,3.375,0) and point K(0,0,0,5,4,0)
(trace lines and intersections of other constraint generic planes on plane x4ox5 are all on the
outside of feasible region,so they are omitted).
Under higher dimensions,methods in plane rectangular coordinate system can be supplement.
Figure 40 represents the trace lines' distribution of every constraint generic plane on plane x2ox3.
It is observed that,only trace lines BG ,DG of the second ,the third constraint generic plane and
their intersections are on the inside of feasible region,other trace lines and intersections are all on
the outside of feasible region.Then it is observed that,graphical method has great advantage,
69
considerable useless calculation can be dispose of by
graphical method.
Trace line S′of objective generic plane S on plane x1ox2 (refer to it in figure 41)is called objective trace line,it is the
cut trace of intersection of objective generic plane and plane
x1ox2.
Along with values of objective function increasing from
zero step by step,S intersects a series of planes parallel to
plane x1ox2,projection of its cut trace on plane x1ox2 is
equivalent to S′translating along the direction of 5e1+6e2
graduallly,during the translating it will intersect every vertex
of feasible region. figure 40
In posterior nine chapters,there are constant distance theorem and optimum point (substitute
a point of feasible region into objective function,t will get the maximum value,the point is called
figure 41
optimum point )theorem:Under all oblique axis transforms which can make S into the shape of
plane,directviewing distance (directviewing distance is the distance that we can directly see with
our naked eyes,it isn't always the actual distance)of every vertex of feasible region to objective
trace line is invariable;The farthest vertex from directviewing distance to objective trace line is
optimum point to be solved. According to the two theorems,the directviewing distance between
vertex L to S′is the farthest(point G takes second place, point A takes third place),so it is the
optimum point to be solved. Substitute coordinates of L into objective function,it gets t=48,this is
the optimal solution to be solved.
Choose proper direction and unit length and make out the figure of upright axis t,lead point
L to point L′.Parallelogram MNPW is the figure of objective generic plane S(figure 41).
Now,we use simplex method which hides slack variable and be introduced in the ninth
chapter to test the above results. (The contents is useful for people who are engaged in operational
research or mathematical statistics,people of other majors need not read it):
Arrange constraint conditions into the following :
70
( )1
22234330
63 25390
4 43210
4 52 330
536 2440
54321
54321
4321
54321
54321
−−−−−≤
−−−−−≤
−−−−≤
−−−−−≤
−−−−−≤
xxxxx
xxxxx
xxxx
xxxxx
xxxxx
In objective function,the coefficient of x2 is maximum,so,choose x2 as swapin variable,objective
function will has great increment . order x1= x3= x4= x5=0,it has
( )′
−≤
−≤
−≤
−≤
−≤
1
3330
2390
4210
2330
440
2
2
2
2
2
x
x
x
x
x
Now only choosing 4
21
3
33,
2
39,
4
21,
2
33,44Min 2 =
=x
can make formula (1)′be true,so the third equation of formula (1)is regarded as equation of the
greatest constraint force,then it is regarded as equality which hides slack variable,translate x2 to
the left end of the equality,it has
2224333
63 5392
4 3214
4 5 332
536244
54312
54312
4312
54312
54312
−−−−≤
−−−−≤
−−−=
−−−−≤
−−−−≤
xxxxx
xxxxx
xxxx
xxxxx
xxxxx
viz. ( )2
8457690
1227570
43214
829450
2082351550
5411
5431
4312
5431
5431
−+−−≤
−−−−≤
−−−=
−+−+≤
−−−−≤
xxxx
xxxx
xxxx
xxxx
xxxx
now a radical feasible solution is acquired(0,5.25,0,0,0,0)this is equivalent to coordinates
of point B in figure 37.
Substitute 43124
1
4
3
4
21xxxx −−−=
into objective function,it gets . 232
5
2
1
2
635431 xxxxt −−++=
In new objective function,the coefficient of x3 is a bit larger,and take x3 as swapin variable.
Order x1= x2= x4= x5=0 ,
it has ( )′
−≤
−≤
−≤
−≤
−≤
2
5 69 0
57 0
21 0
9 45 0
231550
2
3
3
3
3
x
x
x
x
x
71
Now only choosing ,55
69,57,21,
9
45,
23
155Min 3 =
=x
can make formula (2)′be true,so the second equation of formula(2)is regarded as equation of the
greatest constraint force. Regard formula(2)as an equality which hides slack variable,and translate
9x3 to the left end of the equality,it has
8 4 769 5
122 757
4 321 4
8 2 45 9
208515523
5413
5413
4132
5413
5413
−+−≤
−−−≤
−−=+
−++=
−−−≤
xxxx
xxxx
xxxx
xxxx
xxxx
viz. ( )3
1613341980
255161170
419147218
82459
259341800
541
541
5412
5413
541
−+−≤
−−−≤
+−−=
−++=
+−−≤
xxx
xxx
xxxx
xxxx
xxx
then a new radical feasible solution is acquired (0,4,5,0,0,0),it is equivalent to coordinates
of G in figure 37.
Substitute 54139
8
9
2
9
15 xxxx −++=
into the equation of the new objective function,then another new objective function
, 9
38
9
22
9
744 541 xxxt −−+=
In the new equation,the coefficient of x1 is the largest. So,choosing x1 as swapin variable
will make objective function value increase more quickly.Order x2=x3=x4=x5=0 ,it has
683960
644680
281440
45 0
683600
1
1
1
1
1
−≤
−≤
−≤
+≤
−≤
x
x
x
x
Now only choosing 7
36
28
144
68
396,
64
468,
28
144,,
68
360Min 1 ==
−=x
can make the above formulas be true at the same time. So,the third equation of formual(3)again
is regarded as an equality which hides slack variable. Take x2 as swapout variable and change it
with x1,it has
−+≤
−−≤
−+−=
−+=−
+−≤
541
541
2541
5413
541
161319834
25511716
184197214
82459
25918034
xxx
xxx
xxxx
xxxx
xxx
72
viz. ,
6123608283240
2884142344860
184197214
181089702126
612108180720
254
254
2541
2543
254
+−+≤
+−+≤
−+−=
−−+=
+−−≤
xxx
xxx
xxxx
xxxx
xxx
viz. ( )4
17102390
1623132707
9
7
2
14
19
7
367
1
7
6
14
1
7
39
173520
254
254
2541
2543
254
+−+≤
+−+≤
−+−=
+−+=
+−−≤
xxx
xxx
xxxx
xxxx
xxx
then another radical feasible solution is acquired
0 , 0 ,
7
39 , 0 ,
7
36 ,
This is equivalent to coordinates of point L in figure 37. Substitute formula(4)into the posterior
objective function,it gets . 42
748 542 xxxt −−−=
Coordinates of all variables of the new objective function are all less than zero,according to
optimal solution theorem of simplex method introduced in the ninth chapter,it is the optimal
solution to be solved,viz. when
0 , 0 , 7
39 , 0 ,
7
3654321 ===== xxxxx ,
t will take the maximum value 48,it is in accord with the graphic result of figure 41.
73
The Sixth Chapter: Interleaving and distance between two
linear figures.
In the chapter,we begin to discuss ubiety among linear figures.
The ubiety among linear figures is determined by interleaving,distance and included angle. Of
which the third factor,viz.included angle,because its contents are too much,we will discuss it in
the seventh and the eighth chapter.
§§§§1 Interleaving between two linear figures
Assume equations of two linear figures A and B separately are
A: 01
=+∑=
axa ji
n
iji
(j=1,2,…,r1)
B: 01
=+∑=
bxb ki
n
iki
(k=1,2,…,r2)
and assume the rank of coefficients matrixes of A and B are separately r1 , r2,the rank of
coefficients matrixes of simultaneous group(viz. the equation of A∩B)
( )
( )
==+
==+
∑∑
=
=
2
1
1
1
,,2,1 0
,,2,1 0
rk
rj
M
bxb
axa
ki
n
iki
ji
n
iji
L
L
:
is r,the rank of augmented matrix is R. Investigate the relationship between r1,r2,r and R,it is any
more than the following four kinds:
(1) r1+r2≥r;
(2) r1 , r2≤r;
(3) r1, r2, r≥1, R≤n+1;
(4) r≤R, viz. R-r=1 or R-r = 0.
So,A and B have the following kind of ubiety:
(1)coincident .
(2)parallel.
(3)having common point but not coincident,again it can be divided into four kinds :
1) Intersect in a w-dimensional (1≤W≤n-2) linear figure.
2) Both A and B are straight lines and intersect at a point.
3) A and B intersect at a point,but one of A and B is a straight line,the other is a
sufficient-order of deficient-order generic plane,we call A and B singly through. The intersection
point is called through point.
4) A and B intersect at a point,but both A and B are deficient-order(the orders are not always
equal,include zero order) generic plane,under proper oblique axes transform, A can be changed
into the shape of line, B can be changed into the shape of plane,A is through B from direct vies;And under another proper transform,B can be changed into the shape of line and A is changed into
the shape of plane, B is through A from direct view.We call A and B mutually through,intersection
74
point is called through point.
(4) A and B have no common point and aren't parallel,we call A and B mutually interleave.
The above four cases can be generalized into two kinds,of which,coincident can be regarded
as a particular case of intersectant,parallel can be regarded as a particular case of mutually
interleaving .
Theorem 1 When R=r=r1≥r2 or R=r=r2≥r1 ,
A coincides with B.
Proof: Because R=r,then M has solution,it represents that A and B have common point.
and r=r1≥r2,
then M=A∩B=A,
now AA ⊆ and BA ⊆ ,
coordinates of all points on A satisfy equations of B, so,all points on A are on B at the same time,viz. A coincides with B.
It can be proved by the same token,when R=r= r2≥r1,A also oincides with B. Particularly,
when R=r= r1= r2, A=B . ▌
Theorem 2: when R>r= r1≥r2 or R>r= r2≥r1,A is parallel to B.
Proof : because R>r,then M has no solution,it represents that A and B have no common
point.
But because r= r1≥r2,
then M'=(A'∩B')=A'
(A',B',M' are separately leadingout groups of equations A, B,M ), now
'' AA ⊆ , '' BA ⊆
so from theorem1,A'coincides with B'. In geometry,the coincidence of two figures are regarded as
a particular case of parallel of two figures,viz. if two figures are coincident then they must be
parallel,so,A'∥B'. But A',B'
are separately equivalent to be figures of A and B translating to the origin,so
A'∥A, B'∥B
then it has A∥B.
It can be proved by the same token,when r=r2≥r1 ,it also has A∥B. ▌
Theorem 3: when R=r>r1,r2 ,
it has the following four cases:
(1) r≤n-1, A and B intersect at a W-dimensional (1≤W≤n-2) linear figure;
(2) when r=n, r1=r2=n-1,A and B are two straight lines intersecting at a point;
(3) r =n, r1=n-1, r2<n-1 or r2=n-1, r1<n-1,A and B singly through;
(4) r =n, 1<r1, r2<n-1 ,A and B mutually through.
Proof:because R=r,
then M has solution,and r>r1 , r2,
it represents that A and B have common point but not coincident under the following four cases.
(1)because r≤n-1,
then the dimension of linear figure M as intersection of A and B, W≥1,and because
r1, r2≥1 and r>r1,r2,
Then each equation of A and B at least includes an equation of generic plane,Viz. r≥2,then the
75
dimension of M is W≤n-2,so,all common points of A and B filled a W-dimensional (1≤W≤n-2)
linear figure.
(2) Because r1= r2=n-1,then both A and B are straight lines,and because r=n,then M has the
only group of solutions,it represents A and B have the only common point,the solution of M is
coordinates of the common point.
(3) If r1=n-1,r2<n-1,
then A is a straight line, B is a sufficient-order or deficient-order generic plane under proper oblique
axis transform.Contrarily,if
r2=n-1, r1<n-1,
then B is a straight line,A is a sufficient-order or deficient-order generic plane under proper
transform. And because r=n,then A and B have the only common point. When r1>r2,A runs
through B, When r2>r1,B runs through A. The solution of M is through point coordinate.
(4) Because r1,r2<n-1,
Then bothA and B are not straight lines,and because r1,r2>1 ,then both A and B aren't
sufficient-order generic plane(they aren't n-4-dimensional generic solid, yet).But because r=n,then
A and B have the only common point. When change A into the shape of line and change B into the
shape of plane under proper oblique axis transform,A runs through B from direct view;But when
change B into the shape of line and change A into the shape of plane under another proper
transform, B runs through A from direct vies. Then A and B run mutually through,the solution of M
is through point coordinate.
Now let's prove that,the proper oblique axis transform of changing one of A and B into the
shape of line and changing the other into the shape of plane is exist.
Because the actual dimension of A、B can be separately marked as
W1=n- r1, W2=n- r2, Assume we will change A into the shape of line and change B into the shape of plane,viz.
assume Z1=1,Z2=2 (Z1,Z2 is separately the directviewing dimension of A, B),then from dimension
theorem,the order of A、B is separately
J1=W1-1=n- r1-1, J2=W2-2=n- r2-2,
Because J1+J2=n-3+n -( r1+ r2),
then,when r1+ r2=n,J1+J2=n-3,on A, B,except the common point M,separately choose another
J1=p,J2=s points whose coordinates are the same with those of M with relation to xn (or the same
with relation to any other coordinate axis)
M11,…,M1p and M21,…,M2s,
and make vectors MM11,…,MM1p, MM21,… , MM2s
linearly independent and satisfy conditions of theorem 2 of the third chapter. When r1+ r2>n,
J1+J2<n-3,
we can choose another m= r1+ r2-n points M1,…,Mm except A,B and make them satisfy the above
conditions along with the above points. Order MM11…M1pM21…M2s(when r1+r2=n )
or MM11…M1pM21…M2sM1…Mm(when r1+r2>n )
Viz. A can be changed into the shape of line and B can be changed into the shape of plane.
76
It can be proved by the same token,proper transform of changing B into the shape of line and
changing A into the shape of plane is exist. ▌
Theorem 4: when R>r>r1,r2,A and B are mutually interleaving.
Proof:because R>r,then M has no solution,it represents A and B have no common point.
And r>r1 ,r2, then A and B aren't parallel. So A and B are mutually interleaving. ▌
Example 1: Try to judge the interleaving of the two planes in four-dimensional space
=++++
=++++
0222
0322
4321
4321
1xxxx
xxxxI :
=+−+
=−−+
013
04322
421
321
2xxx
xxxI :
Solution: Because R=r=4,r1=r2=2, then we know the two planes are mutually through from
question (4) of theorem 3. The through point to be solved is
−−
19
6
19
26
19
9
19
10,,, First order I1 be changed into the shape of line,I2 keep the shape of plane,so J1=1,J2=0.Choose
a point
−
19
6
19
83
19
47
19
28,,,
on I1,choose a vector 2e1+2 e2-3 e3 through the point and the
through point,and order
3 e3□2 e1+2 e2,
Viz. make I1 run through I2〔refer to it in figure 42,the quadrilateral
of the figure represents I2,the through point is located at
intersection point of the two diagonals of the quadrangle, its
coordinates of the left two vertexes are separately(-2,1,-2,0)
and
−−
36
17,
18
19,
3
5,
4
5, figure 42
the coordinates of the two adjacent vertexes are known,coordinates of the other two vertexes are easily to be
known〕;But,because the rank of coefficients matrix of the
anterior two columns of I2 is 1,the rank of coefficients matrix of
the anterior three columns is 2,then another point which can
change I2 into the shape of line and satisfy theorem 2 of the
third chapter isn't exist except the through point,from theorem 2
of the fourth chapter,we choose axis x2 as upright axis,change
x4 into vertical axis,except the through
point,choose another point on I2 figure 43
−−
19
9
19
20
19
9
19
1,,,
77
choose a vector 3 e1+ e4+2 e3
between the two points and use transform
2 e3□-3 e1- e4,
then a preferred four-dimensional system of keeping I1 in the shape of plane、change I2 into the
shape of line,making I2 run through I1 is acquired〔refer to in figure 43,quadrangle represents I1,the left two vertexes are separately (-3.5,-2.5,0,6.5) and (-3.5,-2.5,1,4.5)〕.
§§§§2 Orthodromic space and normal space
We know,equations of linear figures can be divided into (they are system of linear equations)
the homogeneous and the nonhomogeneous. In linear algebra theory,figures denoted by system of
homogeneous linear equations can be called subspace,the relations between them can be easily
solved. But because figures denoted by system of non-homogeneous linear equations aren't
subspace,relations between them or relations between them and figures denoted by system of
homogeneous linear equations are not easy to solve. So,system of linear equations need to be given
more geometric concepts in order to offer people convenience for further study .
Still assume equations of two linear figures A and B are consistent system of linear equations
A: 01
=+∑=
axa ji
n
iji
(j=1,2,…,r1)
B: 01
=+∑=
bxb ki
n
iki
(k=1,2,…,r2)
and assume the ranks of coefficient matrixes of A and B separately are r1,r2,the rank of
coefficient matrixes of simultaneous group
( )
( )
==+
==+
∑∑
=
=
2
1
1
1
,,2,1 0
,,2,1 0
rk
rj
M
bxb
axa
ki
n
iki
ji
n
iji
L
L
:
is r. And assume α,βare solution vectors of system of non-homogeneous linear equations A,or α,β∈A,but apparently,α-β=γ∉A.
And in the derivative group of A
A′: 01
=∑=
xa i
n
iji
(j=1 ,…,r1) ,
it has γ∈A′,because end points of α,βare on A,and γ is parallel to the line which runs
through end points of αandβ (refer to it in figure 44),so,for the convenience of description ,we give the following defination:
Defination 1: A random solution vector of the derivative group
A′: 01
=∑=
xa i
n
iji
(j=1 ,…,r1)
of system of non-homogeneous linear equations A is called a orthodromic direction vector of A( be
called orthodromic vector for short). All orthodromic vectors of A formed solution space of A′,
78
so,we call A′orthodromic direction space of A(be called orthodromic space for short).
Defination 2: Assume
A′: 01
=∑=
xa i
n
iji
(j=1 ,…,r1)
B′: 01
=∑=
xb i
n
iki
(k=1,…,r2)
separately are orthodromic spaces of A and B ,then we call M′=A′∩B′,Viz.
( )
( )
==
==′
∑∑
=
=
2
1
1
1
21 0
21 0
r,,,k
r,,,j
M
xb
xa
i
n
iki
i
n
iji
L
L
:
is common orthodromic space of A and B (be called
common-orthodromic space for short).
We have known that,so-called sum of subspaces is all
gatherings Q1+Q2
of α1+α2 (α1∈Q1,α2∈Q2)of two random subspaces Q1,Q2,of linear space Vn.
If every decompose formula α=α1+α2 ofα∈Q1+Q2
is unique,then Q1+Q2 is direct sum,marked as figure 44 Q1
•
+ Q2 .
Sum of subspace or direct sum is still subspace.
In linear algebra theory,if Vn is n-dimensional euclidean space,and
Vn =Q1
•
+ Q2,
then we call Q1,Q2 orthogonal complement of each other,marked as
Q1=⊥2Q ,Q2=
⊥1Q .
Because all linear combinations among normal vectors of A also form subspace,and the
subspace and orthodromic space of A are orthogonal complement of each other,so we give the
following:
Defination 3: Assume Vn is euclidean space,A′is orthodromic space of linear figure A of
Vn. If ⊥
•
′+′= AAVn ,
then we call ⊥′
1A normal direction space of A(be called normal space). Especially,for normal
space M′of common-orthodromic space of A and B⊥⊥
(when M has solution, ⊥′
1M is also normal
space of M),we call it common-orthodromic normal space of A and B.
Assume a group of vectors α1,α2,…,αn- r1 of A are linearly independent
and all their linear combinations form orthodromic space of A,of which
79
{ } ( )1 21 21 rn,,,,,, n −== LL ταααα ττττ ,
then the equation of ⊥′
1A is
( )1
1
,,1 0 rnxAn
i
ii −==′ ∑=
⊥Lτατ: .
Defination 4: Assume
( )1
1
,,1 0 rnxAn
i
ii −==′ ∑=
⊥Lτατ: .
( )2
1
,,1 0 rnxBn
i
ii −==′ ∑=
⊥Lδβδ:
are separatly normal space of A and B, then we call T=⊥⊥ ′′
11 BA I ,
viz.
( )
( )
−==
−==
∑∑
=
=
2
1
1
1
,,1 0
,,1 0
rnx
rnx
Tn
i
ii
n
i
ii
L
L
δβ
τα
δ
τ
:
is common normal space of A and B ( be called common-normal space for short).
Because the dimension of linear subspace is n-r(r is the rank of coefficient matrix of equation
of the subspace),so,from defination 2 it canimmediately get:
Theorem 5: Random two linear figures A and B have n-r-dimensional common-orthodromic
space (r is the rank of coefficient matrix of simultaneous group). ▌
From defination 3 it canimmediately get:
Theorem 6: Random two linear figures A and B have r-dimensional common-orthodromic
normal space (r is the rank of coefficient matrix of simultaneous group).
Theorem 7: Random two linear figuresA and B have r1+r2-r-dimensional common-normal
space(r1,r2 are separately the rank of coefficient matrix of system of equations A and B,r is the rank
of coefficient matrix of simultaneous group).
Proof:From defination 3,a random normal vector of normal spaceis always a orthodromic
vector of a linear figure,so⊥′
1A and ⊥′
1B separately have n- r1 and n- r2 normal vectors. Of
which,n-r vectors of common-orthodromic space M′of A and B are common-normal vectors of
⊥′1A and
⊥′1B ,so,in all (n-r1)+(n- r2)normal vectors of
⊥′1A and
⊥′1B ,n-r can be linearly
represented by other normal vectors, so the rank of coefficient matrix of equations of
ommon-normal space T is (n-r1)+(n- r2)-(n-r)=n-r1-r2+r,
so,the dimension of T is WT =n-(n- r1- r2+r)= r1+ r2-r. ▌
For example,when A and B are coincident or parallel,if r1>r2,then the orthodromic space
A′of A is common-orthodromic space of A and B, the normal space ⊥′
1A of A is
80
common-orthodromic normal space of A and B, and the normal space ⊥′
1B of B is common-normal
space of A and B.
Proof:From theorm 1、theorem 2,it has r= r1>r2,so M′=A′∩B′=A′,
viz. A′ is common-orthodromic space of A and B, and ⊥′
1A is common-orthodromic normal space
of A and B;From theorem 7,the dimension of common-normal space T is
r1+r2-r=r2,
so it has T=⊥⊥ ′′
11 BA I =⊥′
1B ,
viz. ⊥′
1B is common-normal space of A and B. ▌
And for example,when A and B are mutually interleaving, the dimension of their
common-orthodromic space is
0≤W≤n-3,
the dimension of common-orthodromic normal space is 3≤r≤n,
the dimension of common-normal space is 1≤WT ≤n-2.
Proof : Assume r1≥r2,from theorem 4 , it has R>r ,and r>r1,r2,so we know by deduction
that r1≤n-1, r2≥2
(if r1 n-1, r2 2,for example r1=n or r2=1,then it has
r= r1 or R=r= r1+ r2,
this is in contradiction with the conditions of theorem 4).When r1= r2=2,it has R=4, r=3,so it has
W=n-3;
When r1= r2=n-1,it may has R=n+1,r=n,then it has W=0,
so it has 0≤W≤n-3.
and,W=n-r,sowe can know by deduction that the dimension of common-orthodromic normal space
⊥′1M is 3≤r≤n. Then from theorem 7,the dimension of common-normal space is r1+ r2-r,when
r1= r2=2, r=3,
then it has r1+ r2-r=1;
and when r1= r2=n-1,r=n,so it has
r1+ r2-r=n-2.
Then we know 1≤WT ≤n-2. ▌
When A intersects with B ,from theorem 3,there are four cases:
(1) it has 1≤W≤n-2-dimensional common-orthodromic space,2≤r≤n-1-dimensional
common-orthodromic normal space and 0≤WT ≤n-3-diemensional common-normal space.
All the following three cases have zero-dimensional common-orthodromic space and
n-dimensional common-orthodromic normal space,but the dimensions of common-normal space
are not the same,of which
(2) When two straight lines interleave, there is n-2-dimensional common-normal space.
(3) When A and B are singly through,0≤WT ≤n-3.
(4) When A and B are mutually through,0≤WT ≤n-4.
81
Under the above four cases,the dimension of common-orthodromic space has been given in
theorem 3,because the rank of common-orthodromic normal space is n-r, then we can know by
deduction that its dimension is r.
So difference of the dimension of its common-normal space is not discussed,readers can try to
write out the proof process by yourselves.
§§§§3 Exterior sumof two linear figures To study the distance between two linear figures,we need extend the concepts of sum of
subspace .
3.1 Concepts of exterior sumof two linear figures
From structural theory of solutions of system of linear equations,assume r0 is a particular
solution vector of system of non-homogeneous linear equations A,r is a random solution vector of
A,when we take all solution vectors ηof derivative group A′, then
r=r0+η
will take all solution vectors of A. Because for every r, r0 is the same,then the decompose formula is unique. So when A, A′are
separately regarded as gatherings of all solution vectors of A and A′,{r0}is regarded as much
multiplicity gathering including countless same elements r0(r0 is A particular solution vector of
A),viz. { } } {
boundless
0000
48476
L rrrr ,, , = ,
then all gathering of r=r0+η (r∈A, r0∈{r0},η∈A′) can be written into the form which likes
direct sum { } ArA ′+=•
0 (1)
it is called exterior direct sum of A′and gathering{r0}.Because 0∉A,then the exterior direct sum
isn't subspace,it doesn't close for addition and amount multiplication of linear space Vn.
From the above analysis we know,relation between linear figure and its orthodromic space is
the relation between exterior direct sum of subspace and the subspace. Then,we extend and apply
concepts and property of subspace to linear figures denoted by system of non-homogeneous linear
equations,then we can learn and study more of their geometric property.
Definition 5 : Assume equations of two linear figures A and B are system of
non-homogeneous linear equations,A∩B isn't empty,then
A+B=S
is called the exterior sumof linear figures A and B . Of which,S is also a system of
non-homogeneous linear equations.
Assume S′=A′+B′is orthodromic space of S(equation of S is non homogeneous,A′and B′are separately orthodromic spaces of A and B),when assume α′∈A′,β′∈B′,then δ′=α′+β′∈S′.
assumeδ∈S ,andδ0 is a particular solution vector of S,{δ0}is a much multiplicity gathering
including countless identicalδ0 .
Because for every solution vectorδ of S,decompose formula δ=δ0+δ′
is unique,then from formula (1),for allδ∈S it has
{ } SS ′+=•
0δ
82
further it has { } ( )BAS ′+′+=•
0δ
viz. { } ( )BABA ′+′+=+•
0δ (2)
This is the formula of relation between exterior sumof A and B and sum of their orthodromic
space . If assume α∈A, β∈B ,then the corresponding decompose formula can be marked as α+β=δ0+α′+β′.
3.2 Property of outer sum Now,we regard δ0 as a particular solution vector of A( it is called particular solution for
short),when β′=0 ,from decompose formula
r=r0+η ,
it has α+β=α∈A;
when δ0+α′∈A∩B
it has α+β=β∈B;
But ,when BAI∉′+≠′′ αδβα 0 and 0, ,it has α+β∉A , andα+β∉B,
But it also has α+β=δ∈S,
viz. α+β∈A+B .
So, A ⊆ A+B, B ⊆ A+B ,
and A∩B ⊆ A+B.
If regardδ0 as a particular solution of B,result is the same. So,as long asδ0 is a particular solution
of one of system of non-homogeneous linear equations A or B ,then it is also a particular solution
of A+B at the same time. Of courseδ0 can be neither a particular solution of A nor a particular
solution of B, but it is still a particular solution of exterior sumA+B .
Exterior sumdoesn't close for addition and amount multiplication. At the same time,exterior
sumis not fit for commutative laws outside of A′+B′ ,viz.
{ } ( ) { }( ) { }( )
{ } ( ) { } ( )ABBA
ABBABA
′+′+=′+′+
+′+′≠′++′≠′+′+••
•••
00
000
but
δδ
δδδ
So,braces of formula (2) can' t be got rid of easily.
Besides,exterior sumstill has the following property(all the following discussion has
assumed that A∩B is not empty):
Theorem 8: Assume one of linear figures A and B is subspace,the other is not subspace A′and B′ are separately orthodromic space of them,then it has
A+B=A+B and A+B=A′+B′,
viz. A+B=A′+B′.
Proof:assume equation of A is non homogeneous, equation of B is homogeneous, viz . B=B′
and assumeα∈A, β∈B , apparently,β=β′∈B′.
And assumeδ0 is a particular solution vector of A,then from decompose formula
r=r0+η
and α+β=δ0+α′+β′
we know α=δ0+α′(α∈A,α′∈A′)
83
viz. α+β=α+β′=α+β
so it has A+B=A+B
and assumeβ0 ∈A∩B , thenβ0∈A , then from decompose formulaβ0=δ0+α0′,at the same
timeβ0∈B ,viz. β0∈B′, assume α0=-α0′, apparentlyα0∈A, so,δ0 is regarded as linear
combination ofα0∈A′and β0∈B ′,
viz. δ0=α0+β0,
so it has α+β=δ0+α′+β′=α0+α′+β0+β′
but α0+α′∈A′,β0+β′∈B′, then it has
A+B=A′+B′,
so from transitive law,it has A+B=A′+B′ ▌
Theorem 9: Assume both equations of linear figures A and B are non homogeneous,A′and
B′ are separately orthodromic spaces of them,if A′∩B or B′∩A is not empty,thenit has
A+B=A′+B′.
Proof: if A′∩B is not empty then B′∩A is also not empty,so it has
A+B=A′+B=A+B′,
but from theorem 8, it has
A′+B=A′+B=A′+B′,
A+B′=A+B′=A′+B′,
then A+B=A′+B′ ▌
3.3 Equation of outer sum From concept of exterior sumwe know,orthodromic space of exterior sumof linear figures is
sum of orthodromic space of linear figures. And from defination 3,we know linear figures have the
same normal space with their orthodromic space,and normal space has nonnegligible
contribution to determine equations of linear figures.So we give : Theorem 10: Assume both Q1 and Q2 are subspace of linear space Vn,⊥
1Q and⊥2Q are
separately their orthogonal complement,thenit has
( ) ⊥⊥⊥=+ 2121 QQQQ I (3)
( )⊥⊥⊥ =+ 2121 QQQQ I (4)
are called De Morgan law of intersection and sum of subspace. Proof:similar to De Morgan law of intersection and union of gathering,assume
u∈(Q1+Q2)⊥,
then u∉Q1+Q2,
so u∉Q1,and u∉Q2,
it has u∈ ⊥1Q and u∈ ⊥
2Q ,
then u∈ ⊥1Q ∩ ⊥
2Q ,
so it has (Q1∩Q2)⊥⊆ ⊥
1Q ∩ ⊥2Q Contrarily,assume u∈ ⊥
1Q ∩ ⊥2Q ,
then u∈ ⊥1Q and u∈ ⊥
2Q ,
it has u∉Q1, and u∉Q2,
and because we have assumed
84
u∈ ⊥1Q ∩ ⊥
2Q ,
so in random decompose formula
u=α+β
of u it also hasα,β∉Q1, and α,β∉Q2,
then u∉Q1+Q2,
so u∈(Q1+Q2)⊥,
then it has ⊥
1Q ∩ ⊥2Q ⊆
( Q1+Q2)⊥.
From the above we know (Q1∩Q2)⊥
= ⊥
1Q ∩ ⊥2Q It can be proved by the same token
⊥1Q +
⊥2Q =(Q1∩Q2)
⊥. ▌ Theorem 10 represents, that common-normal space of two linear figures is normal space of
exterior sumof the two linear figures. From formula (3), it has
Q1+Q2= (Q⊥
1∩Q⊥
2)⊥
So in formula(2) ( )⊥⊥⊥ ′′=′+′ BABA I (5)
viz. { } ( )⊥⊥⊥ ′′+=+ BABA I。
0δ (6)
formula (6) can be regarded as general formula of equation of outer sum.From theorem 7,when
assume the ranks of A , B and coefficients matrix of equations of A∩B are separately r1, r2 and
r, ( )⊥⊥⊥ ′′ BA I of formula (6) are r1+ r2-r generic planes taking a group of linearly independent
solution vectors of common-normal space of A and B ,and coordinates of δ0∈{δ0} are
coordinates of a common point of these generic planes. When both equations of A and B are homogeneous , viz . A=A′,B=B′,thenδ0=0,it
represents these generic planes take the origin as a common point.So,formula (5) can be regarded
as a particular case of formula (6),it is general equation of equation of sum of subspace.
§§§§4 Distance between two linear figures Apparently,when A and B are coincident or intersectant, the distance is zero.So we only discuss
the distance when A and B are parallel or mutually interleaving.
4.1 Distance between two parallel figures Assume A∥B ,and r1≥r2(r1, r2 are separately the ranks of coefficients matrix of equations of
A and B,the following is the same). Because ⊥′
1A is common-orthodromic normal space of A and
B ,then ⊥′
1A ⊥A, ⊥′
1A ⊥B.
order ⊥′
1A separately interleave with A and B ,then
M1= ⊥′
1A ∩A
is a point M1(x11, x21,…, xn1);
85
and M0 =⊥′
1A ∩B
is a r1- r2-dimensional linear figure. Of which,when r1= r2 ,M0 is also a point M0 (x10 , x20 ,…, xn0). Now,the length of line segment M1 M0 is the distance between A and B ; When r1>r2 ,make r1- r2 generic planes taking r1- r2 linearly independent orthodromic vectors of M0 as normal
vector and through point M1,at the same time, it intersects M0 at point M2 (x12 , x22 ,…, xn2), Because M1,M2 are separately on A ,B ,and both M1,M2 are on ⊥′
1A , then line segment
M1M2⊥A and M1M2⊥B,so,the length of line segment M1M2
( ) ( ) ( )xxxxxx nnMM 12
2
2122
2
1112
2
21 −−− +++= L
is the distance between A and B. Besides,because ⊥′
1B is common-normal space of A, B,then coordinates of point M2 can be
acquired by translating ⊥′
1B , making it through M1 and intersecting B,the result is the same.
4.2 Distance between two mutually interleaving figures Assume A and B are mutually interleaving,and r1≥r2,translate A ( be changed into A0 ) and
make it interleave B,then the exterior sum BAS += 0 of A0 and B is parallel to A , then we kan
find out the distance between A and S according to the solution of distance between two parallel
figures,this is the distance between two linear figures A and B . Example 2: Please find out the distance between straight line
=+−−+
=−+++
=+++−
=−++−
=+−++
052
04272
01322
03232
0223
4321
4321
4321
4321
4321
xxxx
xxxxB
xxxx
xxxx
xxxx
A
:
:
of four-dimensional space.
Solution::::Because r = 4, R =5 , then A and B are mutually interleaving.
And because A has a orthodromic vector
18e1+ e2-11 e3- e4
B has two orthodromic vectors 3e1-4e3+11e4 , 3e1-4e2- e4
So,equation of common-normal spaceT is
=−−
=+−
=−−+
04
011
01118
421
431
4321
xxx
xxx
xxxx
T:
and one solution vector of T is
3 e1+2 e2+5 e3+ e4 ,
choose a point (9,-14,0,0)
of B , through the point and taking vector {3,2,5,1}
as normal vector,and then make a generic plane
S:3 x1+2 x2+5 x3+ x4+1=0 ,
It is the equation of exterior sum SBA =+0
of A0 (A0 is the new generic plane formed of intersection of A s translation and B ) and B . Because
86
S∥A and 3>1 (3 is the rank of coefficient matrix of A,1 is the rank of coefficient matrix of
S ),then 01118 4321 =−−+′⊥xxxxA :
is also the common-orthodromic normal space of A and S. And the intersectionsof ⊥′
1A and A, S
are separately
=−−+
=++++
=−−+
=+++−
=−++−
=+−++
01118
01523
01118
01322
03232
0223
4321
4321
0
4321
4321
4321
4321
1
xxxx
xxxxM
xxxx
xxxx
xxxx
xxxx
M
:
:
then M1 is a point , 447
1397,
447
278,
447
391
149
381
−− ,: M
M0 is a plane. Taking two orthodromic vectors 2e2+e3-9e4 of M0 and e1-7e2+11 e4
as normal vector, through point M1 , then make generic planes
2 x2+ x3-9 x4+27=0
and x1-7 x2+11 x4-28=0 ,
they intersect M0 at
=−+−
=+−+
=−−+
=++++
028117
02792
01118
01523
421
432
4321
4321
2
xxx
xxx
xxxx
xxxx
M :
at the same time the intersection point is
−−
5811
17863,
5811
5104,
5811
4487,
1937
7922M
Then | M1 M2|≈0.32
is the distance between straight line A and plane B . Coordinates of the above point M2 can be acquired by translating T through M1 and making
it intersect S (The result is the same with that of the above method).
Exercises 6.1 Please find out interleaving and distance between the two linear figures of four-dimensional
space
=−+++
=−+−+
=++−+
=++++
05223
030432
072255
053234
4321
4321
2
4321
4321
1
xxxx
xxxxL
xxxx
xxxxL
:
:
.
87
The Seventh Chapter: Included angle questions and their
linear solutions between two linear figures
Included angle questions have the most abundant and interesting contents of the position
relations among linear figures.
§§§§1 Diversity of included angle questions bertween two linear figures
of higher space
This question has two meanings,one is the nonuniqueness of included angle amounts between
two linear figures,the other is the nonuniqueness of solutions of included angle questions .
1.1 Nonuniqueness of included angle amounts
1.1.1 Two kind of projection methods
In the first chapter we have referred to:∀α∈Vn , if ασα,then σ is a identical
transform.
When Vn is regard as a n-dimensional euclidean space (the following is the same),then σ
can be regarded as projection(be referred to orthographic projection) of Vn to itself,
viz. σ(Vn)=Vn.
Contents of projection not only include image setσ(Vn) (is also called projection of (Vn ) and
image source set(is also called passive set, space been projected or figures been projected for
short),but also include two contents of set (be called path set、projection direction set、projection
direction space) of projection direction(also be called projection path) and space of accepting or
containing source gatherings(be called contain set、acceptance set or acceptance space).Because of
the nonsingularity of identical transformation(or be called nondegenerate),so,its projection
direction is a zero vector.
When σ is a (orthographic)projection of Vn to one of its subspace Vm (m<n),Viz. when σ(Vn)=Vm ,σis a singular linear transform. For example,when assume
ei∈Vn (i=1 ,2 ,… ,n), e j∈Vm (j=1 ,2 ,… ,m;m<n),
the corresponding transform can be marked as
( )( )
( )( )
( )
=
=
=
=
=
+
0
0
1
22
11
n
m
mm e
e
e
e
e
e
e
e
σ
σ
σ
σ
σ
LL
LL
From singularity of σ(or be called degeneracy),we can know by deduction that projection
directions (can be called degenerate direction)are nonzero vectors. All linear combinations among
projection direction also form subspace of Vn ,viz. projection direction space(also be called path
set).Because under σ,random projection direction is perpendicular(or only satisfy that inner
product is zero) to image (also be called projection of random element)of random element of Vn ,so,projection direction space is normal space
⊥mV of Vm . Then,the whole process of the
88
projection can be described as:
Image set of passive set ....................V.n. from path set ....... ⊥mV to contain set ..........V.m. is ..σ.(.V.n.). Passive set..........、.path ....
set...、.contain set.......... and image set........ are four principal elements of forming projection,every .....element of .........the four should not be absent.........................
For the conveniece,we can apprehend projection concepts like this: σcan be decomposed into n-m unidirectional stagebystage projections,viz. assume σ′(Vn)=Vn-1,
then σ2′(Vn)=σ′〔σ′(Vn)〕=σ′(Vn –1 )= Vn -2,
then σ(Vn)= σn-m′ (Vn)= Vm.
All the above discussions can be regarded as projecion of Vn to one of its subspaces.The
following will discuss the second kind of projection:projection between two different subspaces of
Vn .
Assume nsnp VVVV ⊂⊂ 21 , , and Vp1≠Vs2 , dimensions of Vp1 and Vs2 are separately p
and s,then projectionσ0 from Vp1 to Vs2 is involved in the first kind of projection σ,viz.
because σ(Vn)= Vs2 andσ0 (Vp1) ⊆ Vs2,
then σ0 (Vp1) ⊆σ(Vn) .
When we carry the first kind of projectionσ,we also carry the second projectionσ0 at the
same time,when the first kind is finished,the second kind is finished at the same time.
Assume Vp1 is sufficient-order or difficient-order (include zero-order)generic plane through the
origin and in the ordinary position,Vs2 is sufficient-order or difficient-order generic coordinate
plane,and assume the equation of Vs2 is
( ), : 1-2
0
0
2
1
2 ns
x
x
x
V
n
s
s
s ≤≤
=
=
+
+
LL
and Vp1 is all the linear combinations of linearly independent vectors
r1,r2,…,rp (2≤p≤n-1) ,
of which r i ={ri1 ,ri2 ,…,rin }i=1 ,2 ,… ,p.
then the corresponding expressions of the orthographic projection from Vp1 to Vs2 is
( )( )
( )
,:
2211
22221212
12121111
0
+++=
+++=
+++=
spsppp
ss
ss
rrrr
rrrr
rrrr
eee
eee
eee
L
LLLL
L
L
σ
σ
σ
σ
When we orderly carry unindirectional stagebystage projection taking directions parallel to
xn , xn-1 ,…,xs+1 as projection direction,coordinates of ri orderly become zero from back to front,column vectors of coefficient matrix
89
=
pnpp
n
n
rrr
rrr
rrr
r
L
LLLL
L
L
21
22221
11211
of ri fade away from back to front. By n-s stagebystage projections,the back n-s coordinates of ri
all become zero,at last,matrix r become coefficient matrix of σ0(r)
=′
pspp
s
s
rrr
rrr
rrr
r
L
LLLL
L
L
21
22221
11211
When s>p,r′is row-nonsingular matrix( the rank is p);When s=p ,r′is a p-order square
matrix
(the rank is still p).So,when s≥p,projection or ri in Vs2 is still linearly independent. But when
s<p ,r′is column-nonsingular matrix (the rank is s),image σ0(ri) of vector groupri also
become linearly dependent vector group, so, the dimension of imageσ0(Vp1) of Vp1 also drops to s .
1.1.2 Defination of included angle between two linear figures
In the following,let's first discuss a kind of very important projection—rightangle projection,it is the most important foundation of all included angle questions (include defination and solution
method,etc)between two linear figures of higher space.
Theorem 1:Assume is a plane of n-dimensional space, 1π is a coordinate plane, ⊥= 12 ππ is a n-4-dimensional generic coordinate coordinate plane which is orthogonal
complement with 1π of each other,H1 and H2 are two straight lines ofπ,1π , H21 and H12 , H22
are separately projections of H1,H2 onπ1 and 2π . If
H11⊥H21 and H12⊥H22
(or only satisfy that the inner product is zero),then
H1⊥H2 .
Proof:Assume h1={α11,…,αn1}, h2={α12,…,αn2}
are separately orthodromic vectors of H1,H2,and assume equation ofπ1 is
=
=
=
0
0
0
4
3
1
nx
x
x
L
: π
then equation of 2π is
=
=
0
0
2
1
2x
x: π
so projections of h1,h2 on1π , 2π are separately
h11 ={α11,α21,0…,0}, h21={α12,α22,0…,0}
h12={0,0,α31,…,αn1},h22={0,0, α32,…,αn2}
If H11⊥H21,H12⊥H22
Then h11⊥h21 ,h12⊥h22
So α11α12+α21α22=0
90
and α31α32+…+αn1αn2=0
then α11α12+…+αn1αn2=0
so h1⊥h2
viz. H1⊥H2. ▌
Theorem 1 it can also be called rightangle projection theorem,it is brought forward and
proved by Mr Jian Zhaoquan of Beijng Institute of Technology[13],then it is called Jianshi solution
theorem in this book.
Projection process of theorem 1 is equivalent to interchange path gathering and contain
gathering,viz. first take 2π as path gathering,take 1π as contain gathering,and finish the
projection from πto 1π ;And then take 1π as path gathering ,take 2π as contain gathering,and
finish the projection fromπ to 2π contrarily.
For Jianshi solution theorem,we can extend it like this:
Theorem 2:assume Vm(m<n) is subspace of euclidean space Vn,Vmo is a sufficient-order or
difficient-order generic coordinate plane which has the same dimension with Vm.
H1,H2,…,Hm
is m straight lines of Vm, H11,H21,…,Hm1
and H12,H22,…,Hm2
are separately projections of H1,H2,…,Hm
on Vmo and ⊥
moV . If H11,H21,…,Hm1
and H12,H22,…,Hm2
are perpendicular to each other(or only satisfy that the inner product is zero),then
H1,H2,…,Hm
are perpendicular to each other.
Proof::::divide H1,H2,…,Hm in pairs into ( )
2
1−mm groups,then prove it according to
method of theorem1. ▌
But,the inverse of theorem1, theorem 2 don't exist.
Proof:because theorem 2 is the generalization of theorem 1,so it is ok to
prove that the inverse of theorem1 don't exist .
Assume H3,H4 are anther two straight lines of π which are perpendicular to each other,
h3={β11 ,…,βn1}, h4={β12 ,…,βn2}
are separately their orthodromic vectors. Because H3,H4 and H1,H2 have coplanarπ,then
both h3 and h4 can be linearly represented by h1,h2,viz.
h3=λ11 h1+λ12 h2, h4=λ21 h1+λ22 h2
(λ11,λ21,λ12,λ22 all nonzero numbers).So, β11=λ11α11+λ12α12 , …,βn1=λ11αn1+λ12αn2 β12=λ21α11+λ22α12, …,βn2=λ21αn1+λ22αn2.
because h3⊥h4,
and α11α12+…+αn1αn2=0,
then β11β12+…+βn1βn2=λ11λ21(α211+…+α2
n1)+λ12λ22(α212+…+α2
n2)=0,
viz. λ11λ21(α211+…+α2
n1)= -λ12λ22(α212+…+α2
n2).
And because projectons of h3 ,h4 on 1π are separately
h31={β11,β21,0,…,0}, h41={β12,β22,0,…,0},
91
and α11α12+α21α22=0,
if h31⊥h41,
then β11β12+β21β22=λ11λ21 (α211+α2
21)+λ12λ22 (α212+α2
22) =0,
viz. λ11λ21 (α211+α2
21)= -λ12λ22 (α212+α2
22),
so it must has ,2
2
2
12
2
22
2
12
2
1
2
11
2
21
2
11
nn αα
αα
αα
αα
++
+=
++
+
LL
(now π and 1π are coincident or parallel),but when
,2
2
2
12
2
22
2
12
2
1
2
11
2
21
2
11
nn αα
αα
αα
αα
++
+≠
++
+
LL
it must has λ11λ21 (α211+α2
21) ≠ -λ12λ22(α212+α2
22),
viz. β11β12+β21β22≠0 ,
so now h31 is not parallel to h41 .
By the same token,h3 ,projections of h4 on 2π are not always perpendicular . ▌
It is not difficult to know according to the above reasons that not random two straight lines
perpendicular to each other are useful for included angle questions.
From theorem 2 ,it can immediately get
Theorem 3:Assume Vm(m<n) is a subspace of euclidean space Vn,Vmo is a sufficient-order
or difficient-order generic coordinate plane which has the same dimension with Vm.
H1,H2,…,Hm
are m straight lines of Vm, H11,H21,…,Hm1
and H12,H22,…,Hm2
are separately projections of H1,H2,…,Hm
on Vmo and ⊥0mV . If H1,H2,…,Hm
are perpendicular to each other. And their projections H11,H21,…,Hm1
on Vmo are also perpendicular to each other,then their projections
H12,H22,…,Hm2
on ⊥0mV must be perpendicular to each other(or only satisfy that the inner product is zero).
Contrarily,if H1,H2,…,Hm
are parallel to each other. And their projections
H12,H22,…,Hm2
on ⊥0mV are also parallel to each other(or only satisfy that the inner product is zero),then their
projections H11,H21,…,Hm1
on Vmo must be parallel to each other. ▌
Theorem 2: and theorem 3 can be called generalized rightangle projection theorem by a
joint name .
We know:zero vector is parallel to random vectors,it is linearly dependent with random
vectors at the same time.Verticality can determine the changelessness of projection dimensions ,
92
and correlativity can determine the change of projection dimensions. In the proof of the above three
theorems,we stipulate that projection of two vectors can only satisfy the inner product is zero,in
order to emphasize its verticality and ignore its correlativity.
Because the inverses of theorem1, theorem 2 aren't exist,then vector group h1,h2,…,hm
parallel to each other of Vm which satisfy theorem 3 is uniquely determined.
According to principle of the above theorem 3,we give the following:
Defination 1:Assume the dimension of two subspaces Vp1 and Vs2 is W=p ,if projections σ(ri ) of a group of vectors ri (i=1 ,2 ,… ,p) parallel to each other of Vp1 on Vs2 are also parallel
to each other(aslo including some or all projections that only satisfy the inner product is zero to
each other),then these vectors are called vectors that can determine included angles between Vp1
and Vs2,be called angle vectors for short.
In fact,to a considerable degree,the process of determining included angles φi between two
linear figures is the process of determining the group of angle vectors ri or projectionsσ(ri )
(i=1 ,2 ,… ,p) of the group of angle vectors .
Defination 2: assume the angle vectors between Vp1 and Vs2 are
r1,r2,…,rp
and ri∈Vp1 (i=1 ,2 ,…,p),then the ratio
( )
pir
ri
i
i,,2,1 cos
L == ϕ
σ
of module |σ(ri ) | of their projections on Vs2 to module | ri | of their own is cosine of included
angle φi between Vp1 and Vs2 .When Vp1 and Vs2 are separately regarded as orthodromic spaces of
linear figures A and B (equations of A and B can be non homogeneous),because
A≌Vp1, B≌Vs2
(“≌”represents congruence[12]
),then included angle between Vp1 and Vs2 is the included angle
between two linear figures A and B.
When all |σ(ri ) | = | ri | (i=1 ,…,p )
are true at the same time,the included angle between two linear figures A and B is zero,so A and B
are coincident or parallel,only at this time,the zero included angle between two linear figures is
considered to be meaningful;But when |σ(ri ) | = | ri | (1≤i≤p)
aren't true at the same time,the zero included angle between two linear figures is considered to be
meaningless,we only choose those nonzero included angles φi (1≤i≤p) to define the included
angle between two linear figures. The purpose of making the rule is to ensure the result consistency
of included angles under different solution methods.
The projection process of the above defination1, defination 2 can be described as:
Passive set Vp1⇒ path set ⇒⊥2sV contain set Vs 2⇒ image setσ(Vp1).
1.1.3 Common vector and uncommon vector————————sameness of two linear figures' dimension
In n-dimensional euclidean space,system of orthogonal vectors formed of n vectors are called
a group of orthogonal bases;Orthogonal bases formed of unit vectors are called orthonormal basis,be called orthogonal-unit vector.
Because the subspace Vm(m≤n) of euclidean space Vn is also a euclidean space under the
defined inner product,so,we can also choose m unit orthogonal vectors ε1,ε2,…,εm of
Vm to form orthogonal-unit vector of Vm,and this group of orthogonal-unit vectors can be extend
93
to the orthogonal-unit vectorsε1,ε2,…,εm of Vn.
Defination 3::::Assume both Vp1 and Vs 2 are subspaces of euclidean space Vn,εi1∈Vp1 andεj2∈Vs2 (i=1 ,…,p; j=1 ,…,s) are separately orthogonal-unit vectors of Vp1 and Vs2. If εi1 ,εj2∈Vp1∩Vs2(1≤i≤p; 1≤j≤s) ,
viz. εi1 ,εj2 are also orthogonal-unit vectors of Vp1∩Vs2, thenεi1 ,εj2 are also called common
orthogonal-unit vectors of Vp1 and Vs2, be called common vector for short.
Otherwise,if εi1 ,εj2 ∉Vp1∩Vs2 (1≤i≤p;1≤j≤s) ,
then they are called uncommon vectors of Vp1 and Vs2.
We know,when two linear figures are parallel or coincident (r=p or r=s,r is the rank of
coefficient matrix of equations' simultaneousness of two figures,refer to it in theorem 1 and
theorem 2 of the sixth chapter),their included angle is zero(it is known),if order their dimensions
be same,it is ok to substitute common-orthodromic space for figures of high dimensions. So we
only discuss how to change the dimensions of two linear figures from different to same when they
are neither coincident nor parallel(viz. r>p, s).
when two linear figures are neither coincident nor parallel,those angle vectors only determine
included angles between two linear figures can be changed into common vectors (from defination 3,
we can infer that the number of common vectors included in the two subspaces are the same)of
orthodromic space of the two linear figures by unit change ,and those orthogonal-unit vectors
that determine nonzero angles between two linear figures are uncommon vectors of orthodromic
space of the two figures.
Assume the dimensions of two linear figures A and B are separately p and s,and p>s,p-s=k,and assume they have r-dimensional common-orthodromic space M,and A′, B′, M′are
separately orthodromic spaces of A , B , M . Then ,A′and B′separately have r common vectors,p-r and s-r uncommon vectors. From defination 3 and defination of orthodromic space,we know
M′is all linear combinations of the r(r>p, s) common vectors. Assume A0′and B0′ are
separately all linear combinations of p-r and s-r uncommon vectors,then all nonzero angles
between A′and B′are the same with those between A0′and B0′.
But how to separately change A′ and B′ into A0′ and B0′? It is ok to separately delete r
common vectors from A′and B′. The method is separately taking A′and B′as passive set,taking M′as path set,taking M
⊥′as contain set,then finishing the projection from A′, B′through M′to M
⊥′,and then image sets A0′and B0′are separately acquired;Or,make A and
B separately intersect M⊥,it also can get
A∩M⊥
=A0, B∩M⊥
=B0,
apparently,the included angle between A0′and B0′is also the included angle between A0 and B0 .
This step can be called deleting common vector for short(If the dimension of common-orthodromic
space is zero,then this step can be omitted.).
From defination of the included angle,the dimensions of two linear figures should be equal,but,although by deleting common vectors ,dimensions of A0′and B0′separately been changed
into p-r=p0, s-r=s0,it still p0>s0,and p0-s0=k,then the included angle between A0′and B0′can't be determined yet.
Because p0-s0=k,it shows that there are k unwanted uncommon vectors in A0′, they
should be deleted from A0′. How these k uncommon vectors appeared unwantedly? Originally,in
normal space⊥′
0B of B0′,there are k vectors linearly dependent with orthodromic vectors of A0′,
94
make ⊥′
0B intersect A0′, it gets A0′∩ ⊥′0B =M0′. Because the dimension of B0′is s0 , then
the rank of ⊥′
0B is s0 ,then the rank of M0′is s0+n-p0=n-k ,viz. the dimension of M0′is k,and M0′is on A0′,k orthogonal-unit vectors of M0′are just the unwanted uncommon vectors of
A0′ .
Taking A0′as passive set,M0′as path set, ⊥′0M as contain set,finish image set A1′of
projection from A0′through M0′to ⊥′
0M , or,make A0 intersect ⊥0M ,then it gets
A0∩ ⊥0M =A1.
Now,because the rank of A1 changes from n-p0 of A0 to n-p0+k=n-s0 ,then,then dimension
of A1 changes from p0 of A0 to n-(n-s0)=s0 which is the same with that of B0 . Apparently,the
included angle between A1′ and B0′ is equal to the included angle between A1 and B0,it is
equal to the included angle between A and B at the same time.
From the above description,we have the following
Theorem 4::::Random two linear figures of different dimensions can be changed into linear
figures of same dimensions in the case of keeping included angle changeless. ▌
Inference: After two linear figures A and be changed into A1 and B0 of same dimensions, the
amount of uncommon vectors they include has the following two cases:s0≤1;s0>1. ▌
From defination of included angle between two linear figures,when s0>1,the amount of
nonzero angles between A and B is also more than 1.This is so-called nonuniqueness of the amount
of included anges.
1.2 Nonuniqueness of the solution of included angle questions
Because of the nonuniqueness of the amount of included angles,it determines that methods of
solving included angle questions are not unique.
There are two kinds of methods of solving included angle quesions among linear figures,one
is called linear solution,the other is called nonlinear solution.
Definition 4: Adopting methods of solving system of linear equations,separately find out one
of their orthodromic vectors(or normal vector) on two linear figures(or on their normal space),and
method that use included angles (or cosine of the included angles)between the two vectors to define
included angles between the two linear figures are called linear solutions of included angle
questions.
Definition 5: Method that need analyzing the relation between nonlinear equation and system
of linear equations of two linear figures to determine included angles between the two linear figures
are called nonlinear solution of included angle questions.
In graph field,Mr Jian Zhaoquan of Beijing Institute of Technology takes the lead in using unit
circle to determine included angles [13]
between two planes of higher space,he opened the door of
using nonlinear solution. So,nonlinear solution is also called Jianshi solution solution or JianShi
method. Linear solution is comparatively easy,but it can't solve all included angle questions;Jianshi
solution solution can solve all included angle questions,but the method is very complex.So,the two
methods can learn from others's strong points to offset one's weakness and supplement each other,
95
make included angle questions among linear figures of high space that trouble people for a long
time be readily solved.
For the range of application of linear solution and Jianshi solution solution,we have the
following:
Theorem 5:Two linear figures A and B have the same amount of uncommon vectors, when
s0≤1, they are fit for linear solutions,when s0>1, they aren't fit for linear solutions.
Proof:when s0≤1,the amount of nonzero angles between two linear figures is zero or one 1,
of which,when s0=0, the included angle between two linear figures must be zero,so the two linear
figures are parallel or coincident. From theorem 1 and theorem 2 of the sixth chapter,this kind of
questions are already fit for linear solutions.
When s0=1,the amount of nonzero angles between two linear figures is also 1,from theorem
4,both uncommon vectors of two figures are uniquely determined,their included angles (must be
nonzero) are included angles between the two linear figures.In fact, the proof process(belonging
to linear operation process)of theorem 4 has preliminarily solved methods of determining the two
uncommon vectors.
If s0>1,then the amount of nonzero angles of two linear figures is also more than 1.From
structure questions of solutions of system of linear equations we know,the amount of solution
vectors involved in a random system of elementary solutions of equations of A1 and B0 is more than
1. The method of theorem 4 can only determine the amount of solutions in their elementary solution
analysis,but it can't ultimately give respective and uniquely determined solution. That is to say,directions of each uncommon vector can't be uniquely determined by methods of solving system of
linear equations. So,included angle between the two figures A and B aren't fit for linear solution.
▌
Included angle questions aren't fit for linear solution can all be solved by Jianshi solution
solution.
The contents of Jianshi solution solution are too many,we will speciallly introduce them in the
eighth chapter,here we only introduce linear solution.
§§§§2 Linear solution of included angle questions between two linear
figures
Linear solution can be divided into positive angle method and complementary angle method.
2.1 Positive angle method
1°The included angle φ.between two generic planes
F1:a11x1+…+an1xn+a1=0
F2:a12x1+…+an2xn+a2=0
Because direction of normal vector of generic plane in n-dimensional space is uniquely
determined,then included angle questions between two generic planes is generalization of dihedral
angel questions in three-dimensional space. So,we use included angle between normal vectors of
F1,F2 to define included angle between two generic planes.Then
2
2
2
12
2
1
2
11
211211cos
nn
nn
aaaa
aaaa
++++
++=
LL
L
ϕ (1)
this is the cosine formula of included angle φbetween two generic planes F1,F2.
96
(1)If two generic planes are perpendicular,then their normal vectors are also perpendicular.
Then the necessary and sufficient conditions of F1⊥F2 is
a11a12+…+an1an2=0;
(2)If two generic planes are parallel ,then their normal vectors are parallel.So the necessary
and sufficient conditions of F1∥F2 is
2
1
12
11
n
n
a
a
a
a==L .
2°The included angle φ. Between two straight lines
( )
( )1,,2,1 0
1,,2,1 0
1
2
1
1
−==+
−==+
∑∑
=
=
njbxbH
njaxaH
n
i
jiij
n
i
jiij
L
L
:
:
From linear algebra theory,one system of elementary solution of derivative group H1′only
includes one solution,assume the solution is xi=αi(i=1 ,…,n),of which
( )
1,1,11,11,1
11,11,111
11
−−+−−−
+−
+−=
nnninin
nii
i
i
aaaa
aaaa
LL
LLLLLL
LL
λα
(i=1 ,…,n. λ are nonzero numbers,the following is the same),so,the orthodromic vector(also
be called direction vector) of H1 is
1,1,1,1
1111
1
1
−−−
=
nnnin
ni
ni
aaa
aaa
LL
LLLLL
LL
LL eee
H λ
={α1 , … ,αn} (2)
The orthodromic vector H2 = {β1 , … ,βn}
of H2 can be got by the same method,Then
22
1
22
1
11 ||cos
nn
nn
ββαα
βαβαϕ
++++
++=
LL
L
(3)
this is cosine formula of included angle between two straight lines H1, H2.
(1)If two straight lines are perpendicular, then their orthodromic vectors are also
perpendicular. So,the necessary and sufficient conditions of H1⊥H2 is α1β1+…+αnβn = 0 ;
(2)If two straight lines are parallel,then their orthodromic vectors are also parallel.So,the
necessary and sufficient conditions of H1∥H2 is
n
n
β
α
β
α==L
1
1 .
3°The included angle between two intersectant or mutually interleaving planes
97
( )
( )2,,2,1 0
2,,2,1 0
1
2
1
1
−==+
−==+
∑∑
=
=
njbxbI
njaxaI
n
i
jiij
n
i
jiij
L
L
:
:
which have one-dimensional common-orthodromic space is φ.
Apparently,the rank of coefficient matrix of common-orthodromic space
( )
( )
−==+
−==+
∑∑
=
=
221 0
221 0
1
1
n,,,jbxb
n,,,jaxa
Hn
i
jiij
n
i
jiij
L
L
:
is n-1. Assume the equation of common-orthodromic space H⊥
=F is
F:c1x1+…+cnxn=0
Apparently,F is a generic plane.
Because F∩I1=H1 and F∩I2=H2 are separately straight lines
( )
( )
=
−==+
=
−==+
∑∑∑∑
=
=
=
=
0
1,,2,1 0
0
1,,2,1 0
1
12
1
11
n
i
ii
n
i
jiij
n
i
ii
n
i
jiij
xc
njbxb
H
xc
njaxa
H
L
L
:
:
Then,this question is changed into question 2°.The included angle between H1,H2 is
included angle φ.Between two intersectant or mutually interleaving planes I1,I2 .
4°Included angle questions between two n-r+1-dimensional linear figures which have
n-r-dimensional common-orthodromic space.
Assume ( ) ; : 111,,2,1 01
−<<−==+∑=
nrrjaxaAn
i
jiij L
and ( ) ; : 111,,2,1 01
−<<−==+∑=
nrrjbxbBn
i
jiij L
have n-r-dimensional common-orthodromic space
( )11 1,,2,1
0
0
1
1 −<<−=
=
=
∑∑
=
= nrrj
xb
xa
Mn
i
iij
n
i
iij
;: L
98
and find out the equation of common-orthodromic normal space M⊥
is
( )∑=
⊥ −==n
i
iik rnkxcM1
,,2,1 0 L: ,
because intersections of M⊥
and A,B are separately straight lines
( )
( )
( )
( )
−==
−==+
−==
−==+
∑∑∑∑
=
=
=
=
rnkxc
rjbxb
H
rnkxc
rjaxa
H
n
i
iik
n
i
jiij
n
i
iik
n
i
jiij
,,1 0
1,,1 0
,,1 0
1,,1 0
1
12
1
11
L
L
L
L
:
:
So,this question can be changed into question 2°.The included angle between H1,H2 is
included angle between A,B .
Example 1: Please find out the included angle between generic plane
F: 2x1+x2+2x3+x4=0
of four-dimensional system and generic coordinate plane
x4=0 .
Solution::::From formula (1),the cosine of included angle between normal vectors {2,1,2,1}
and {0,0,0,1} of two generic planes is
, 3162.010
1
21212
1101020102cos
22222≈=
+++
×+×+×+×+×=ϕ
so φ≈71°33′54″.
Example 2: It is known that two planes of four-dimensional system
=
=−++
=
=+++
0
06453 and
0
04375
4
321
2
4
321
1
x
xxxI
x
xxxI
:
:
intersect on straight line
=
=−++
=+++
0
06453
04375
4
321
321
x
xxx
xxx
H: ,
find out their included angle φ.
Solution:From formula (2),it gets H={13,-11,4,0},
then the normal space of H is a generic plane
F:13x1-11x2+4x3=0.
Make F intersect I1 , I2 , then it separately gets straight lines
99
=
=+−
=−++
=
=+−
=+++
0
0 41113
0645 3
0
0 41113
0437 5
4
321
321
2
4
321
321
1
x
xxx
xxx
H
x
xxx
xxx
H
:
:
and from formula(2),it separately gets H1={61,19,-146,0}; H2={32,20,-49,0},
then,the cosine of included angle between two planes I1,I2 is
( ) ( )
, 9624.097147350
9486
4920321461961
004914620193261cos
222222≈=
++++
×+−×−+×+×=ϕ
so φ≈15°45′42″.
In three-dimensional space,equations of I1、I2 don't include x4=0,so the normal vector of the
two planes is uniquely determined. In three-dimensional space,we take included angle between
normal vectors of the two planes as included angle between the two planes,the result is identical
with the above result.
Example 3: Find out the included angle between planes
=++++
=++++
=++++
=++++
033423
0332
0122
013
4321
43212
4321
43211
xxxx
xxxxI
xxxx
xxxxI
:
:
of four-dimensional system.
Solution:Because the rank of coefficient matrix of simultaneous group is 3,the rank of
augmented matrix is 4,then I1,I2 are mutually interleaving. And,their common-orthodromic
space is straight line
=+++
=+++
=+++
0 32
02 2
03
4321
4321
4321
xxxx
xxxx
xxxx
H:
(“3x1+2x2+4x3+3x4=0” is already be regarded as unwanted equation,it is deleted from H ).Then
find out the normal space H⊥
=F of H is generic plane F:x1-x4=0 ,
and order F separately intersect I1,I2 at straight lines
=−
=++++
=++++
=−
=++++
=++++
0
033423
0332
0
0122
013
41
4321
4321
2
41
4321
4321
1
xx
xxxx
xxxx
H
xx
xxxx
xxxx
H
:
:
find out H1={1,-5,1,1,}; H2={1,5,-4,1},
100
Then ,the included angle between I1,I2 can be denoted by vectors H1,H2,viz.
, 7781.04328
27cos ≈
−=ϕ
so φ≈38°54′47″.
2.2 Complementary angle method
5°The included angle between straight line
( )1,,1 01
−==+∑=
njaxaHn
i
jiij L:
and generic plane F:b1x1+…+bnxn+b=0
is φ.
Included angle questions between straight lines and generic planes is generalization of those of
three-dimensional space. So, the complementary angle of included angle between normal vectors of
straight line and generic plane is included angle between straight line and generic plane.
Assume from formula (2),we find out orthodromic vector of H is
H={α1,…,αn}
then ( )4 sin22
1
22
1
11
nn
nn
bb
bb
++++
++=
LL
L
αα
ααϕ
this is the sine formula of included angle between straight line H and generic plane F.
(1)if straight line is parallel to generic plane,then the normal vectors of the straight line and
the generic plane are perpendicular. Then the necessary and sufficient conditions of H∥F is α1b1+…+αnbn=0 ;
(2)If straight line is perpendicular to generic plane,then the normal vectors of the straight line
and the generic plane are parallel. Then the necessary and sufficient conditions of H⊥F is
n
n
bb
αα==L
1
1 .
6°Included angle questions between generic palne and a random linear figure.
Assume the common-orthodromic space of a generic plane
F:a1x1+…+anxn+a = 0
and a linear figure(can be regarded as a difficient-order generic plane)
( )11,,1 01
−<<==+∑=
nrrjbxbLn
i
jiij ; : L
is
( )
−<<−==
=
∑∑
=
=
11 1,,2,1 0
0
1
1
nrrjxb
xa
Mn
i
iij
n
i
ii
;
:
L
and the rank of coefficient matrix of M is r+1 , and assume the equation of common-orthodromic
normal space M⊥
to be found out is
101
( )1,,1 01
−−==+∑=
⊥rnkcxcM
n
i
kiik L: ,
From theorem 3 of the sixth chapter ,we know M⊥∩L=H is a straight line
( )
( )
−−==
==+
∑∑
=
=
1,,1 0
,,1 0
1
1
rnkxc
rjbxb
Hn
i
iik
n
i
jiij
L
L
:
Then,included angle questions between generic plane F and difficient-order generic plane L
can be changed into included angle questions between generic plane F and straight line H,viz. the
included angle between F and H is equal to the included angle between F and L. According to
method of the above question 5°,the included angle φ. Between F and L can be aquired.
7°The included angle φ.between straight line ang a random linear figure.
Assume the equations of straight line H and difficient-order generic plane L are separately
( )1,,1 01
−==+∑=
njaxaHn
i
jiij L:
( )11,,1 01
−<<==+∑=
nrrkbxbLn
i
kiik ; : L ,
and assume the equation of normal space H⊥
=F of H to be found out is
F:c1x1+…+cnxn= 0
Apparently,F is a generic plane. Then according to method of question 6, we can first find out
the included angle ϕπ
−2
between F and L ,its complementary angle φ is the included angle
between H and L.
8°Included angle questions between plane
( )2,,1 01
−==+∑=
njaxaIn
i
jiij L:
which has one-dimensional common-orthodromic space and difficient-order generic plane
( )11,,1 01
−<<==+∑=
nrrkbxbLn
i
kiik ; : L .
Apparently,common-orthodromic space is a straight line
( )
( )
−<<==
−==
∑∑
=
=
11,,1 0
2,,1 0
1
1
nrrkxb
njxa
Hn
i
iik
n
i
iij
;
:
L
L
Assume we find out H⊥
=F is generic plane
F:c1x1+…+cnxn= 0 ,
102
because F∩I=H0 is also a straight line
( )
=
−==+
∑∑
=
=
0
2,,1 0
1
1n
i
iik
n
i
jiij
xc
njaxa
H
L
:
then this question is changed into question 7°.The included angle between H0 and L is the
included angle between plane I and difficient-order generic plane.
9°Included angle questions between n-r+1-dimensional linear figure
( )1,,1 01
−==+∑=
rjaxaAn
i
jiij L:
which has n-r-dimensional common-orthodromic space and a random difficient-order generic
plane ( )21,,1 01
−<<==+∑=
nsskbxbLn
i
kiik ; : L .
Apparently,the rank of coefficient matrix of common-orthodromic space
( )
( )
−<<==
−==
∑∑
=
=
21,,1 0
1,,1 0
1
1
nsskxb
rjxa
Mn
i
iik
n
i
iij
;
:
L
L
of A , L is r,assume the equation of M⊥
to be found out is
( )rncxcMn
i
ii −==+∑=
⊥ ,,1 01
Lτττ: .
then because M⊥∩A=H is a straight line
( )
( )
−==
−==+
∑∑
=
=
,1, 0
1,,1 0
1
1
rnxc
rjaxa
Hn
i
iik
n
i
jiij
L
L
τ:
then the question is changed into question 7°.The included angle between H and L is included
angle between A and L.
In fact ,if we cancel the restrict to r,s,then from theorem 4,question 9°can generalize
all contents of linear solution. For example, in 9°, ofder r=2,s=1 ,viz. it is question 1°;Order
r=n, s=n-1 , viz. it is question 2°;order r=n,s=1 ,viz. it is question 5°etc.
Example 4: Find out the included angle φ.Between generic plane
F:3x1-5x2+2x3+x4-2x5-6 = 0
of five-dimensional space and plane
=++−−+
=−−+++
=++−−+
013 25
0233
035432
54321
54321
54321
xxxxx
xxxxx
xxxxx
I:
103
Solution::::Because common-orthodromic space is a straight line
=+−−+
=−+++
=+−−+
=−++−
0325
033
05432
02253
54321
54321
54321
54321
xxxxx
xxxxx
xxxxx
xxxxx
H:
from formula (2) it gets H={73,8,-407,329,-153}
So,the equation of H⊥
=F0 is
F0:73x1+8x2-407x3+329x4-153x5 = 0
make F0 intersect I at straight linez
=
=
=
=
++−−+
−−+++
++−−+
−+−+
0
0
0
0
1325
233
35432
153329407873
54321
54321
54321
54321
0
xxxxx
xxxxx
xxxxx
xxxxx
H :
And from formula (2) ,it gets
H0={15201,-14920,2325,-8201,-17347}
from formula(4) sin 802447.043827257236
151346sin ≈
×=ϕ
then we get the included angle between F and I is φ≈53°21′52″.
Example 5: find out the included angle between straight line
=++−−
=+
=−−
=+
0354352012
0
03453
035
54321
53
421
21
xxxxx
xx
xxx
xx
H:
and plane
=++−−+
=−−+++
=++−−+
01325
0233
035432
54321
54321
54321
xxxxx
xxxxx
xxxxx
I:
of example 4.
Solution::::the direction of H is vector
=
−−
−−=
354352012
10100
034053
00035
2380
1
54321 eeeee
H
={3,-5,2,1,-2},
then H⊥
is the derivative group of equations of generic plane of example 4
F′:3x1-5x2+2x3+x4-2x5 = 0
So,included angle between H and I is the complementary angle of included angle between F and
104
I ,viz. φ≈90°-53°21′52″≈36°38′8″.
Exercises 7.1 Please give voice to the four elements that compose projection in turn. 7.2 Please give voice to contents and the relations of two kinds of projections. 7.3 Please give voice to what's rightangle projection? 7.4 Prove:If a plane of 4-dimensional space ang a generic plane have no common point,then
they must be parallel(clue:use theorem 2 of the sixth chapter). 7.5 Prove:If a plane of 4-dimensional space and a straight line have no common point, then
they are not always parallel(clue:make out another plane through the straight line interleaving with
the known plane,…,use theorem 2 and theorem 4 of the sixth chapter). 7.6 Please find out the included angle φ.Between two generic planes F1:7x1+4x2-5x3+2x4+2x5-x6+3 = 0
F2:4x1-3x2+6x3-5x4-2x5+3x6-5 = 0 7.7 Please find out the included angle φ.Between two straight lines
=++−+
=+−++
=+−−+
=+−++
=−+−+
=++−−
012322
02535
03543
014235
023433
03322
4321
4321
4321
2
4321
4321
4321
1
xxxx
xxxx
xxxx
H
xxxx
xxxx
xxxx
H
:
:
7.8 Please find out the included angle φ.Between two planes
=++−−
=++−−
=−+++
=+−−+
06246
054353
0233
035432
4321
43212
4321
43211
xxxx
xxxxI
xxxx
xxxxI
:
:
7.9 Please find out the included angle φ.Between the plane
=−+++
=+−−+
0233
035432
4321
43211 xxxx
xxxxI :
of exercise 7.8 and the straight line of exercise 7.7
=+−++
=−+−+
=++−−
014235
023433
03322
4321
4321
4321
1
xxxx
xxxx
xxxx
H :
105
The Eighth Chapter: Jianshi solution solution of included
angle questions between twolinear figures
Jianshi solution solution is the nonlinear solution referred to in the seventh chapter.
According to graphic principle, Mr Jian Zhaoquan of Beijing Institute of Technology use unit
circle method to solve included angle questions [13]
between two planes of n-dimensional space.
This is a breakthrough of great significance in studying included angle questions between two
linear figures of higher space. Now,we introduce this method(also be called Jianshi solution
solution) by principle of analytical geometry,and extend this method to included angle questions
between two random linear figures.
§§§§1 Orthogonal transform and principal axis questions
1.1 Orthogonal matrix and orthogonal transform We know,assume Q is n-order square matrix of real number field,if
Q′=Q-1 ,
viz. inverse matrix of Q is equal to its transposed matrix,then Q is called a orthogonal matrix. If the
matrix of linear transform σis a orthogonal matrix,then σ is called a orthogonal transform. Orthogonal transform can be divided into orthogonal-unit vector transform and coordinate
transform of vectors(sometimes it includes coordinate transform of points). Of which,when assume
e1,…,en are a group of orthogonal-unit vectors of Vn,and α={x1,…,xn} are nonzero vectors of Vn,then
′
′
=
=
nnn e
e
e
e
Q
e
e
MMM
111
σ (1)
(Q is a orthogonal matrix) is called orthogonal transform of orthogonal-unit vector e1 ,… ,en to
e1′,…,en′, when keep α motionless and only rotate coordinate system
=
nn x
x
Q
y
y
MM
11
: σ (2)
is called orthogonal transform of coordinates of vectorα,it is also called orthogonal linear
transform of x1,…,xn to y1,…,yn.
The second kind of orthogonal transform is used the most in this chapter,it is mainly used for
some linear figures to be changed from general position to particular position.But this time we
always use its inverse transform,so,we write inverse transform of formula (2) underneath:
′=
−
nn y
y
Q
x
x
MM
111: σ (3)
The core of orthogonal transform is orthogonal matrix,but for constructing orthogonal
106
matrix,we usually choose a group of linearly independent vectors and orthogonalize them,and then
extend them to orthogonal basis of Vn , last unitize them. This process is called
orthogonal-unitization of vectors.
1.2 1.2 1.2 1.2 Concepts and property of exterior product among vectors————
orthogonal-unitization of vectors
Assume α1,…,αm (m≤n) are linearly independent,then there are two approaches of
separately find out a group of orthogonal vectors β1,…,βm .
The first approach is called Schmidt orthogonal process(Schmidt),the second is called exterior
product method(accordingly,the first can be called inner product method).
Let's introduce the first approach.
Order β1=α1,
and assume β2=α2+kβ1,
in order to determine k, makeβ1 intersectβ2,and it need to make inner product
(β1,β2)=(β1,(α2+kβ1)=(β1,α2)+k(β1,β1)=0.
becauseβ1 ≠0, (β1,β1)≠0,then only choose
( )( )11
21
,
,
ββ
αβ−=k
it can get ( )( )11
2122
,
,
ββ
αβαβ −= ,
similarly, it can get ( )( )
( )( ) 2
22
32
1
11
31
33,
,
,
,β
ββ
αββ
ββ
αβαβ −−= , … … … …,
( )( )
( )( ) 1
11
1
1
11
1
,
,
,
,−
−−
−−−−= m
mm
mmm
mm βββ
αββ
ββ
αβαβ L , β1,…,βm are not only intersect with each other,but also can be mutually linearly
represented withα1,…,αm.
Now let's introduce exterior product method.
Defination: assume αi={αi1 ,αi2 , … ,αin }, i=1 ,… ,n .
Then the exterior product of n-1 linearly independent vectorsα1,…,αn-1 can be represented as
,12,11,1
22221
11211
21
121 n
nnnn
n
n
n
n λα
ααα
αααααα
ααα ==×××
−−−
−
L
LLLL
L
L
L
L
eee
(4)
in the formula,λis a nonzero number,λαn is called the exterior product
of α1,…,αn-1 ,and α1,…,αn-1 are all called multiplication factors of αn.
Theorem 1:Exterior productαn separately intersect its multiplication factorsα1,…,αn-1 .
107
Proof::::because the coordinates of αn={αn1,…, αnn}
are algebraic cofactors of the first row's elements of determinant (it is also called exterior product
determinant )of formula(4),viz.
( )
nnjnjnnn
njj
njj
j
jn
,11,11,12,11,1
21,21,22221
11,11,11211
1
11
−+−−−−−
+−
+−
+−=
ααααα
αααααααααα
λα
LL
LLLLLLL
LL
LL
(j=1,…,n ),and the inner product of a random vectorαi(i=1,…,n-1)αn ofα1,…,αn-1 can
be represented as
( ) 0
1,
,12,11,1
................................................................21
11211
...............................................................21
2211 ==+++=
−−− nnnn
inii
n
inii
nninninini
ααα
ααα
ααα
ααα
λαααααααα
L
LLLL
L
LLLL
L
L
L
(i=1 ,… ,n-1)viz. αn and αi are orthogonal. ▌
It should be pointed out that,direct addition and multiplication operation are unsuitable for exterior product determinants,viz. when both determinants |A| and |B| of n×n matrixes A and B are
exterior product determinants,it has
|A|+|B|≠|A+B| and |A|·|B|≠|A·B|
So,when more than two exterior product determinants add or multiply,it should first carry
internal arithmetic of exterior product determinants ,then add or multiply.
As will be readily seen,formula (4) is generalization of defination of exterior product between
two vectors in three-dimensional space. In the past,when determine normal vector (refer to it in example 4 of §2 in the third chapter)of
generic plane F:3x1+2x2+x3+3x4+5x5+x6+6=0
and orthodromic vector (formula (2) of §2 in the seventh chapter)of straight line,we have used
this kind of exterior product of vectors.
Now let's discuss the process of orthogonalizing linearly independent vectors α1,…,αm (2≤m≤n-1)
by exterior product method.
Because αi={αi1 ,αi2 , … ,αin }( i=1 ,… ,m) ,
then when 0
21
22221
11211
≠
mmmm
m
m
ααα
αααααα
L
LLLL
L
L
we can assume α1×…×αm×em+1×em+2×…×en-1=
108
( ) 11
1
1111
1
1
1,1,1
11,11,1111
1111
1
1
1
++
−−
−+
−+
−+−
=−=
==
mm
mnmmm
nm
nm
mn
mnnmmmmmm
nnmm
nnmm
βλ
ααα
ααα
ααααα
ααααα
L
LLLL
L
L
LLLLLL
LLOLLLL
LLLLLL
LL
LLLLLLL
LL
LL
eee
eeeee
(both λm+1 and the followingλ1,…,λm, λm+2,…,λn are nonzero numbers). Assume αi∈Vm( i=1 ,… ,m),Vm ⊂ Vn,
then βm+1∈ ⊥mV .
When separately regard α1,…,αm
as normal vectors of m generic planes,the intersection of orthodromic spaces of the m generic
planes is ⊥
mV . So,βm+1 is a common-orthodromic vector of the m generic planes. Assume βk={β1k ,…,βnk }k=m+1,m+2,…,n
in turn.Then we can separately find out all the orthodromic vectors which is perpendicular to each
other λm+2βm+2=α1×…×αm×βm+1×em+2×…×en-1 , λm+3βm+3=α1×…×αm×βm+1×βm+2×em+3×…×en-1 , … … … … … … , λnβn =α1×…×αm×βm+1×…×βn-1 ,
of ⊥
mV Separately substitute βn ,β1,β2,…,βm-1 for e1,…,em in turn,then λ1β1=βn×e2×e3×…×em×βm+1×…×βn-1, λ2β2=βn×β1×e3×…×em×βm+1×…×βn-1, … … … … … … , λmβm =βn×β1×β2×…×βm-1×βm+1×…×βn-1
can be separately found out,When 0
21
22221
11211
=
mmmm
m
m
ααα
αααααα
L
LLLL
L
L
,
order λm+1βm+1=e1×e2×…×en-m-1×α1×…×αm, … … … … … … , λmβm =βm+1×…×βn-1×βn×β1×β2×…×βm-,
then allβ1,β2,…,βm can be found out in turn.
It can be verified that,β1,β2,…,βn are not only orthogonal with each other,but also β1,β2,…,βm can be linearly represented with:α1,…,αm (butβ1,β2,…,βm here
109
are not required to be consistent with the group of orthogonal vectors aquired by inner product
method).
Last,use formula nii
i
i 1 ,,, L==β
βε
and separately unitize β1,β2,…,βn,
then a group of orthogonal-unit vectorsε1,ε2,…,εn are acquired.
As will be readily seen,exterior product method also offer another method to solve system of
homogeneous linear equations. In one of system of elementary solutions of system of
homogeneous linear equations acquired by this method,when the amount of solution vectors is
more than one,every solution vector is orthogonal with each other.
Example 1: Please change the two planes
=++++
=++++
0222
0322
4321
4321
1xxxx
xxxxI :
=+−+
=−−+
013
04322
421
3212 xxx
xxxI :
of four-dimensional space from general position to particular position.
Solution:::: translate I1,I2 to the origin,it gets
=+++
=+++′
022
022
4321
43211 xxxx
xxxxI :
=−+
=−+′
03
0322
421
3212 xxx
xxxI :
first find out two solution vectors
{ } { }3,0,1,13,0,1,1
121
112
0100
1221
12123 =−−−=−==
eeeeeee
β ,
{ }4,11,6,6
3011
1221
12124 −=
−
=
eeee
β
of I1′, order λ1β1=β4×e2×β3= -β4×β3×e2=
{ }1,2,0,311
0010
3011
41166=
−
−−−=
eeee
(λ1=11) and order -11β2=β4×β1×β3=
110
{ }. 3,6,14,511
3011
1203
41166−−−−=
−
−=
eeee
assume
{ }
{ }
{ }
{ }4,11,6,6209
1
3,0,1,111
1
3,6,14,5266
1
1,2,0,314
1
4
44
3
33
2
22
1
11
−==′
−==′
−−−==′
==′
β
β
β
β
β
β
β
β
e
e
e
e
then it gets orthogonal matrix
−
−
−−−
=
209
4
209
11
209
6
209
611
30
11
1
11
1266
3
266
6
266
14
266
514
1
14
20
14
3
Q
use fourmula (3),it gets
+−−=
−−=
++−=
+++=
−
43214
4213
4322
43211
1
209
4
11
3
266
3
14
1209
11
266
6
14
2209
6
11
1
266
14209
6
11
1
266
5
14
3
yyyyx
yyyx
yyyx
yyyyx
: σ
chooseβ1,β2 as new normal vector of I1′,then it gets
=−−−
=++′
036145
023
4321
4311 xxxx
xxxI :
and two normal vectors of I2′are orthogonal with two normal vectors of I1′,then I2′is also
changed into
=+−+
=−+′
041166
03
4321
4212 xxxx
xxxI :
111
substitute the above transforms into new equations of I1′,I2′,it separately gets
=
=′′
=
=′′
0
0
0
0
4
3
2
2
1
1
y
yI
y
yI
:
:
this is the equations when they are in particular position.
Under most cases,we can only change one of two linear figures from general position to
particular position,and the other is still on general position after orthogonal transformation.
Because orthogonal transformation keep the inner product changeless,then the included angle
betweeen two figures is changeless.
1.3 Principal axis questions [14]
Principal axis questions can be generalized in a ward,that is change quadratic form to
principal axis .
So-called change quadratic form to principal axis is to find out a orthogonal transformation(3)
for the quadratic form
( ) ( )
=
n
nn
x
x
Axxxxf MLL
1
11 ,, (5)
given in real number field. Of which,A=(aij )n×n a real symmetric matrix,and substitute
formula(3) and its transpose into formula(5),make
( )
′
n
n
y
y
QQAyy ML
1
1 (6)
has the simplest form.
22
22
2
11 nn yyy λλλ +++ L (7)
In the first chapter we have introduced concepts of characteristic root and characteristic vector
of matrix A in linear algebra theory,but the contents it introduced are not integral ,so give
complementary introduction as the following:
Characteristic roots of real symmetric matrix are all real numbers.
If ξ,η are characteristic vectors correspondent with two different characteristic roots of
real symmetric matrix A,thenξ,ηare orthogonal.
If matrix B=QAQ′,
of which Q is orthogonal matrix,A and B are real symmetric matrix,then we call A and B
contragradient similar. If A and B are contragradient similar,characteristic polynomials of A and B
are the same,then A and B have the same characteristic roots. For a random real symmetric matrix
A,there must be a orthogonal matrix Q making QAQ′be a diagonal matrix,viz.
=′
n
QQA
λ
λO
1
(8)
112
So in formula(3),formula(6),as long as find out proper orthogonal matrix Q and change A into
diagonal matrix,then quadratic form(5) is changed into the form of quadratic sum of formula(7).
But,finding out orthogonal matrix must first find out all characteristic roots of A,then we can
find out all characteristic vectors belonging to different characteristic roots ,last unitize them,it is
very troublesome to do this.In fact,all characteristic roots (include multiple roots)of A are just
elements of principal diagonal of diagonal matrix in formula(8).So,as long as all characteristic
roots(include multiple roots) of A are found out ,and be used as coefficients of corresponding
variables of formula(7),this solved questions of changing quadratic form into quadratic sum and
relieved the trouble of finding out orthogonal matrix.
Using elementary transform,change characteristic matrix AI −λ of real symmetric matrix A
into diagonal matrix,elements of principal diagonal are called invariant divisors of AI −λ . Divide
invariant divisors of AI −λ into different products of one-order factor powers of λ,all one-order
factor powers (the same are counted by times they appeared ) are called elementary divisor of
AI −λ .
By elementary divisors of AI −λ , we can find out all characteristic roots(include multiple
roots) of A.
Example 2: change quadratic form
1484525 323121
2
3
2
2
2
1 =+++++ xxxxxxxxx
into the form of quadratic sum (it is also called normalized form).
Solution::::the characteristic matrix of the matrix
=
524
222
425
A
of the quadratic form is
−−−
−−−
−−−
=−
524
222
425
λ
λ
λ
λ AI
make elementary transform
( ) ( )( ) ( )( )
( ) ( )( )( )
( ) ( )( )( ) ( )( )
−−
−⇒
−−
−−−⇒
−−
−−−
−−−
⇒
−−−−
−−−
−−−
⇒
−−−
−−−
−−−
⇒
−−−
−−−
−−−
10100
010
001
10100
1120
001
10100
1120
524
91120
1120
524
425
222
524
524
222
425
λλ
λ
λλ
λλ
λλ
λλ
λ
λλλ
λλ
λ
λ
λ
λ
λ
λ
λ
onλΙ-A so,the three characteristic roots of A are separately 1,1,10,from formula(6) and
formula (8), it gets .110 2
3
2
2
2
1 =++ yyy
113
§§§§2 Projective generic elliptic cylinder and projection of generic circle
2.1 Projection of sufficient-order generic circle
In example 5 of the third chapter,we have introduced generic cylindrical surface
22
3
2
2
2
1 Rxxx =++
the generic cylindrical surface is also called projective generic cylindrical surface. When a generic
sphere ( ) 22
404
2
3
2
2
2
1 Rxxxxx =−+++
intersect generic plane x4=x40 ,the cut trace is a generic circle.What about the projection of the
generic circle on generic coordinate plane x4=0 ? Substitute x4=x40 into equation of the generic
sphere,then a generic cylindrical-surface equation through the generic circle is acquired,its
generating line is perpendicular to the generic coordinate plane,and make it simultaneous with
generic coordinate plane,then
=
=++
04
22
3
2
2
2
1
x
Rxxx
is the proetion equation of the generic circle on the generic coordinate plane.
When generic plane which through generic centre of sphere and intersect generic sphere is on
general position,the cut trace is a generic circle. The following is an example of finding out
projection of the generic circle on generic coordinate plane.
Example 3: Assume center of a generic phere of four-dimensional space is on the origin,its
radius is 1. A generic plane
F:2x1+x2+2x3+x4=0
intersect the generic sphere at generic circle
=+++
=+++
022
1
4321
2
4
2
3
2
2
2
1
xxxx
xxxxK:
Please find out the projection of the generic circle with relation to generic coordinate plane x4=0.
Solution:::: Apparently,the generic circle is on an oblique position with relation to generic
coordinate plane,so,its projection on the generic coordinate plane is a generic ellipse. First find out
equation of projective generic elliptic cylinder (in the following it is called projective generic
cylinder surface by a joint name or be called generic cylinder surface for short)which includes the
generic circle and the generating line is perpendicular to generic coordinate plane.And make the
equation of the projective generic cylinder surface with the equation of generic coordinate plane,viz. projective equation of the generic circle on generic coordinate plane is acquired.
Equation of projective generic cylinder surface can be determined in this way:arrange F and
transpose it,then it gets F0:x4= -2x1-x2-2x3 , apparently, F0 and F have the same solutions,viz. F=F0. Sustitute F0 into the equation of the
generic sphere,it gets
1484525 323121
2
3
2
2
2
1 =+++++ xxxxxxxxx
Apparently,figures denoted by the quadratic form equation include the generic circle K,or
generic circle K is on the figures denoted by the the quadratic form . We also observe that,the
114
quadratic form doesn't include variable x4,so,the quadratic form is equation of projective generic
cylinder surface which includes generic circle K and its generating line is perpendicular to generic
coordinate plane x4=0 .
figure 45
Make equation of the projective generic cylinder surface simultaneous with equation x4=0 of
generic coordinate plane,it gets
=
=+++++′
0
1484525
4
323121
2
3
2
2
2
1
x
xxxxxxxxxK :
this is the projective equation of generic circle K to be found out on generic coordinate plane(refer
to it in figure 45,the four-dimensional system in figure is generated by transform 2e3 □ e1+2e2,the ratio of directviewing unit length of every axis is |σ(e4)|∶|σ(e1)|∶|σ(e2)|=1∶1∶1).
Usually,for random generic circle
=+++
=+++
0
1
2211
22
2
2
1
nn
n
xaxaxa
xxK
L
L
:
its generic plane equation can be arranged as
( ) ( )0 1
1111 ≠++−= −− nnn
n
n axaxaa
x L
substitute it into generic sphere equation of K, then it gets equation of projective generic cylinder
surface
( ) 12
111
121211112121
2
12
2
12
22
2
22
12
2
1
=+++++
+
+++
++
+
−−−−−−
−−
nnnnnn
n
n
n
n
nn
xxaaxxaaxxaaa
xa
ax
a
ax
a
a
LL
L
or ∑ ∑∑−
=
−
=
−
= ≠
=+
+1
1
1
1
1
1
2
2
2
, 12
1n
i
n
i
n
j ji
jiji
n
i
n
i xxaaa
xa
a
115
and make it simultaneous with xn=0 ,then it gets projective equation of the generic circle K on
generic coordinate plane
=
=+ +′ ∑ ∑ ∑−
=
−
=
−
= ≠
0
, 12
11
1
1
1
1
1
2
2
2
n
n
i
n
i
n
j ji
jiji
n
i
n
i
x
xxaaa
xa
a
K :
(an≠0).
2.2 Projection of difficient-ordergeneric circle
Difficient-order generic circle is the cut trace got by difficient-order generic plane cutting
generic sphere.
Example 4: Assume the cut trace of generic sphere
12
6
2
5
2
4
2
3
2
2
2
1 =+++++ xxxxxx
of six-dimensional space cut by a difficient-order generic plane
=
=
−−−++
−+−−+
0
0
2223
22
654321
654321
xxxxxx
xxxxxx
is a two-order generic circle
=
=
−−−++
−+−−+
=+++++
0
0
2223
22
1
654321
654321
2
6
2
5
2
4
2
3
2
2
2
1
xxxxxx
xxxxxx
xxxxxx
L:
Please find out projection of L on difficient-order generic coordinate plane
=
=
0
0
6
5
x
xI: .
Solution::::From the question,equation of difficient-order projective generic cylinder surface
found out should not include variable x5,x6,but it should include the given difficient-order generic
circle L. Regard the front four terms of every equation of difficient-order generic plane which L is
onas constant,then,the posterior two variables can be found out according to Cram's rule. Because
the determinant of coefficient of the posterior two variables is
, 411
22−=
−−
−=D
( )( )
( )( )
4321
4321
4321
4321
4321
4321
535-7x
22231
2
3535
12223
2 and
6
5
xxx
xxxx
xxxxD
xxxx
xxxx
xxxxD
x
x
+−−=
−++−−
−−+−=
+−−−=
−−++−
−−−+−=
then it gets
−++=
−++=
43216
43215
4
5
4
3
4
5
4
74
3
4
5
4
3
4
5
xxxxx
xxxxx
116
apparently,the new equation has the same solution with equation of difficient-order generic plane.
Substitute it into equation of generic sphere and make it simultaneous with equation of I,then
projective equation of L on difficient-order generic coordinate plane is
=
=
=−−+−
−+++++
′
0
0
830343050
465025252545
6
5
43423241
3121
2
4
2
3
2
2
2
1
x
x
xxxxxxxx
xxxxxxxx
L :
Usually,for random difficient-order generic circle
( )
−≤≤==
=+++
∑=
n
i
iij
n
nrrjxa
xxx
1
22
2
2
1
22,,1 0
1
; L
L
(difficient-order generic plane∑=
=n
i
iij xa1
0 doesn't include unwanted equation) , when the
determinant of coefficient 0
,2,1
22,22,1
11,21,1
≠=
+−+−
+−+−
+−+−
nrrrnrrn
nrnrn
nrnrn
aaa
aaa
aaa
D
L
LLLL
L
L
of the posterior r variables of difficient-order generic plane
( )∑=
−≤≤==n
i
iij nrrjxaL1
1 22,,1 0 ; : L
we find out
nrrkrn
rn
i
iirrkenrrn
njjkrn
rn
i
iijjkrnjrn
nkrn
rn
i
iikrnrn
k
aaxaaa
aaxaaa
aaxaaa
D
LL
LLLLLLL
LL
LLLLLLL
LL
,1
1
,1,1
,1
1
,1,1
11,1
1
11,11,1
++−
−
=−+−+−
++−
−
=−+−+−
++−
−
=−+−+−
∑
∑
∑
−
−
−
=
(j,k=1 ,… ,r;2≤r≤n-2),then it gets
,, ,
D
Dx
D
Dx
D
Dx r
nrnrn === +−+− L2
21
1
or
+++=
+++=
+++=
−−
−−+−
−−+−
rnrrnrrn
rnrnrn
rnrnrn
xdxdxdx
xdxdxdx
xdxdxdx
L
,2211
2,2221122
1,2211111
10
L
LLLLL
L
L
: (9)
117
in the formula ( )∑=
+−−=
r
j
ijjk
kj
ik aAD
d1
11
(i=1 ,… ,n-r;k=1 ,… , r),of which,Ajk is algebraic cofactor of elements of the j-th row、the k-th column of determinant D (1≤j ,k≤r),or
D
Dd ik
ik = ,
of which
nrrkrnirrkrnrrn
nkrnikrnrn
ik
aaaaa
aaaaa
D
LL
LLLMLLL
LL
,1,1,1
11,111,11,1
++−−+−+−
++−−+−+−
−
−
=
(i=1 ,… ,n-r;k=1 ,… , r),apparently ,L10 and L1 have the same solutions.
Substitute formula(9) into equation of generic sphere
122
2
2
1 =+++ nxxx L
it gets , 11 1 1 1
2 =+∑ ∑∑∑−
=
−
=
−
= =
rn
i
rn
i
rn
j
r
k
jijkiki xxddx
Pick up i=j terms of the formula and separately incorporate them with every terms of ∑−=
rn
i
ix1
2then it
gets 111 1 1 1
2
1
2 =+
+∑ ∑∑∑∑−
=
−
=
−
= = ≠=
rn
i
rn
i
rn
j
r
k ji
jijkiki
r
k
ik xxddxd (10)
From formula(10),equation of projective generic cylinder surface doesn't include variable
xn-r+1 , xn-r+2 , … , xn ,
So,the generic generating line of the generic cylinder surface is perpendicular to generic plane
xn-r+1=0 , xn-r+2=0 , … , xn=0 ,
So,projection of the difficient-order generic circle on difficient-order generic coordinate plane
=
=
=
+−
+−
0
0
0
2
1
2
n
rn
rn
x
x
x
LL
:
is system of equations acquired by simultaneousness of formula(10) and equation of L2
=
=
=
=+
+
+−
+−
−
=
−
=
−
= = ≠=
∑ ∑∑∑∑
0
0
0
11
2
1
1 1 1 1
2
1
2
n
rn
rn
rn
i
rn
i
rn
j
r
k ji
jijkiki
r
k
ik
x
x
x
xxddxd
LL
(11)
Before the operation of the above projection,we should first examine dimensions of the two
linear figures L1, L2 are equal or not. If dimensions of the two linear figures are not equal,then
uncertainty of included angles in operation of solving included angles will appear,such solution
results are nonsensical. Now,we can adopt method of theorem 4 of the seventh chapter,first make
the dimensions of the two figues identical .
118
When determinant of coefficient of posterior r terms of L1 is D=0,we can choose normal
space ⊥1L of L1 and make it intersect generic sphere,find out projection of the cut difficient-order
generic circle on L2 according to the above method,and then solve included angle questions
between L1 and L2 by complementary angle method.But,the dimension of L2 is not equal to the
dimension of ⊥1L ,then we should find out projection of difficient-order generic circle of
⊥1L
on ⊥2L ,then the question become using included angle between normal spaces
⊥1L , ⊥
2L of L1,L2 to define included angle between L1,L2.
§§§§3 Jianshi solution solution of included angle questions between two
linear figures
3.1 Principle and steps of Jianshi solution solution
From Jianshi solution theorem and generalized rightangle projection theorem(refer to it in the
seventh chapter),m orthogonal-unit vectors of Vm can be projected into right angles in another Vmo,but,the m orthogonal-unit vectors satisfying the above conditions are not always easy to uniquely
etermined. So,when Jianshi solution study included angle questions between two planes,he first
make a circle whose radius is 1 on one of the plane,it is called unit circle,its projection on the
other plane is usually a ellipse(sometimes it is a circle,sometimes it is a line segment,sometimes it
is even a point). The unit circle always has such two radii perpendicular to each other,their
projections on another plane are not only perpendicular to each other ,but also uniquely become
semi-major axis and semi-minor axis (both are principal axes)of the ellipse. Then,of two random
radii perpendicular to each other of the unit circle,only one of projections of the two radii is the
longest,the other is the shortest.Jianshi solution use the length of the two semi-major axis and
semi-minor axis as cosine of the included angle between two planes.This principle is also called
Jianshi solution principle.
Jianshi solution principle can be made the following generalization::::the projection of
difficient-order unit circle denoted by quadratic equation with s unknown of a s-dimensional
linear figure of n-dimensional space on another s-dimensional linear figure is usually a
difficient-order generic ellipse. The difficient-order generic ellipse altogether has s radii
perpendicular to each other,their projections on another linear figure are not only perpendicular to
each other but also uniquely become s semiaxes (all are principal axes)of the difficient-order
generic ellipse.
According to the above principle,Jianshi solution method can be divided into the following
steps:
First,examine two linear figures through the origin or not and their dimensions are the same
or not.If they are not through the origin and their dimensions are not the same,then translate them
to the origin ang make their dimensions the same.
Second,assume equations of the two linear figures has already separately been system of
homogeneous linear equations
∑∑
=
=
=
=
n
i
iij
n
i
iij
xbL
xaL
1
2
1
1
0
0
:
:
( j =1 ,… ,r )
119
(the ranks of coefficient matrixes of L1,L2 are r),find out a orthogonal transformation σby
method of the first section of this chapter,and separately change L1,L2 into
0
0
2
1
1
=′
=′′
+−
=
∑jrn
n
i
iij
yL
yaL
:
:
( j =1 ,… ,r )
Third,arrang equation of L1′into the form of formula (9),substitute it into equation
122
2
2
1 =+++ nyyy L
of unit generic sphere,equation(10) of projective generic cylinder surface which included
difficient-order generic circle
( )
==′
=++
∑=
n
i
iij
n
rjya
yy
1
2
2
2
1
,,1 0
1
L
L
and its generic generating line is perpendicular to difficient-order generic coordinate plane L2′is
acquired,and make it simultaneous with equation of L2′,then equation(11) of difficient-order
generic circle is acquired.
Fourth,find out a real symmetric matrix A from formula(11) ,and find out all characteristic
roots λ1 ,λ2,…,λn-r (includemultiple roots) of A.
Fifth,change (10) into the form of quadratic sum
122
22
2
11 =+++ −− rnrn zzz λλλ L
from formula(6) and formula(8).
Sixth,it is known by deduction from ellipse questions,halves of every principal axis length of
generic ellipse separately are
,1
,,1
,1
21 rn−λλλL
this is cosines of all angles φ1, φ2,…,φn-r between L1,L2,viz.
,1
cos
,1
cos
,1
cos
2
2
1
1
rn
rn
−
− =
=
=
λϕ
λϕ
λϕ
LLL
Whenλ1 ,λ2,…,λn-r are not equal to 1 at the same time,chooseλi≠1 (1≤i≤n-r) viz.
angles of φi≠0 to define included angles between L1,L2.
3.2 Included angle questions between two planes
According to the priciple and steps of Jianshi solution method,assume two planes I1,I2 are
separately changed into
120
( )
( )nkyI
njyaI
k
n
i
iij
,,4,3 0
2,,1 0
2
1
1
L
L
==′
−==′ ∑=
:
:
after translation and rotation.make generic sphere (unit generic sphere)whose centre on the origin
and the radius is 1 intersect I1′, it gets
( )
−==
=+++
∑=
n
i
iij
n
njya
yyy
L
1
2
2
2
2
2
1
2,,1 0
1
L
L
:
then the question become question of finding out projection of unit circle. assume projection of unit
circle Lon I2′is ellipse
( )
==
=++′
nky
NyyPyMyL
k L,4,3 0
12
221
2
1:
it can be changed into normalized form (viz. the form of quadratic sum) after a orthogonal
transformation
( )
==
=+′′
nkz
BzAzL
k ,,4,3 0
12
2
2
1
L
:
From ellipse questions,we know lengths of its semi-major axis and semi-minor axis are separately
BA
1 and
1,
then BA
1cos ,
1cos 21 == ϕϕ
this is cosine of included angle to be found out between I1,I2.
For the convenience of the latter operation,we give the following:
Theorem 2:(also be called coefficient theorem):
there is the following relation
( )
( )
−+++=
−+−+=
22
22
2
1
2
1
MNPNMB
MNPNMA
(12)
between quadratic form ( ) 1, 2
221
2
121 =++= NxxPxMxxxf
and its normalized form ( ) 1, 2
2
2
121 =+= ByAyyyf .
Proof:::: Because the real symmetric matrix of f(x1 , x2 ) is
NP
PM
2
2
assume its two characteristic roots are separately A and B ,viz. it has
121
( )
−+±+=2
2
2
1, MNPNMBA ▌
In order to save length of writing,in the following examples both the two planes are assumed
through the origin ,and one plane has become horizontal coordinate plane x1ox2.
Example 5: Please find out included angles φ1,φ2 between two planes of four-dimensional
space
=+++
=+++
022
022
4321
43211 xxxx
xxxxI :
=
=
0
0
4
3
2x
xI :
Solution::::because the determinants of coefficient of the posterior two variables of I1 are
zero,then the relation between x3,x4 and x1,x2 can't be uniquely determined.But,the included
angle between normal space ⊥
1I of plane I1 and I2 is the cosine of the included angle between I1
and I2. So,the cosine of included angel between ⊥
1I and I2 is the sine of included angle between I1
and I2.
Choose two orthodromic vectors {2,2,-3,0}and {1,1,0,-3}of I1 as normal vector
of ⊥
1I ,then it gets
=−+
=−+⊥
03
0322
421
3211 xxx
xxxI :
because the determinant of coefficient of two posterior terms is
930
03=
−
−=D
( )( )
( )( ) 21
21
21
2
21
21
21
1
330
223
663
022 but
xxxx
xxD
xxxx
xxD
+=+−
+−−=
+=−+−
+−=
then substitute . 3
1
3
1 ,
3
2
3
2214213 xxxxxx +=+=
into the equation 12
4
2
3
2
2
2
1 =+++ xxxx
of unit generic sphere,and make it simultaneous with I2,then the projection is a ellipse
=
=
=++
0
0
9141014
4
3
2
221
2
1
x
x
xxxx
From coefficient theorem〔viz. formula(12)〕it gets
( )
19514
951414141014142
1 22
=+=
=−= −+−+=
B
A
so,the equation of projection is changed into
122
=
=
=+
0
0
19
19
4
3
2
2
2
1
y
y
yy
so , 6882.019
3sin , 1sin 21 ≈== ϕϕ
then φ1=90°, φ2≈43°29′
viz. the maximum included angle between plane I1 and I2 is 90°,the minimum included angle is
about 43°29′.
Example 6: Please find out the included angles φ1,φ2 between two planes
=
=
=+++
=+++
0
0
022
022
4
3
4321
4321
1
x
xI
xxxx
xxxxI
:
:
of four-dimensional space.
Solution:::: the determinant of coefficient of two posterior terms of I1 is
, 321
12==D
and ( )( )
( )( ) , 3
21
22 , 3
22
122
21
2121
21
211 x
xx
xxDx
xx
xxD −=
+−
+−=−=
+−
+−=
then substitute x3= - x1 , x4= - x2
into equation of generic sphere and make it simultaneous with I2
=
=
=+
0
0
122
4
3
2
2
2
1
x
x
xx
of which,the generic cylinder surface has been normalized form,so
, 2
1coscos 21 == ϕϕ φ1=φ2=45°
Example 7: In example 1,I1,I2 are separately changed itno
=
=′
=
=′
0
0
0
0
4
32
2
11
y
yI
y
yI
:
:
please find out their included angle φ1,φ2.
solution::::Because ⊥′
1I =I2′,then substitute the equation of I2′into the equation of generic
sphere and make them simultaneous according to method of example 5,then it gets that the
projection of unit circle on its personal plane is also a unit circle
123
=
=
=+
0
0
1
4
3
2
2
2
1
y
y
yy
so sinφ1= sinφ2=1;φ1=φ2=90°
In the above examples,we all writen out the projective equation of unit circle on another plane
(or the plane of its own). In fact,as long as directly using formula(12) to change equation of
projective generic cylinder surface into normalized form,then questions will be solved.So in the
following discussions,under most cases,we can only consider transform of equation of projective
generic cylinder surface,and needn't write out the simultaneous equations of them intersecting a
plane.
Example 8: Please find out the included angle between two planes of five-dimensional space
=+−+
=−++
=−++
022
02_3
023
5431
54321
5421
1
xxxx
xxxxx
xxxx
I :
=
=
=
0
0
0
5
4
3
2
x
x
x
I :
solution::::determinant of coefficients of three posterior terms of I1 is
, 1
122
111
230
−=
−
−
−
=D
( )( )
( )( )
( )( ) , 16x8
222
2311
30
, 115
122
1231
20
, 3
122
1123
23
but
21
1
21
21
3
21
1
21
21
2
2
1
21
21
1
+−=
−−
−−
+−
=
+−=
−
−−−
−+−
=
=
−−
−−−
−+−
=
x
x
xx
xx
D
xx
x
xx
xx
D
x
x
xx
xx
D
then substitute x3= -3x2 ,x4=5x1-11x2,x5=8x1-16x2,
into equation of generic sphere,it gets
138736690 2
221
2
1 =+− xxxx
from formula(12)
124
( )
( ) 1719.4742221654772
1
8281.22221654772
1
≈+=
≈−=
B
A
then it approximately gets
, 0459.01719.474
1cos , 5946.0
8281.2
1cos so
11719.4748281.2
21
2
2
2
1
≈≈≈≈
=+
ϕϕ
xx
then φ1≈53°31′, φ2≈87°22′.
From example 2,example 3 of the seventh chapter and examples of this chapter,It is observed
that ,included angles betweeen two planes can be divided into the following three kinds:
(1) quasi-dihedral angle kind:two planes either mutually interleaving or intersect at a straight
line,their maximum angle is 0°<φ1≤90°,the minimum angle isφ2=0°.This kind of
questions are equal to dihedral angle questions (include dihedral angle questions)in
three-dimensional space. Solving this kind of questions always adopt linear solution,sometimes
adopt Jianshi solution solution. At the same time,we only use their maximum included angleφ1 to
define included angles between them.
(2) equal-oblique kind:the maximum angle is equal to the minimum angle,viz. φ1=φ2 ,and 0°≤φ1,φ2≤90°. Of which,whenφ1=φ2=0°,two planes are parallel or coincident(fit
for linear solution);whenφ1=φ2=90°,random vectors of one plane is perpendicular to all vectors
of the other plane,then we call the two planes are hyper perpendicular.When they have common
point(the common point is unique), we call them perpendicularly mutual-through.
(3) unequal-oblique kind:the maximum angle is not equal to the minimum angle,viz.
φ1>φ2 orφ1<φ2,and 0°<φ1,φ2≤90°.
For example,example 2,example 3 of the seventh chapter are examples of quasi-dihedral
angle kind;example 6,example 7 of this chapter are examples of equal-oblique kind,of which,two
planes of example 7 are hyper perpendicular;example 5 and example 8 are examples of
unequal-oblique kind.
Two concentric unit circles on hyper perpendicluar plane have a very interesting phenomenon.
Take I1′,I2′of example as an example,this pair of concentric unit circles are separately
=
=
=+
0
0
1
2
1
2
4
2
3
1
y
y
yy
L:
=
=
=+
0
0
1
4
3
2
2
2
1
2
y
y
yy
L :
Of which,a random point of L1 is vertex of a right-circular cone taking L2 as the base line,and a random point of L2 is vertex of a right-circular cone taking L1 as the base line. That is to say,the distance of a random point of L1 to L2 is the same,at the same time, the distance of a random
point of L2 to L1 is the same. This phenomenon also exists in higher dimensional space.Now prove it as the following:
125
proof::::Assume M ( x1 , x2 ,0,…,0), N ( 0 , … , 0 , xn-1 , xn )
are separately random points of two unit circles
=
=
=+
0
0
1
3
2
2
2
1
1
nx
x
xx
LL
:
=
=
=+
−
−
0
0
1
2
1
22
1
2
n
nn
x
x
xx
LL
:
because vectors
{ }
2 so
1 but
then
,,0,,0,,
22
1
2
2
2
1
22
1
2
2
2
121
=
=+=+
+++=
−−=−=
−
−
−
MN
xxxx
xxxxMN
xxxxOMONMN
nn
nn
nnL
By the same token,for another two points M0,N0 of L1,L2,it also has
200 == MNNM ▌
In three-dimensional space,two different circles of the same circular cylinder(or right-circular
cone) are perpendicular to each other. This is not only because of two planes the two circles on
paralleling to each other ,but also on the two circles,the distance between every two corresponding
points( it is referred to point intersecting the same generating line of the circular cylinder or the
circular cone) is the same. So,when people judge whether two circles are parallel or not,sometimes they need not consider whether the plane they on is parallel or not,only consider
whether the distances between corresponding points are equal or not. In other words,when
distances between corresponding points of two circles are equal, the two circles must be parallel.
But in n-dimensional space,distances between two random points of two concentric unit circles on
two perpendicular mutual through planes are identical. So,a random section of arc of one circle can
be regarded to be parallel to a random section of arc of anothercircle,we call this phenomenon
hyper parallel phenomenon (or be called pseudo parallel phenomenon)between two circles.
When diameters of two concentric circles aren't the same,hyper parallel phenomenon also
exists. At the same time ,it also exists on difficient-order (the orders are nonzero)generic circle of
same or different diameters of two random hyper perpendicular linear figures,be limited to the
length of writting,here we will not introduce it detailedly.
3.3 Included angle questions between other linear figures
According to principle and steps of Jianshi solution method,we assume equations of two
linear figures has been changed into system of homogeneous linear equations,and at least one
126
linear figure has been changed into a difficient-order(or sufficient-order )generic coordinate plane. Example 9: Please find out included angle φ.Between generic plane
F:2x1+x2+2x3+x4=0
of example 3 and generic coordinate plane x4 = 0.
Solution::::From example 3 and example 2,projective generic cylinder surface
1484525 323121
2
3
2
2
2
1 =+++++ xxxxxxxxx
has been changed into the form of quadratic sum
110 2
3
2
2
2
1 =++ yyy
so , 10
1cos ,1coscos 321 === ϕϕϕ
then φ1=φ2=0°,φ3≈71°34′. Choose nonzero angle φ3=φ to define included angle between Fand x4=0 ,the result is the
same with that of example 1 of the seventh chapter.
Example 10: Find out included angle between two two-order generic planes
=−−−++=−+−−+
02223
022
654321
6543211 xxxxxx
xxxxxxL:
==
00
6
52 x
xL :
of six-dimensional space.
Solution:::: From example 4,equation of projective generic cylinder surface is
830343050465025252545 434232413121
2
4
2
3
2
2
2
1 =−−+−+++++ xxxxxxxxxxxxxxxx
characteristic matrix of the quadratic form is
−
−−−
−−−
−−−
25151725
15251523
17152525
25232545
λ
λ
λ
λ
its invariant divisor are 1,1,λ-8, (λ-8)(λ2-104λ+1004),
so its elementary divisor is
. 171052 , 171052 ,8 ,8 −−+−−− λλλλ
so the quadratic form is changed into
( ) ( ) 817105217105288 2
4
2
3
2
2
2
1 =++−++ yyyy
viz. 14
17526
4
17526 2
4
2
3
2
2
2
1 =+
+−
++ yyyy
127
so , 2929.017526
2cos , 8619.0
17526
2cos , 1coscos 4321 ≈
+=≈
−=== ϕϕϕϕ φ1=φ2=0°,φ3≈30°28′, φ4≈72°58′.
Choose nonzero angleφ3,φ4 as included angle between L1,L2.
Like this kind of questions,because the order of the quadratic form is very higher,then
amount of calculation is much. In fact,included angle between normal space ⊥1L of L1 and
⊥2L
of L2 is also equal to included angle between L1 and L2 ,so,we can also use included angle between
⊥1L and
⊥2L to define included angle between L1 and L2.
For example,first determine four linearly independent solution vectors in L1 {3,-5,0,0,0,-1}, {0,5,-3,0,0,4}, {0,0,3,5,0,-4}, {0,0,0,4,-3,-5},
then the normal space of L1 is
=−−
=−+
=+−
=−−
⊥
0534
0453
0435
053
654
643
632
621
1
xxx
xxx
xxx
xxx
L :
and the normal space of L2 is plane x5ox6,viz.
=
=
=
=
⊥
0
0
0
0
4
3
2
1
2
x
x
x
x
L :
because the determinant of coefficient of four anterior terms of ⊥1L is D=180≠0
then we find out
+=
−−=
−−=
−−=
654
653
652
651
4
5
4
3
4
3
4
54
5
4
34
7
4
5
xxx
xxx
xxx
xxx
substitute it into the equation of unit generic sphere,it gets the projective generic cylinder surface
equation 4403121 65
2
6
2
5 =++ xxxx
from coefficient theorem,it gets
17526 , 17526 +=−= BA
viz. , 14
17526
4
17526 2
6
2
5 =+
+−
yy
so , 17526
2cos ,
17526
2cos 21
+=
−= ϕϕ
this is completely in accord with the above results.
128
Example 11: Find out included angle between difficient-order generic plane
=−−−++
=−+−−+
02223
022
654321
654321
xxxxxx
xxxxxxL:
of example 10 and plane
=
=
=
=
0
0
0
0
6
5
4
3
x
x
x
x
I:
solution::::The included angle betweeen plane I and normal space L⊥
of difficient-order generic
plane L is complementary angle of included angle between I and L,so,the cosine of included angle
betweeen I and L⊥
is sine of included angle between I and L.But in example 10,determinant of
coefficient of four posterior terms of L⊥
is
, 45
5340
4053
4003
1000
−=
−−
−−
−
−
=D
Then it gets
−=
+−=
=
−=
216
215
24
213
533
295
54
xxx
xxx
xx
xxx
substitute it into equation of unit generic sphere,then projective generic cylinder surface equation is
, 13
298
9
130951 21
2
2
2
1 =−+ xxxx
from coefficient theorem it gets
126.1489038.492222.98
18
44827
9
884
9
4591309
9
88804
9
1309459
2
1
≈+≈
≈+=
−
+++
=A
B≈98.2222-49.9038≈48.3184 ,
then, , 14390318448
1sin , 08220
126148
1sin 21 .
..
.≈≈≈≈ ϕϕ φ1≈4°43′, φ2≈8°16′.
Example 12: Find out included angle between one-order generic plane
=+++++
=+−++−
=+++++
022
0
032322
654321
654321
654321
1
xxxxxx
xxxxxx
xxxxxx
G:
of six-dimensional space and
=
=
=
0
0
0
6
5
4
2
x
x
x
G :
Solution::::Because determinant of coefficient of three posterior terms of G1 is D = 9,then we
129
find out , 3
1 ,
9
1 ,
9
731621514 xxxxxxxx +=+==
substitute it into the equation of unit generic sphere,then equation of projective generic plane is
815418162162140 3121
2
3
2
2
2
1 =++++ xxxxxxx ,
So,the real symmetric matrix is
=
162027
01629
279140
G ,
the characteristic root of G found out is
, 931151 , 931151 , 162 321 −=+== λλλ
so equation of projective generic cylinder surface is changed into the form of quadratic sum
, 8199.0931151
9cos , 668.0
931151
9cos ,
2
1cos Then,
. 181
931151
81
9311512
321
2
3
2
2
2
1
≈−
=≈+
==
=−
++
+
ϕϕϕ
yyy
So, φ1=45°, φ2≈48°05′, φ3≈34°55′.
Kinds of included angles between two linear figures of n-dimensional spacen,similiar to
included angle questions between two planes,they can be divided into three kinds:
(1) quasi-dihedral angle kind,like example 9. This kind of questions usually adopt linear
solutions ;
(2) equal-oblique kind:in example 12,if coefficient of x1 of the first equation becomes 1,it
will hasφ1=φ2=φ3=45°;
(3) unequal-oblique kind,like example 10、example 11、example 12.
If discussions of the seventh chapter are added,it also includes
(4) line angle kind(it is only referred to included angles between two straight lines);
(5) line-plane angle kind(it is referred to included angles between straight lines and
sufficient-order or difficient-order planes).
3.4 Other questions of Jianshi solution principle————————exterior product method and
included angle
In the above discussions,we have accepted such a fact: in unit difficient-order generic circles
denoted by s-order quadratic equation of a s-dimensional figure in n-dimensional euclidean space
Vn, there must be such s radii lines perpendicular to each other,their projections on another
s-dimensional linear figure are not only perpendicular to each other but also uniquely become s
semiaxes of a difficient-order generic ellipse. But ,what are these s radii lines on earth?How about
their directions(whether they are uniquely determined or not)?Now let's answer this question.
Assume equations of two linear figures L1,L2 have been changed into homogeneous,and
their dimensions also been changed into the same,of which, L2 has been changed into a
difficient-order generic coordinate plane,viz. assume
( )
( )22,,1 0
22,,1 0
2
1
1
−≤≤−==
−≤≤−==
+
=∑
nssnjxL
nssnjxaL
js
n
i
iij
; :
; :
L
L
and assume the projective generic cylinder surface equation of unit diffcient-order generic circle
130
( )
−≤≤==
−+++
∑=
n
i
iij
n
nssjxa
xxx
1
22
2
2
1
22,,1 0
1
; L
L
with relation to L2 found out is ( ) , 11
21 =
nx
x
Gxx ML
Of which,G is s×s a real symmetric matrix. Apparently,at the time of projective generic
cylinder surface equation being changed into the form of quadratic sum,L1 and L2 also go through
orthogonal transformations.
Assume characteristic roots of matrix G (apparently, G is a real symmetric matrix)of the
quadratic form found out are separately λ1,…,λs ,and characteristic vectors belonging to λ1,…,λs of G are separately α1,…,αs (be a system of orthogonal vectors),all their linear combinations
form a s-dimensional subspace( be called a characteristic subspace of G ). Unitize α1,…,αs and
make become orthogonal-unit vectors e1′,…,es′of the characteristic subspace ,viz. order
, , , 1
1
1
s
s
sα
α
α
α=′=′ ee L
of which e1′={β11 ,…,βs1 },
e2′={β12 ,…,βs2 }, … … …,
es′={β1s ,…,βss },.
and assume es+1′= es+1,es+2′= es+2,…,en′= en, then they are expanded into orthogonal-unit
vectors e1′,…,en′of Vn. So it gets orthogonal matrix
,
1
11
111
=
O
L
L
L
sss
s
Qββ
ββ
Assume the rank of coefficients matrix of L1 is A,then it has
=
=
′
0
011
1 MMM
nn x
x
A
y
y
QAL:
Order AQ′=A0 (Q′= Q-1
is the transpose of Q ),and assume row vectors of A0 are separately
r1′={r11 ,…, rn1},
r2′={r12,…, rn2 }, … … … …,
rn-s′={r1,n-3,…,rn,n-s },
then the equation of L1 become
( )∑=
−==′
=
′
n
i
iij
n
snjyrL
y
y
AL
1
1
1
01
,,1 0 or
0
0
L
MM
:
:
131
Accordingly,the equation of L2′become L2′:ys+j = 0 ( j = 1 ,… ,n-s )
At this time, because equations of difficient-order (sometimes it can be sufficient-order)projective
generic cylinder surface have been changed into the form of quadratic sum,so,according to Jianshi
solution principle (include extended Jianshi solution principle),directions of s radii lines
perpendicular to each other of difficient-order generic circle cut by L1′can be denoted by s
solution vectors of L1′,and projections (are directions of s semi principal axes of difficient-order
generic ellipse )of the s solution vectors on L2′are separately coincident with axes y1,y2,…, ys.
So,when solve the s solution vectors using exterior product method, in multiplication factors,one
of new orthogonal-unit vectors e1′,e2′,…,es′should be deleted in turn,viz. when assume the
s solution vectors(from defination1 of the seventh chapter,they are also angle vectors of L1′,L2′)
are P1′,PPPP2′,…,PPPPs′in turn,it has
k1′P1′= r1′×r2′×…×rn-s′×e2′×e3′×…×es′,
k2′P2′= r1′×r2′×…×rn-s′×e1′×e3′×…×es′, … … … … … … … …,
ks′Ps′= r1′×r2′×…×rn-s′×e1′×e2′×…×es-1′
(k1′,k2′,…,ks′are numbers). Then the cosine of included angle between L1 and L2 can be
denoted as , |}0,,0,1{|
cos 1
1
1
11
11
1P
P
eP
eP
′
•′=
′•′
′•′=
−876
L
n
ϕ
, |}0,,0,1,0{|
cos 2
2
2
22
22
2P
P
eP
eP
′
•′=
′•′
′•′=
−876
L
n
ϕ
, |}0,,0,1,0,,0{|
cos
1
s
sns
s
ss
ss
sP
P
eP
eP
′
•′=
′•′
′•′=
−−876
L
876
L
LLLLLLLL
ϕ
Because P1= P1′Q,P2= P2′Q,…,PS = Ps′Q
are directions of the s radii lines of the difficient-order generic circle before transformation,so,
P1= P1′Q,P2= P2′Q,…,PS = Ps′Q
can also be found out by exterior product method. Because
ki′Pi′= r1′×…×rn-s′×e1′×…×ei-1′×ei+1′×…×es′,
but ( )s,1,2,i , L==′=′i
i
iii QQα
αerr ,
then when assume rj= {a1j , a2j ,…, anj }
( aij are elements of coefficient matrix of L1;i =1 ,2 ,… ,n; j = 1 ,2 ,… ,n-s), it also has
k1P1= r1×r2×…×rn-s×α2×α3×…×αs,
k2P2= r1×r2×…×rn-s×α1×α3×…×αs, … … … … … … … … …,
ksPs= r1×r2×…×rn-s×α1×α2×…×αs-1
( k1, k2,…, ks are numbers),so it also has
132
. cos
, cos
, cos
22
22
2
11
11
1
ss
ss
sα
αϕ
α
αϕ
α
αϕ
•
•=
•
•=
•
•=
P
P
P
P
P
P
LLL
Because P1′,PPPP2′,…,Ps′are perpendicular to each other,and orthogonal transformation
changes vectors perpendicular to eache other into vectors that perpendicular to each other,then,P1′,PPPP2′,…,Ps′are also perpendicular to each other.
According to the above analysis we know,s angles vectors P1,PPPP2,…,Ps of L1 that
perpendicular to each other and their projections on L2 also perpendicular to each other (or
only satisfy that the inner product is zero)are not only exist ,but also their directions are also
uniquely determined;Angular degrees of included angles between L1 and L2 are also uniquely
determined.
This kind of solution method can be called (in Jianshi solution solution)exterior product
method,accordingly,the method introduced in the above section can be called projection method.
Example 13: Find out included angle between plane
=−+
=−+
03
0322
421
3211 xxx
xxxI :
of four-dimensional space and coordinate plane
=
=
0
0
4
3
2x
xI : .
Solution::::Because here I1 is just ⊥
1I of example 5,then the projective generic cylinder surface
equation of I1 with relation to I2 is 9141014 2
22
2
1 =++ xxxx
(refer to it in example 5),its real symmetric matrix is
=
145
514
9
1 G
( )19
19
9
14
9
59
5
9
14
but −
−=
−−
−−=− λλ
λ
λλ GI
(I is a two-order unit matrix),then two characteristic roots of G are separately
, 1 , 9
1921 == λλ
Regard every element of determinant |λI-G| as coefficients of the following system of equations,it has
=
−+−
=−
−
09
14
9
5
09
5
9
14
21
21
xx
xx
λ
λ
First substitute λ1 into the system of equations,it gets
133
=
−+−
=−
−
09
14
9
19
9
5
09
5
9
14
9
19
21
21
xx
xx
we find out a characteristic vector of λ1 is α1={1,1,0,0},
secondly substituteλ2 into the above system of equations,it also gets
=
−+−
=−
−
09
141
9
5
09
5
9
141
21
21
xx
xx
we find out a characteristic vector of λ2 is
α2={-1,1,0,0},
because r1={2,2,-3,0}, r2={1,1,0,-3},
then it gets { }2,4,3,33
0011
3011
03223
4321
2211 =
−
−
−=××=
eeee
rrP α ,
{ }0,0,1,19
0011
3011
03229
4321
1212 −=−
−=××=
eeee
rrP α ,
Because ( ) , ,,2,1 cos siii
ii
i L=•
•=
α
αϕ
P
P
then it gets , 19
3
4169911
33cos 1 =
++++
+=ϕ
, 122
11cos 2 =
−−=ϕ
then φ1≈46°31′, φ2=0°.
(φ1 , φ2 are just complementary angles of included angles between the two planes of example 5)
Choose nonzero angleφ1≈46°31′to define included angle between I1 and I2 .
Now let's see the result after orthogonalization:Unitize α1,α2, it gets
, 0,0,2
1,
2
10,0,
2
1,
2
121
−=′
=′ e , e
order e3′= e3 , e4′= e4 , Then it gets orthogonal matrix
−=
1
12
1
2
12
1
2
1
Q
134
because
−
−
−=′
1
12
1
2
12
1
2
1
321
322QA
then it gets { } { }23,0,0,20,23,0,4 21 −=−= r , r ,
So, { }2,22,0,36
0010
23002
023046
4321
1211 =−
−
′′′′
=′××=
eeee
errP
{ }0,0,1,018
0001
23002
0230418
4321
2211 −=−
−
′′′′
=′××=−
eeee
errP
[under new coordinates, of P1, e1′={1,0,0,0},of P2 , e2′={0,1,0,0}]. it gets
, 1cos , 19
3cos 21 == ϕϕ
this is in coincident with the result of the above.
The purpose we introducing exterior product method is to give an account of the existence of
the s solution vectors PPPP1,PPPP2,…,PPPPs. But in actual application,we still adopt simple and practical
projection method.
Exercises 8.1 Is exterior product operation of vectors in n-dimensional space carried between two
vectors?Why? 8.2 Try to use exterior product method to orthogonalize system of vectors α1={3,2,0,5,3}, α2={5,3,2,1,4}, α3={2,1,2,7,5}, α4={8,-1,2,4,3} . 8.3 Try to use exterior product method to find out a system of elementary solutions of
one-order generic plane . 0322
0235
54321
54321
=+−+−
=++−+
xxxxx
xxxxx 8.4 Use Jianshi solution solution to find out the included angle between generic plane 3x1+5x2-4x3-2x4=0
of four-dimensional space and generic coordinate plane
x4 = 0 ,
and check it by linear solution. 8.5 Find out the included angle between plane
=++−
=−−+
0453
0423
4321
4321
xxxx
xxxx
of four-dimensional space and coordinate plane
135
=
=
0
0
6
5
x
x. 8.6 Find out the included angle between two planes
==
=++−=−−+
00
02532
04254
4
32
4321
43211 x
xI
xxxx
xxxxI : :
of four-dimensional space. 8.7 Find out the included angle between one-order generic plane
=
=
=
=+−++−
=−+−−+
=−+−−+
0
0
0
022322
042332
0652326
6
5
4
2
654321
654321
654321
1
x
x
x
K
xxxxxx
xxxxxx
xxxxxx
K
:
:
.
8.8 Change the plane I2 of example 5 into
=
=⊥
0
0
2
1
2x
xI :
then it become to find out the included angle between I1 and⊥2I . And check whether the result
is complementary angle of the included angle of example 5 or not. 8.9 Change the coefficient of the first equation x1 of G1 in example 12 into 1,check
whetherthe included angle between new G1 and G2 is equal-oblique kind or not. 8.10 Find the included angle between two planes
.02433
0225
02532
04254
4321
43212
4321
43211
=+−−
=+++
=++−
=−−+
xxxx
xxxxI
xxxx
xxxxI
:
:
136
The Ninth Chapter: The application of High Dimensional
Euclidean Geometry
High Dimensional Euclidean Geometry has very extensive application prospect. But be
limited to the length of writing,we only introduce some applications of High Dimensional
Euclidean Geometry in operations research.
§§§§1 Application of High Dimensional Euclidean Geometry in linear
programming
Methods of solving linear programming mianly had two kinds in the past:one is graphical
method,the other is simplex method. Of which,when the amount of variables of graphical method
is not less than 4,people once tried to use methods of high dimensional descriptive geometry [8]
. In
these years,many people are engaged in the study of non-linear programming,of which,(of
constrained extreme value questions)interiorpoint method is regarded as a kind of approximately
perfect method,and be attempted to use to solve linear programming questions [15]
. In the following we will introduce new methods,they are methods of using oblique axes
transform 、oblique axes draughting to solve linear programming,this kind of new method is also
called graphical method of preferred n-dimensional system. The main characteristics of the new methods are rigorousness and integrity of theory,concision
and directviewing of figures,it is easy to learn and master,convenient and practical,accurate and
reliable,it has very extensive application prospect. To verify its correctness and reliability,sometimes the acquired results should be compared to
some above methods .
1.1 Example one
A certain factory has three kinds of products of A,B,C,according to technological
requirements,each product should be manufactured on three equipments of the first、the second、the third in turn,processing time of each product on every equipment ,running time of every
equipment in every day and profit of each product can be achieved can be referred to in the
following table. Three kinds of products are all salable. How to arrange the output of this three
kinds of products,then it can make the factory achieve the maximum profit?
product
time
equipments
A
B
C
running time of every equipment in
every day(h)
the first 1 1/2 1 ≤23
the second 1/2 1 1 ≤23
the third 1 2/3 2/3 ≤22
profit (yuan/piece) 25 20 15
Solution::::According to conditions given in the question,assume we should arrange
the amounts of A,B,C to be x, y, z,the objective function is t,then it has the
following linear programming model
137
( ) ( )1 0,,
223
2
3
2
232
1
232
1
condition constraint
152025max function objective
≥
≤++
≤++
≤++
++=
zyx
zyx
zyx
zyx
zyxt
:
:
Now,change coefficients of each equation into integers
( )( )( )
( )2
66223
4622
4622
R
Q
P
zyx
zyx
zyx
≤++
≤++
≤++
first use simplex method to compute it. Order
x = x1, y = x2, z = x3,
and introduce slack variables x4,x5,x6,then the model is changed into normalized form:
objective function:Max t =25x1+20x2+15x3+0x4+0x5+0x6 constraint condition:
=+++
=+++
=+++
66 223
46 22
4622
condition constraint
6321
5321
4321
xxxx
xxxx
xxxx
:
so it has
−−−=
−−−=
−−−=
3216
3215
3214
22366
22 46
2 246
xxxx
xxxx
xxxx
When order x1 = x2 = x3 = 0,
it gets t = 0,now it gets a radix feasible solution
X (0)
= (0,0,0,46,46,66)
because the coefficient of objective function x1 is larger,then t will increase much after taking x1 as
swapin variable. Order
x2 = x3 = 0
it has
≥−=
≥−=
≥−=
0366
0 46
0246
16
15
14
xx
xx
xx
(3)
Now only choosing 223
66
1
46
2
46min 1 =
= ,,x
can make formula(3) be true. Then take x6 as swapout variable and transpose it with x1,
6321
3215
3214
22663
2246
2462 getsit
xxxx
xxxx
xxxx
−−−=
−−=+
−−=+
138
viz.
−−−=
+−−=
+−+=
6321
6325
6314
3
1
3
2
3
222
3
1
3
4
3
424
3
2
3
2
3
12
xxxx
xxxx
xxxx
substitute it into objective function,it gets
, 3
25
3
5
3
10550 632 xxxt −−+=
When order x2 = x3 = x6 = 0 ,it can get another radix feasible solution
X(1)
= (22,0,0,2,24,0).
In the new objective function,only the coefficient of x2 is more than zero,so again take x2
as swapin variable. In order to determine swapout variable,we can order x3 = x6 = 0 , then it gets
( )4
03
222
03
424
03
12
21
25
14
≥−=
≥−=
≥+=
xx
xx
xx
now only choosing 18
3
2
22,
3
4
24, Min2 =
−=x
can make formula(3) be true,so,take x5 as swapout variable and transpose it with x2 ,
6312
5632
6324
3
1
3
222
3
23
1
3
424
3
43
2
3
22
3
1 getsit
xxxx
xxxx
xxxx
−−=+
−+−=
+−=−
viz.
+−=
−+−=
−+−=
561
5632
5634
2
1
2
110
4
3
4
118
4
1
4
38
xxx
xxxx
xxxx
substitute it into objective function ,it gets
. 2
15
2
55610 653 xxxt −−−=
There is "discriminant theorem of optimum solution "in linear programming,its contents are:substitute operation results of swapin variables into objective function,a newe objective function is
acquired,if variables of the new objective function are all less than zero,the radix feasible solution
found out is optimum solution[15]
. According to the theorem,variables of the new objective function
are all less than zero.
139
So when order x3 = x5 = x6= 0 ,the radix feasible solution
X(2)
= (10,18,0,8,0,0)
found out is optimum solution. Viz. A should be arranged to be producted 10 pieces every day,B
should be producted 18 pieces,C should be producted 0 piece,the maximum profit achieved is 610
yuan.
Although the above computation process is very cockamamie,the results are correct. But,trouble occurred when we use interior point method
[15] to compute(the detailed computation
process is omitted),the optimum solution found out is 10 pieces of A,ten pieces of B,eight pieces
of C,"maximum profit" be achieved is 570 yuan,it is 40 yuan less than that of simplex method!
What's the cause? It turns out that,when we compute by equations,the system of equations is
consistent and the coefficients matrix is nonsingular,from linear algebra theory,it has uniquely
determined solution(but it is not always optimum solution of linear programming).
In fact, considering generalized case,this kind of questions can only aquired analogous results.
For example,we use “interior point method” to solve the following models objective function: ( ) ∑−=
=1
1
Max n
i
ii xAXf
constraint condition: ( ) ∑−=
=≤=1
1
,,2,1 n
i
jijij mjaxaXg L
(x1 , x2 ,…, xn-1≥0 )
Order F(X)=-f (X), qj (X)=-gj (X)+aj≥0 ,
then we can change the modes into the following minimization questions of unconstraint property
( ) , , Min rXP
Of which ( ) ( ) ( )[ ]∑=
−+=
m
j
j XqrXFr,XP1
1
viz. ( ) ∑ ∑ ∑−
= =
−−
=
−−−=
1
1 1
11
1
,n
i
m
j
n
i
jijiii axarxArXP
solve partial derivative for ( ) , r,XP it gets
∑∑=
−
=
−
+−=∂
∂ m
jn
i
jiji
ji
i
iaxa
raA
x
P
12
1
1
order mj
axa
rR
n
i
jiji
j ,,1 2
1
1
L=
−
=
∑−=
it gets the system of linear equations
1,,1 1
−==∑=
njARa j
m
i
iji L
If we don't cosider the restrictionof x1 , x2 , …, xn-1≥0 (viz. don't rank them as constraint
condition or gj(X) doesn't include them), then we can at least find out a set of
R1=b1,R2=b2,…,Rm=bm
140
when m = n-1 and coefficients matrix is nonsingular. extract square root for
mjR
r
j
,,1 L=
and order mjcb
j
j
,,1 1
L==
then it gets a new system of linear equations
mjrcaxa jj
n
i
iji ,,1 1
1
L==−∑−=
viz. ( ) mjrcaXg jjj ,,1 0 L==−−
So it has ( )[ ] ( ) 0 lim0
=−=−−→
jjjjjr
aXgrcaXg
viz. when take limit for obstacle factor r ,it gets system of linear equations
g j (X) = aj j =1, 2, …, m . Goning in a roundabout way only to change sign of inequality of constraint condition into equal
sign,then change inequality into equality!
1.2 Theory 、、、、 steps and assumption of graphical method in preferred
n-dimensionalsystem
1.2.1 Theory The theory of graphical method in preferred n-dimensional system (be called preferred
n-dimensional system method for short) includes graphic rules of preferred n-dimensional system
and two fundamental theorems. Of which,we have detailedly introduced the graphic rules of
preferred n-dimensional system from the first to the fifth chapter,here we only introduce two
fundamental theorems.
Assume the model of linear programming is objective function
S:Max t = f (X)
constraint condition: g j (X) ≤aj,X≥0
in the formula,X = x1, x2,…, xn-1, j =1, 2 , …, m;aj are nonnegative integers. Assume objective generic plane S has been changed in the shape of plane by proper oblique axes
transformation,and Ci (i =1 , 2 ,…, I ) are other vertexes enclosing convex polyhedron of feasible
region except the origin,the trace of generic point that Ci is on is Ci′.
The cut trace of S cut by generic coordinate plane is marked as S t =0 .
When order t = t0 and substitute it into objective function,the equation represents a
generic plane f (X) = t0
parallel to t axis.When introduce leading-axis method in the fifth chapter,we call the generic
plane projective generic plane. Its top intersects objective generic plane S,the cut trace is generic
straight line
( )
=
=
0
0
tt
tXfS : π
its underside intersects generic coordinate plane,the cut trace is generic straight line
141
( )
=
=′
0
0
t
tXfS : π
cut trace πS ′ is just the projection of cut trace
πS on generic coordinate plane. Assume the
projectionπS ′ just intersects a certain vertex Ci of convex polygon,then the projection also
intersects trace point Ci′of generic point that Ci is on.
Make a parallel line of upright axis t through Ci ,it intersects objective generic plane S at point
Li,then Li is located atπS ,the length of line segment |CiLi| is just the corresponding value of t of
objective function when constraint condition satisfies the coordinates of Ci ,and make a parallel
line of upright axis t through Ci′, it also intersects S at generic straight lineπS ,and the
intersection point Li′is projective
trace of generic point that Li is on.So it is easy to know that
. and , iiiiiiii LCLCLLCC ′′′′ ∥∥ The trace line (it is also the trace line of St=0 )of S on plane x1ox2 is called objective trace line,marked as S{1,2},its equation is
{ }
( )
=
=
=
=
−
−
0
0
0
00,,,,
1
3
121
2,1
n
n
x
x
t
xxxf
S
L
L
:
And make perpendicular line Ci′Ki′(Ki′is foot point) to objective trace line S{1,2} through
point Ci′,then Ci′Ki′is parallel to perpendicular line CiKi (Ki is foot point) of St=0 made
through Ci and along the direction parallel to x1ox2,Ki′are projective traces of generic point that
Ki are on,so,
. and , iiiiiiii KCKCKKCC ′′′′ ∥∥
(so, iiii KLKL ′′∥ ,
viz. iiiiii KLCKLC ′′′≌△△ ).
Because figures of Ci Ki are completely coincident with those of Ci′Ki′,under
generalized cases,we can only make diagrammatic representations (Ci′and Ki are usually not
marked out) for point Ci and point Li′of figures be made out,so regard the connecting line
between Ci and Ki′as the connecting line between Ci′ and Ki′, and call |Ci′Ki′|=ki′directviewing distance of Ci to objective trace line S{1,2}.
From theorem 1 of the fourth chapter,we can always change objective generic plane in the
shape of plane by proper oblique axes transform. The question is ,the proper oblique axes
transform is not unoque. Then,under different oblique axes transforms,the trace of the same vertex
142
Ci is different because of the difference of the generic point,are these distances of these different
trace points to objective trace lines(it will not change with the change of oblique axis transform)the
same? If they are not completely identical,it is nonsensical to determine it optimum point or not
by directviewing distance of a certain vertex to objective trace line. So,we give the following:
Constant distance theorem::::On the premise of keeping objective generic plane S in the shape
of plane,no matter what different oblique axes transform we take,the directviewing distance of
each vertex of convex polyhedron to objective trace line S{1,2} is changeless. Proof::::Assume the objective generic plane is S:A1x1+A2x2+…+An-1xn-1- t = 0 ,
C(x10, x20, …, xn-1,0,0) is a random vertex of convex polyhedron,when C is zero with relation to
oblique axes x30, x40,…, xn-1,0,C is the point of plane x1ox2 ,so no matter what different oblique
axes transform we take,its position on plane x1ox2 is always changeless,so,its distance to objective
trace line is also changeless.
When x30, x40,…, xn-1,0 are not zero at the same time,then under different oblique axes
transforms,the directviewing position of C with relation to plane x1ox2 also has some change,the
changing trace should be a straight line. if we want to prove that on the premise of keeping S in the
shape of plane,no matter what different oblique axes transforms we take ,the directviewing
distance of point C to S{1,2} is always changeless,we can only prove this straight line is parallel to
S{1,2} . Therefor we can assume
S11=α11e 1 +α12 e 2 +α3 e 3 ,S21=α21 e 1+α22 e 2 +α4 e 4 ,… ,
Sn-3,1=αn-3,1 e 1 +αn- 3,2 e 2 +αn-1 e n-1 ;
S12=β11 e 1 +β12 e 2 +β3 e 3 ,S22=β21 e 1 +β22 e 2 +β4 e 4 ,… ,
Sn-3,2=βn-3,1 e 1 +βn-3,2 e 2 +βn-1 e n-1
are two different sets of solution vectors of S,they are regarded as principal overlapping directions
of the two different oblique axes transforms. S is in the shape of plane under transform
∑
∑
−−
−−
−−
−−
−−
−−
−−−
−−−
22,311,321-n1
222121244
212111233
2
22,311,311-n1
222121144
212111133
1
eee
eee
eee
eee
eee
eee
nnn
nnn
βββ
βββ
βββ
ααα
ααα
ααα
□□□ □□□
:
:
LLLLLLLLLL
LLLLLLLLLL
and vectorOC (O is the origin) has the following change:
,
,
2
1
0,12,3
3
3012201
1
0,11,3
3
301110
2
1
0,12,3
3
3012201
1
0,11,3
3
301110
2
1
ee
ee
−−+
−−
−−+
−−
−
−−
−
−−
−
−−
−
−−
n
nn
n
nn
n
nn
n
nn
xxx
xxxOC
xxx
xxxOC
β
β
β
β
β
β
β
β
α
α
α
α
α
α
α
α
LL
LL□□
so point C has common generic point with point
143
, 0,,0,, 1
0,12,3
3
301220
1
0,11,3
3
3011101
−−−−−
−−
−
−−LLL
n
nn
n
nn xxx
xxxC
α
α
α
α
α
α
α
α :
of plane x1ox2 and
, ,0,0, ,1
0,12,3
3
301220
1
0,11,3
3
3011102
−−−−−
−−
−
−−LLL
n
nn
n
nn xxx
xxxC
β
β
β
β
β
β
β
β :
successively,and the vector between the two points is
( ) ( )
( ) ( )2
11
2,3112,30,1
33
12331230
1
11
1,3111,30,1
33
1133113021
e
e
−++
−+
+
−++
−=
−−
−−−−−
−−
−−−−−
nn
nnnnn
nn
nnnnn
xx
xxCC
βα
βαβα
βα
βαβα
βα
βαβα
βα
βαβα
L
L
Because the trace line of objective generic plane on x1ox2 is
{ }
=
=
=
=+
−
0
0
0
0
1
3
2211
2,1
t
x
x
xAxA
S
n
L:
its orthodromic vector (also be called direction vector) is
A2e1-A1e2
If ( )211221 ee AACC −= λ
(λ is nonzero number),then the changing trace of C with relation to x1ox2 is parallel to objective
trace line,now we will prove it.
Because S11,S21,…,Sn-3,1 and S12,S22,…,Sn-3,2 are all solution vectors of S,and {A1,…,An-1,-1} is normal vector of S,it is perpendicular to all solution vectors of S,then it has
A1α11+A2α12+A3α3=A1β11+A2β12+A3β3= 0
A1α21+A2α22+A4α4=A1β21+A2β22+A4β4= 0 … … … … … … …
A1αn-3,1+A2αn-3,2+An-1αn-1=A1βn-3,1+A2βn-3,2+An-1βn-1= 0
, ,
, ,
, , so,
1
4422221
1
4422221
1
3312211
1
3312211
LLLLLLLL
A
AA
A
AA
A
AA
A
AA
βββ
ααα
βββ
ααα
+−=
+−=
+−=
+−=
. , 1
112,32
1,3
1
112,32
1,3A
AA
A
AA nnn
n
nnn
n
−−−
−
−−−
−
+−=
+−=
βββ
ααα
substitute vector 21CC between the above points C1, C2 into them and multiplied by
11
2,3112,30,1
33
12331230
1
)()(
−−
−−−−− −++
−−
nn
nnnnnxx
A
βα
βαβα
βα
βαβαL
144
then the direction of the straight line through points C1, C2 is A2e1-A1e2 ,but in
( )211221 ee AACC −= λ
.
)()(
1
11
2,3112,30,1
33
12331230
A
xx
nn
nnnnn
−−
−−−−− −++
−
−=βα
βαβα
βα
βαβα
λ
L
This proves that the straight line is parallel to objective trace line S{1,2} (it also proves that the
changing line of Ci on plane x1ox2 is indeed a straight line).Then the theorem is proved. ▌
Because the length of line segment |Ci′Li′| is just the value of t of objective function,and △CiLiKi≌△Ci′Li′Ki′,
So . ,,2,1, jimjiLC
LC
KC
KC
jj
ii
jj
ii ≠=′′
′′=
′′
′′;L
That is to say,the directviewing distance Ci′Ki′of vertex Ci to objective trace line S{1,2} is
longer,then the value of the corresponding objective function will be larger.And from constant
moment theorem,it has
Optimum point theorem::::Of every vertex of convex polyhedron,the point whose
directviewing distance to objective trace line S{1,2} is the longest is optimum point to be found out.
▌
1.2.2 Method、、、、steps
a. choose proper preferred n-dimensional system
Choose a set of solution vectors
S11=α11e 1 +α12 e 2 +α3 e 3 ,
S21=α21 e 1+α22 e 2 +α4 e 4 , … … … … … … …,
Sn-3,1=αn-3,1 e 1 +αn- 3,2 e 2 +αn-1 e n-1 ;
of objective function S:f (X) = t
as principal overlapping direction,make use of transform
−−
−−
−−
−−−− 22311311
22212144
21211133
eee
eee
eee
,n,nnn ααα
ααα
ααα □□□ LLLLL
then we can get a preferred n-dimensional system that can make S in the shape of plane.
b. determine coordinate of every vertex of convex polyhedron and their directviewing position
in preferred n-dimensional system.
For the convenience of discussion,we regard equations P,Q,R of constraint condition of
example 1 as the same with general equations for the moment,we call them constraint generic
plane. Because P,Q,R don't include variable t,and these three equations are consistent(have no
unwanted equations),so,they intersect at a point on generic coordinate plane,coordinate of the
intersection point is the unique solution
x1=10, x2=10, x3= 8 ,
of system of linear equations with three unknown acquired when equation of constraint condition
145
takes equality .
From the above discussion we have known,this set of solutions is not optimum solution.
The optimum solution is x1=10, x2=18, x3= 0 .
Lead a straight line parallel to axis t through the corresponding optimum point of optimum
solution,coordinate of the intersection point of the straight line with S with relation to axis t is the
optimum solution to be found out. Finding out the intersection point of the straight line with S,the
best method is to change S in the shape of plane.So ,find out a solution vector -2e1+e2+2e3
among S and take it as principal overlapping direction,make use of transform
2e 3 □ 2e1-e 2
then we get the generic coordinate plane of preferred four-dimensional system that makes S in the
shape of plane(figure 46).
figure 46
First find out trace lines of each constraint generic plane on plane x1ox2,then separately
determine coordinates of intersection points of these trace lines according to principle of three kind
of graphic methods and principle of descriptive geometry,we determine them vertexes of convex
polyhedron or not according to whether coordinates of these points satisfy each constraint condition
or not.
Refer to it in figure 46,three trace lines P1P2 , Q1Q2 , R1R2 of P , Q , R on plane x1ox2
separately intersect at three points
( ) ( ); 0,0,18,10 , 0,0,6,26 , 0,0,3
46,
3
46111 CBA −
Three trace lines P2A2 , Q2A2 , R2R3 on plane x2ox3 separately intersect at two points
146
B2(0,20,13,0) and A2 (0,0,23,0)
( Q2A2∥R2R3 ,it has no intersection point in euclidean space);Three trace lines A2P1 , A2Q1 , R3R1
on plane x3ox1 separately intersect at
A2,B3 (20,0,3,0) and C2 (10,0,18,0) .
Connect A1A2 and B1B2 (or C1C2) it gets the intersection point D .Because the projection of
A2A1 on plane x1ox2 is OA1,so,make a parallel line of axis x3 through point D to OA1, it
intersects OA1,coordinate of the intersection point is just coordinate of D with relation to axis x1,
axis x2 (they are omitted in the figure).Coordinate of D(10,10,8,0) is the unique solution when P,
Q, R take equality. The position and coordinate of the above points are all acquired by methods of
elementary descriptive geometry.
Under general cases,the figure of objective function S needn't be made,optimum point can be
directly found out by method of step c,then it is ok to substitute coordinate of optimum point into
equation of S . But in our discussion,for the convenience of illustrating questions,this step is
indispensable. In the following ,we will make out the directviewing figure of S .
In equation of S,order t =200,it gets a projective generic plane
S′:25x+20 y +15 z = 200 ,
its cut trace on generic coordinate plane is a generic straight line,the generic straight line is
coincident with trace line Sxy′. Now assume trace line of generic straight line that OC1 is on on
plane xoy is OJc ,it is directviewing-coincident with OC1,then trace line OJc intersects Sxy′at
point Jn′,we regard Jn′as projective trace of a generic point N of S .
Choose proper direction and unit length, then make out the figure of upright axis t,lead Jn′to
generic point N,connect ON,and make C1M∥Ot ,it intersects the directviewing extension line of
ON at M . Then separately make EG and FH through two generic points E and F of traces Je,Jf on Sxy,make OMJJJJ hfge ∥∥
( Jg , Jh are separately opposite traces of G,H ),connect GH ( M is on GH)),
generic parallelogram EFGH is the figure of generic plane S.
As will be readoly seen,all points that satisfy constraint condition are located at convex
hexahedron OR1C1Q2A2B3D enclosed by quadrilaterals DC1R1B3 ,DC1Q2A2 ,OR1C1Q2 ,OR1B3A2
and △DA2B3 , △OQ2A2 ,because points A1,B1,B2,C2 are separately located at the outside out
of triangle or quadrilateral of plane x1ox2 , plane x2ox3 and plane x3ox1 ,their coordinates don't
satisfy all constraint conditions,then they are not on the convex polyhedron.
c.determine optimum point and optimum solution
1) According to method of drawing parallel lines,put down one right angle side of triangular
rule clinging to objective trace line, put another ruler clinging to the other right angle side of
triangular rule,fix the ruler,slide the triangular rule along direction of crossing feasible region
clinging to the ruler,when the right angle side parallel to objective trace line slide on the border of
being away from feasible region,the last vertex of convex polyhedron pressed by the right angle
side is optimum point to be found out.
2) Order t = t0 ( t0 = 0 , 1 , 2 , …) and substitute them into objective function in turn,then
make them simultaneous with x3 = 0, x4 = 0, … , xn-1= 0, t = 0, it gets
147
{ }
( )
=
=
=
=
′
−
−
0
0
0
0,,,,
1
3
0121
2,1
t
x
x
txxxf
S
n
n
L
L
:
Then S{1,2}′is equivalent to translate objective trace line along direction of crossing feasible
region. When S{1,2}′is about to or just be away from feasible region,the distance between S{1,2}′and the last vertex of convex polyhedron is the nearest,arrange every value of t0 , make its
increase (or decrease)amplitude go down,until the distance between S{1,2}′and the vertex is small
enough to satisfy people,then the value of t0 is optimum solution .
Take figure 46 as an example,apparently optimum point is located at near the other six
points (except the origin)of convex hexahedron OR1C1Q2A2B3D. From constant distance theorem
and optimum point theorem,among the six vertexes ,the distance between C1 and the objective
trace line is the farest ,so it is the optimum point to be found. Substitute coordinate of C1 into
objective function S , it gets t = 610,this is the maximum value of objective function under the case
of meeting constraint conditions.
Or,make a parallel line of axis t through C1 to S, it intersects S at point M, because we
measure that 41 610eMC = ,then we know t = 610 .
1.3 An assumption
By above new graphic methods and steps,we associate the defect of simplex method. At the
same time,we find that ideas of preferred n-dimensional system method happen to have the same
view with that of simplex method in some questions,this idea of happening to have the same view
may help us get over the defect of simplex method. In linear programming,we may as well call every equation served as constraint condition
constraint equation,and deem that every equation has some certain constraint force. Magnitude of
the constraint force is determined by vertexes' amount and vertexes' quality of convex polyhedron
of constraint generic plane denoted by the constraint force. In other words,a constraint generic
plane has more vertexes ,its quality will be better,the constraint force of its equation is greater,contrarily it is less. Of which, the quality of vertex of convex polyhedron is :substitute coordinates
of these vertexes into objective function, values found out is satisfying.
It is observed from figure 46,generic plane R has C1,D,R1 and B3 four vertexes,according
to the distance between them and objective trace line S{1,2}′,sorting order of quality is 1,2,3,4;Q has C1,D,Q2,A2 four points,sorting order is 1,2,5,6;P only has D,B3,A2 three
points,sorting order is 2,4,6. So,constraint force of equation (R) is the greatest,(Q) takes second
place,constraint force of (P) is the least.
At least computation procedure of simplex method happens to have the same view with
constraint force of constraint equations .
But,at least the introduction of slack variable increases much unnecessary work in solving this
kind of questions. The original idea of introducing slack variable is to change unequal relations into
equal relations.But coefficient of slack variable in objective function is zero ,then slack variable is
out of function for the last operation results. So,hiding slack variable will not influence the
correctness of operation results. Then we regard some constraint conditions as equalities hiding
148
slack variable.Still use the above example:
Arrange the constraint conditions as the following:
( )( )( )
( )′
−−−≤
−−−≤
−−−≤
1
223660
22460
22460
321
321
321
R
Q
P
xxx
xxx
xxx
still take x1 as swapin variable,and order x2 = x3 = 0 it gets
( )′
−≤
−≤
−≤
2
3660
460
2460
1
1
1
x
x
x
Because only choosing
223
66,
1
46,
2
461 =
=x
can make formula(2)′be true,then,(R) is regarded as equation of the greatest constraint force, it
is of the corresponding formula (1)′of the third inequality of formula (2)′. Regard (R) as
an equality hiding slack variable,and change the coefficient of x1 into 1 then move it to the left
end of equal mark,it has
( )′
−−=
−−≤+
−−≤+
3
3
2
3
222
22460
24620
321
321
321
xxx
xxx
xxx
Further arrange formula (3)′into
( )″
−−=
−−≤
−+≤
3
3
2
3
222
3
4
3
424 0
3
2
32 0
321
32
32
xxx
xx
xx
Substitute x1 into objective function ,then it gets
. 3
5
3
10550 32 xx
t −+=
When order x2 = x3 = 0 it gets the initial radix feasible solution X(1)
= (22,0,0,0),This is just
coordinate of vertex R1 of convex polyhedron in figure 46.
According to differentiation theorem of optimum solution ,coefficient of variable x2 of new
objective function is still more than zero,so the corresponding radix feasible solution X(1)
isn't the
optimum solution.
And take x2 as swapin variable and order x3 = 0,then it gets
149
( )′
≥−=
−≤
+≤
4
03
222
3
4240
320
21
2
2
xx
x
x
Because only choosing 18
3
2
22,
3
4
24, Min2 =
−=x
can make formula (4)′be true,so,(Q) is regarded as equation of the greatest constraint force
except (R),it is of the corresponding formula (1) of the second unequality of formual (4)′. And
regard the second equation of formula (3)" as an equality hiding slack variable,change
coefficient of x2 into 1,transpose other equations and further arrange them into
( )5
10
183
2
320
1
32
32
=
−=
−+≤
x
xx
xx
substitute formula (5) into objective function,it gets
t = 610-5x3,
from differentiation theorem of optimum solution,variable coefficients of new objective function
are all less than zero. So when order x3 = 0, it gets radix feasible solution X(2)
= (10,18,0,0)
viz. the optimum solution. Apparently,it is just coordinate of C1 of figure 46,now objective
function will take its maximum value
t = 610 .
If simplex method can be improved according to the above ideas,this age-old method will
become more perfect. So,we extend this method to a generalized case:
assume objective function is
∑−+=
=1
1
Max n
ki
ii xAt
constraint condition is m,,,jaxan
i
jiji L21 1
1
=≤∑−=
xi≥0 i =1 , 2 , …, n-1
In order to change constraint condition into standard model,we first introduce slack variable
and make it become
11,22111
11,222212121
11,12121111
0
0
0
−−−+
−−−
−−
−−−−=+
−−−−=+
−−−−=+
nnmmmmmn
nnn
nnn
xaxaxaax
xaxaxaax
xaxaxaax
L
LLLLLL
L
L
it becomes
150
11,2211
11,22221212
11,12121111
0
0
0
−−
−−
−−
−−−−=
−−−−=
−−−−=
nnmmmm
nn
nn
xaxaxaa
xaxaxaa
xaxaxaa
L
LLLLLL
L
L
after hiding slack variable Apparently,this is still a standard model(onlly at the beginning, basis
vector is a zero vector to all appearances, with the introduction of swapin variable, it becomes a
nonzero vector). So for linear programming,all definations、theorems are still efficient here. But,because initial radix feasible solution changes from representing the origin to representing a vertex
of convex polyhedron located at axis x1, x2 , …, xn-1,proof methods of these theorems have some
small changes:
Differentiation theorems of optimum solution and infinite optimum solutions
1.Assume after several iterations
X(0)
= ( c1, c2 , …, ck , 0 , … , 0 ) ( 1≤k≤m )
become a corresponding feasible basis vector,at the same time the objective function becomes
( )∑−+=
≤≤>+′=1
1
1 0 n
ki
kkii mkbbxAt ;
If for all i = k+1 , k+2 , …, n-1 it has Ai′≤0 ,then X(0)
is optimum feasible basis vector or be
called optimum solution,Ai′is called check number.
2. If X(0)
= ( c1, c2 ,…, ck , 0 , … , 0 ) ( 1 ≤ k ≤ m )
is a optimum solution,for all i = k+1 , k+2 , …, n-1 it has Ai′≤0 ,at the same time there is
check number Al′= 0 ( k+1≤l≤n-1 ) of a certain non-radix variable, then the programming has
infinite optimum solutions.
Proof::::Use reduction to absurdity. If
X(0)
= ( c1, c2 , …, ck , 0 ,…, 0 ) (1≤k ≤m )
is not optimum solution,there must be a non-radix variable xi ( for example assume it is xk+1 ) that
can be swaped in radix variable,then it gets a new radix feasible solution
X(1)
= ( c1′, c2′, …, ck′, ck+1 , 0 ,…, 0 ) (1≤k ≤ m )
then when substitute radix variable xk+1 = ck+1 newly swaped in into new objective function,it gets
∑ ∑−
+=
−
+=+++ +′+′′=+′′=
1
2
1
2
111
n
ki
n
ki
kkkiikii bcAxabxat
viz. bk+1 = Ak+1′ck+1+bk
Because Ak+1′≤0 , ck+1≥0,then Ak+1′ck+1≤0 .
When 11 ++′
kk cA = 0 and ck+1>0 ,both X(1)
and X(0)
are optimum solutions(now the
programming has infinite optimum solutions,refer to it in the second section of the theorem). When
11 ++′
kk cA <0,it has bk+1<bk ,so X(1)
is not optimum solution,but X(0)
is optimum solution. Then
the first section of the theorem is proved. ▌
Proofs of the second section of the theorem and differentiation theorem of unbounded
solution (is omitten) are similar to this,forgive not giving uncecessary details.
§§§§ 2 Application of High Dimensional Euclidean Geometry in
non-linear programming
In the above we have introduced method of using preferred n-dimensional system method to
151
solve linear programming,now let's introduce method of using preferred n-dimensional system
method to solve non-linear programming.
Although traditional methods of solving non-linear programming take extreme value theory as
basic theory,in concrete operating process,on one hand they painfully pursue optimal value(viz.
minimum value) of t,on the other hand they don't definitely regard t of objective function t = f (X)
as function of X ,instead they always omit the variable t. A variable is cut down to all
appearances,but both actual amount of calculation and computing results are not satisfying.
In fact,even t is regarded as function of X,and extremum of t exists,the extremum is not sure
to be optimal value to be found out. In example one,we find that mistake occured in interior point
method in linear programming. In fact,extremum questions have no direct、inevitable(only
indirect、accidnetal )relations with optimal value of objective function.
Based on the above reasons,preferred n-dimensional system method doesn't begin with
extremum problems,but deal with some problems according to projective theory of the seventh
chapter in this book.
2.1 Example 2
In order to keep the continuity of ideas,we will discuss it on the basis of example 1. At the
same time, for the conveniece of description,we omit English signs' subscripts of every vertex of
convex hexahedron in example 1,of which ,quadrilateral
DCRB,DCQA and △DAB
are separately formed of simultaneousness of
equations of constraint generic plane and generic
coordinate plane :
=
=++
=
=++
=
=++
0
66223
0
4622
0
4622
321
321
321
t
xxxR
t
xxxQ
t
xxxP
:
:
:
The two quadrilaterals and a triangle and three
coordinate planes x1ox2, x2ox3, x3ox1 form an enclosed
hexahedron(figure 47).
Just like curve of second order
y = ax2
figure 47
(in plane analytic geometry, it is an equation of parabola taking axis y as principal axis)when a→0,
it can be regarded as a straight line,the cut trace of objective generic plane
25x1+20x2+15x3-t = 0
of example one on generic coordinate palne can be regarded as a generic parabola whose radius of
curvature is infinite. These two terms can be mutually transformed under some certain cases,viz.
this quadratic function can be transformed into linear function:
first rewrite this quadratic function into
02
2
1 =′−′ xxaf:
152
and regard it as trace line of S′on plane x1ox2. And the cut trace ( be marked as S0′ for the
moment)of S′on generic coordinate plane Ox1x2x3 is a parabolic cylinder of three-dimensional
space,its directrix is f,and its generating line is parallel to vector
321 2641
6
41
13eee ′+′+′ , its equation is
=
=
′−′−
′
−′′
0
041
3
412
13 32
2
31
0
t
xx
xxa
S :
then the equation of S′is
041
3
412
13 32
2
31 =
′
−′+
′−′−′
xx
xxatS :
after rotation
=′
=′
−−=′
+−=′
tt
xx
xxx
xxx
33
212
11
41
4
41
5
241
5
41
4
S′becomes
( ) 034541
1
2
1354
41321
2
321 =++−
−+−− xxxx
xxa
tS:
Apparently,when a→0 and t = 0,here S will be coincident with Sπ(figure 48)of example
one graduallly.
Choose a = 0.05,it gets
figure 48
153
( ) 034541
1
2
1354
820
1321
2
321 =++−
−+−− xxxx
xxtS:
Now we make explanations of simulative economic issues for the above discussions:
A certain factory produces three kinds of products A,B,C. Assume the optimal production
plan is to planthe productions of A,B,C to be x1, x2, x3,and need operating the following three
kinds of costs and price system (unit: thousand yuan,the following is the same)at the same time:
the first: 2x1+ x2+2x3 ≤46 (P)
the second: x1+2x2+2x3 ≤46 (Q)
the third: 3x1+2x2+2x3 ≤66 (R)
x1, x2, x3≥0
The amount of profit produced ( assume it is t) is in obedience to equation of the above
objective generic curved surface S. Then it has the following model:
objective function:
( )321
2
321 345
41
1
2
1354
820
1Max xxx
xxxt +++
−+−=
constraint condition:
≥
≤++
≤++
≤++
0,,
66223
4622
4622
321
321
321
321
xxx
xxx
xxx
xxx
How to plan productions(viz. how x1, x2, x3 take values) of the three kinds of products, then it
can acquired maximum profit ?
In the example,objective function is a quadratic function,so it is a very typical non-linear
programming (belong to quadratic programming [15]
)in substance.
2.2 Methods and steps
It is generally agreed that,the core of solving non-linear programming is to determine search
direction and search range. Although it is not always necessary,but in order to illustrate questions
step by step,we also begin with these two questions.
2.2.1 Determination of search direction Assume objective function is
S:t = f (X)(X = xj ,j = 1 , 2 , … , n-1 ),
and when assume α is principal axis direction of generic curved surface t = f (X),it has
1 and , 0 ≠⋅≠⋅ αα OtOt (viz.:α is neither perpendicular nor parallel to upright axis t ,of
which O is the origin).
Order t= t0, substitute it into the equation of S and make it simultaneous with t = t0,then it gets
the cut trace ( )
=
=−=
0
0
0
0 tt
tXfS tt :
of S cut by generic plane t = t0 Of which,
f (X) - t0 = 0
because it doesn't include variable t ,then it is a projective generic cylindrical surface (the concept
of projective generic cylindrical surface can be found in §2 of the eighth chapter)including
154
0ttS = , its generating line is parallel to axis t . And make equation of the projective generic
cylindrical surface simultaneous with t = 0,it gets projection
( )
=
=−′
0
00
t
tXfS : π
of 0ttS = on generic coordinate plane In the projection process,
0ttS = is passive set, f (X)- t0 =0 is
path set,generic coordinate plane is contain set,0ttS = is image set. This process can be descripted
as:0ttS = is through direction of axis t and be projected onto generic coordinate plane,then it
gets πS ′ .
When determining search direction,it is very simple for cored generic curved surfaces:when
πS and πS ′ are cored generic curved surfaces,of which,
πS is the cut trace of S on generic coordinate plane,then direction of connecting line between centre of πS
and centre of πS ′ can be called search direction(and length
of the connecting line can be regarded as search range,in
figure 49, “d-c-t”is the abbreviation of "directviewing
common tangent"; “ c-c-l ” is the abbreviation of
"center-connecting line";“c-t-p-c-l”is the abbreviation of
"corresponding trace point connecting line";“c-t-p”is the
abbreviation of "corresponding trace point").
But under many cases, objective generic curved
surface is coreless,it can be divided into two different kind figure 49
of methods according to different cases——translation method and inflated translation method . we
will introduce them in several:
When πS ′ is parallel and congruent the cut trace πS of S on generic coordinate plane,use
ruler to make out trace line of common generic tangent line of πS ′ and Sπon x1ox2, we call this
trace line directviewing common tangent of πS and πS ′ .Direction of this directviewing common
tangent can be regarded as search direction(figure 50).
Because this method is equivalent to translating πS to πS ′ along search direction,then this
method is called translation method (or be called simple translation method).
When πS is similar to but not congruent with πS′ ,
we can add a constant term t0 to S and make it become
S +:t + t0 = f (X) or S
+:t = f (X)- t0
then make it simultaneous with t = 0,it gets projection
( )
=
=−+
0
00
t
tXfS : π
of S +
on generic coordinate plane. Now two cases may
appear:one is +πS coincident with πS′ and
+πS =
πS′ (search direction can't be determined, refer to it
in figure 52);the other is +πS not coincident with πS′ , figure 50
but +πS ≌ πS′ ,now we can use ruler to make out their directviewing common tangent ,direction
of the common tangent can be regarded as search direction(figure 50).
This method is equivalent to inflating the directviewing figure of πS to +πS ,then translate
155
it and make it become πS′ ,so this method is called inflated translation method〔if directviewing
common tangent is not existent,we can also take direction of connecting line of corresponding
trace points of πS and πS′ (or πS′ and πS ) as search direction. Solution of corresponding trace
points:first make out tangent of trace line of +πS or πS′ through a random point of trace line of
+πS or πS′ on plane x1ox2,then make out a parallel of this tangent and make it tangent to trace line
of πS′ or +πS on plane x1ox2,the two points of tangency are the corresponding trace points of
+πS and πS′ .Solution of corresponding trace points of simple translation method is similar to this
(figure 49)〕.
But when +πS is coincident with πS′ ,search direction can't be determined,the above
discussion naturally doesn't include this kind of questions.
2.2.2 Determination of search range
So-called search range is the directviewing distances of every vertex of feasible region to the
corresponding points of Sπ(when objective generic plane is in the shape of plane,the directviewing
distance is equal to the length of line segment between trace point of generic point that the vertex is
on and foot point of perpendicular line of objective trace line). Here,we may as well look back
example one,there,the directviewing distances (can be regarded as search range) between every
vertex of convex polyhedron and objective trace line are always different in different n-dimensional
system. But if ensure that objective generic plane is in the shape of plane,then the directviewing
distances between every vertex to objective trace line are changeless. But in non-linear
programming,many objective generic curved surface can't be transformed into the shape of
hollow curved surface,this bring some difficulty to solutions of this kind of questions,so it must be
treated distinctively.
First discuss cases of objective generic curved surfaces being transformed into hollow ones. It
is divided into two cases
a.The case that search direction can be determined Assume the dimension of generic
generating line of objective generic curved surface
S:Max t = f ( xi) i =1 , 2 , … , n-1
is n-3,and it can be denoted by the following n-3 vectors
Xi1 e1 +Xi2 e2 +Xi+2 ei+2 i =1 , 2 ,… , n-3
by proper orthogonal transformation,use transform
Xi+2 ei+2 □ -Xi1 e1 - Xi2 e2 i =1 , 2 ,… , n-3
then it gets a generic coordinate plane of preferred n-dimensional system that make objective
generic curved surface in the shape of plane. Optimum point theorem of example one can be extend
to this:through trace point of generic points that every vertex of feasible region is on and make
out a parallel of search direction of objective generic curved surface ,it intersects objective trace
line(here objective trace line is a curve) at a point,then,length of line segment between the two
points is called search range from objective trace line to the vertex. When length of search range
from objective trace line to a vertex is the longest,the vertex is optimum point.
But,this method is only suitable for cases of using simple translation method to determine
search direction,but it is not suitable for cases of using inflated translation method to determine
search direction.
Under the case of inflated translation method,assume search direction has been determined,and πS′ intersects a vertex C of feasible region. Through trace of generic point that C is on and
make tangent of trace line of πS′ on plane x1ox2,then make parallel of the tangent and make it
156
tangent to trace line of +πS . The length of line segment (be called connecting line of the
corresponding trace points )between two tangent points( be called corresponding trace points) is
just the search range with relation to point C (figure 49).
Example 1.Find out optimum point of example 2.
Solution::::From figure 48,because when a ≠ 0 ,the figure of S is very complicated,translation is also very troublesome,so we may as well choose the figure of generic straight line
0 : =aSπ when a = 0 and translate it. The method is:first order 41
205=t (≈32.0156),it gets
5x1 + 4x2 + 3x3 - 205 = 0
and make it simultaneous with equation 4x1-5x2=0 of trace line of generic principal axis (be called
axis trace for short, refer to it in figure 51)of Sπ,equation x3=0 and t = 0,it has
=
=
=−
=++
0
0
054
205345
3
21
321
t
x
xx
xxx
Solve the equation set , it gets x1=25 , x2 = 20 , x3 = 0 , t = 0 . This is equal to translating πS
along direction of its axis trace ,solution of the equation set is equal to coordinate of vertex Jn′(25,20,0,0) of Sπafter being translated to new position. And make the figure of πS′ (viz. when
a = 0.05 ) taking Jn′as vertex,then the translation from πS to πS′ is finished. As will be readily
seen,direction of nOJ is the search direction to be found out.
Vertex Jn′is regarded as projection trace of a generic point N of S. Choose proper direction
and unit length and make figure of upright axis t,lead Jn′to generic point N. And continuously
translate πS along direction of axis trace,then points of πS leaved in feasible region will lessen
gradually,until at last only one point left is coincident with A (continuously translate,then all points
of πS will be translated out of feasible region). From direct view,in search range,the
figure 51
157
distance between πS and vertex A is the longest,so A is the optimum point to be found.
Coordinate of a vertex Jm′of πS in the new position is
x1≈29.698, x2≈23.759 , x3 = 0 , t = 0.
Through Jm′and make parallel of axis t ,it intersects the extension line of OJn ( Jn is on
generic point N,it is opposite trace of N ) at Jm,then Jm is the opposite trace of generic point M,so the intersection point is marked as M. Because directviewingly measure | Jm′Jm |≈38.032 ,
then it gets t ≈38.032,this is in accord with the result of making directview AA′∥Ot through
A and it intersecting S at point A′.
Then it gets the optimum solution
x1 = 0 , x2 = 0 , x3 = 23 , t ≈38.032.
Viz. we shouled plan production of A 0,production of B 0,production of C 23 pieces,then we
can get the maximum profit about 38.032 thousand yuan.
b. The case that search direction can't be determined
For objective generic curved surface whose search direction can't be determined,every vertex
quality of convex polyhedron can be judged by determining expansivity. So-called expansivity is
length ratio of length of connecting line between a vertex and centre of πS to length of line
segment between intersection point of the connecting line and πS and centre of πS .
Expansivity is greater,equality is better.
When πS is in the shape of hollow,this expansivity is easy to find out in the figure. But
when πS is solid, becasuse the difficulty is very great,then it is difficult to adopt.
Example 2: Assume constraint condition is the same with the example,and objective function
is 125
541
3
41
4
41
5
400
41
13
41
5
41
4
Max
2
321
2
321
−
+++
+
+−
=
xxxxxx
t
find out optimum solution.
Solution::::Substitute t = 3 into objective function and make it simultaneous with t = 0,it gets the
directview figure of
=
=−
+++
+
+−′
0
0425
541
3
41
4
41
5
400
41
13
41
5
41
4
S
2
321
2
321
t
xxxxxx:π
it is coincident with
=
=−−
+++
+
+−′+
0
0425
541
3
41
4
41
5
400
41
13
41
5
41
4
S
2
321
2
321
t
t
xxxxxx:π
(figure 52),so this is an example that search direction can't be determined. The centre of πS is
located at generic E ,trace of the generic point is
. 0,0,41
20,
41
25
−−eJ
Connecting lines EA,EB,EC,ED,ER,EQ between E and every vertex separately intersect
158
πS at points A′,B′,C′,D′,R′,Q′,and from direct view it has
, AE
EA
QE
EQ
BE
EB
RE
ER
DE
ED
CE
EC
′>
′>
′>
′>
′>
′
So,C(10,18,0,0) is the optimum point to be found out. Substitute coordinate of C into equation
of S,it gets t ≈ 22.29.
figure 52
In the above examples,we determine search range or expansivity by comparison ,linear
programming of example one also adopt this method. We call this method comparison method by
a joint name.
2.3 The case that objective generic curved surface is solid
Let's introduce solution process of this kind of questions by a concrete example,then extend it
to a generalized case.
Because objective function t is really regarded as variable,then case that objective function
is implicit function has come into consideration. Then ,we design the following objective
function into implicit function,and call it objective implicit function.
Example 3:Assume constraint condition still the same with that of example one,then
objective implicit function is
04003
32096128
3
64
15
184
15
262
25
360
3
115716
25
160
25
265
2132
121
22
3
2
2
2
1
=+−++−−
−−+−++
txxtxtx
txxxtxxxS:
try to find out optimum solution.
Solution::::arrange objective implicit function into quadratic function of t
159
04009612825
36016
25
160
25
265
3
320
3
64
15
184
15
262
3
1157
2121
2
3
2
2
2
1
321
2
=
++++++−
−
++++
xxxxxxx
txxxt
from root expression
a
c
a
b
a
b
a
acbbt −±−=
−±−=
2
22
422
4
of quadratic equation with one unknown at2 + bt + c = 0 find out
1414000102403391044526725
5888
5
8384
25
127366456560
25
563824
25
936976
1157
1
1157
160
1157
32
5785
92
5785
131
321
32
31212
3
2
2
2
1
321
+++++
++++
±
+−−−−=
xxxxx
xxxxx
xx
xxxt
Here,objective function t has two roots, it represents that πS has two intersection
opportunities with every vertex of convex polyhedron in translation. But because of solving
maximum value,then we choose
1414000102403391044526725
5888
5
8384
25
127366456560
25
563824
25
936976
1157
1
1157
160
1157
32
5785
92
5785
131
321
32
31212
3
2
2
2
1
321
+++++
++++
+
+−−−−=
xxxxx
xxxxx
xx
xxxt
Here,the directview figure of S is a solid generic paraboloid,so the above method looks
helpless. Then we borrow ideas of traditional nonlinear programming problem using limit method
(note:not extremum method) to solve to solvethis kind of questions.
Take t11= 0.89, t21= 2.19, t31= 3.47 in turn,substitute them into S successively,choose two
among six inequalities of constraint condition each time and change them into equalities (the other
four don't change ,but bracket them and put them under all equalities) then make them separately
simultaneous with S,choose the following equation sets according with constraint condition (all
solution are feasible solutions),divide them into groups of three each according to t ,they are
called feasible groups:
[ ]
≥≤++≤++≤++
=
===
+++−−
=
=
;; ;
:
06622346224622
0
47.3,19.2,89.0
141400010240565601157
1
1157
160
1157
32
0
0
group
feasible
row
th-
1 the
3321321321
3
2
3
3
2
1
1
xxxxxxxxxx
t
xxx
x
x
160
[ ]
≥≤++≤++≤++
=
===
+++−−
=
=
06622346224622
0
47.3,19.2,89.0
141400045267225
936976
1157
1
1157
160
5785
131
0
0
group
feasible
row
th-
2 the
1321321321
1
2
11
3
2
1
xxxxxxxxxx
t
xxx
x
x
;; ;
:
[ ]
≥≤++≤++≤++
=
===
+++−−
=
=
06622346224622
0
47.3,19.2,89.0
14140003391045638241157
1
1157
160
5785
92
0
0
group
feasible
row
th-
3 the
2321321321
2
2
22
3
1
1
xxxxxxxxxx
t
xxx
x
x
;; ;
:
these three feasible groups represent intersection points of three axes x1,x2,x3 of three edges of
convex hexahedron and gradually translatory πS′ ,feasible solutions (refer to it in table 1)found out
separately represent coordinates of these intersection points .
The same variable's value of the same t of different feasible groups is usually different,we call
their non-negative difference value absolute difference of corresponding solutions of the feasible
groups. But,there are only three non-negative differences of the same variable of the same t ,we
only choose the maximum and call it maximum absolute difference.
Table 1 gives maximum absolute difference of corresponding solutions of the above feasible
groups:
table 1(accurate to 0.01)
Value
of t
the orre-
sponding
solution
the 11
-th feasible
group
the 21
-th feasible
group
the 31
-th feasible
group
maximum absolute
difference
x1 0 0 0 0
x2 0 0 0 0
0.89
x3 0 0 0 0
x1 0 9.1 0 9.1
x2 0 0 11.6 11.6
2.19
x3 11.8 0 0 11.8
x1 0 17.9 0 17.9
x2 0 0 23 23
3.47
x3 19.5 0 0 19.5
It can be observed from table 1,with the increase of value of t ,absolute differences of the
corresponding solutions will increase gradually. But we expect these absolute differences not only
decrease gradually but also tend to zero quickly.
161
And choose t12= 4.1 , t22= 4.2 , t32= 4.22 , t42= 4.24 , t52= 4.2588
in tur, and substitute them into objective function S successively,choose two among six inequalities
of constraint condition each time and make them simultaneous with S which is substituted into
new t value ,and choose solutions found out which are all equation sets of feasible solutions,divide them into three groups of five each according to t:
[ ]
≥≥≤++≤++
=
=====
+++
++++
+−−−
=
=++
0066223 4622
0
2588.4,24.4,22.4,2.4,1.4
1414000339104452672
25
1273664
25
563824
25
936976
1157
1
1157
160
5785
92
5785
131
0
4622
group
feasible
row
th-1 the
21321321
21
21
2
2
2
1
21
3
321
2
xxxxxxxx
t
xx
xxxx
xx
x
xxx
,,,
:
[ ]
≥≥≤++≤++
=
=====
+++
++++
+−−−
=
=++
004622 4622
0
2588.4,24.4,22.4,2.4,1.4
1414000339104452672
25
1273664
25
563824
25
936976
1157
1
1157
160
5785
92
5785
131
0
66223
group
feasible
row
th-2 the
21321321
21
21
2
2
2
1
21
3
321
2
xxxxxxxx
t
xx
xxxx
xx
x
xxx
,,,
:
[ ]
≥≥≥≤++
=
=====
+++++
+++++
+−−−−
=++
=++
000 4622
0
2588.4,24.4,22.4,2.4,1.4
14140003391044526725
5888
5
8384
25
127366456560
25
563824
25
936976
1157
1
1157
160
1157
32
5785
92
5785
131
0223
6622
group
feasible
row
th-3 the
321321
213231
212
3
2
2
2
1
321
321
321
2
xxxxxx
t
xxxxxx
xxx
xx
xxx
xxx
xxx
,,,
:
these three feasible groups represent intersection points of three edges QC,RC,DC of convex
162
hexahedron separately intersecting πS′ five times,coordinates of these intersection points are just
solutions of these three feasible groups. Table 2 give all feasible solutions of these three feasible
groups.
table 2(accurate to 0.0001)
Value
of t
the orre-
sponding
solution
the 12
-th feasible
group
the 22
-th feasible
group
the 32
-th feasible
group
maximum
absolute
difference
x1 8.078 17.77 10 9.692
x2 18.961 6.345 16.7656 12.616
4.1
x3 0 0 1.22 1.22
x1 9.2926 12.342 10 3.0494
x2 18.3537 14.487 17.5554 3.8667
4.2
x3 0 0 0.442 0.442
x1 9.5336 11.504 10 1.9704
x2 18.2332 15.744 17.702 2.4892
4.22
x3 0 0 0.2597 0.2597
x1 9.7742 10.71 10 0.9358
x2 18.1129 16.935 17.8591 1.1779
4.24
x3 0 0 0.14 0.14
x1 10 10 10 0
x2 18 18 18 0
4.2588
x3 0 0 0 0
It can be observed from table 2,absolute differences of the corresponding solutions decrease
gradually,it represents that the distances between intersection points of πS′ and three edges QC,RC,DC also decrease gradually. At last, πS′ intersects these three edges at optimum point C.
This method is also applicable for the case that objective generic curved surface is "hollow",it is also applicable for some linear programmings.For example in example one,when t separately
choose 570、590 and 610 , S separately intersect DC1, Q2C1 and R1C1 at three points,tabel 3
lists coordinates of these intersection points:
table 3
Value
of t
the orre-
sponding
solution
DC1
Q2C1
R1C1
maximum absolute
difference
x1
10
3
22
18
3
32
x2
10
3
58
6
3
40
570
x3 8 0 0 8
x1
10
3
26
14
3
16
x2
14
3
56
12
3
20
590
x3 4 0 0 4
x1 10 10 10 0
x2 18 18 18 0
610
x3 0 0 0 0
163
As will be readily seen from table 3,when t is far from maximum value ,the maximum
absolute difference is also much more than zero;When t takes maximum value 610,maximum
absolute differences of the corresponding solutions are all zero.
In order to differentiate it from comparison method introduced before,we call this method
absolute difference method.
In non-linear programming field,because of lacking of ideal tools in the past ,so a kind of
all-purpose common algorithm hasn't been formed. But,after preferred n-dimensional system
appeared ,this situation will change much. Here “absolute difference method” introduced will be
general all-purpose algorithm of non-linear programming. aithough this method is lack of use value
because of its complexity,but by people's studious exploration,this method will become
mainstream algorithm of linear programming and non-linear programming questions ultimately.
Now,we will extend example 3 to a general case:In general quadratic programming,constraint conditions are most linear inequalities. In order to reduce discussion difficulty ,we also
assume constraint conditions as linear inequalities.
Assume a generic curved surface
S:t = f (X) ( X = xj;j = 1 , 2 , …, n-1)
of n-dimensional space is regarded as figure of objective generic curved surface,each equation of
system of linear equations
gi ( X )≤ai ( i =1 , 2 ,…, r )
X≥0
is regarded as each generic plane forming simplex(viz. convex polyhedron). And assume cut trace
of S on generic coordinate plane is ( )
=
=
0
0
t
XfS : π
Choose a proper value of t as initial value(this is similar to interior point method determining initial
point,but they are also quite different two things)and substitute it into S then make it simultaneous
with t = t0,then it gets the cut trace
( )
=
==
0
0
0 tt
tXfS tt :
of S cut by generic plane t = t0 Of which,f (X) = t0 is a projective generic cylinder surface including
0ttS = and its generating line parallel to axis t .Make equation of the projective generic cylinder
surface simultaneous with t = 0 ,it gets projection
( )
=
=′
0
0
t
tXfS : π
In each feasible group, πS′ intersects one edge of convex polyhedron. Because under normal
conditions,the number of constraint equations is not more than n-1,so, πS′ will at most separately
intersect k edges(k≤n-1) of convex polyhedron at one point
( xk1 , xk2 , … , xk,n-1 , 0 ) ( k = 1 , 2 , … , r ;r≤n-1 ).
Then make t take t1 , t2 , … , tδ in turn , make t0<t1<t2<…<tδ ,and make
πS′ change position constantly and separately intersect the k edges of convex polyhedron at one
point ( xl k1 , xl k2 , … , xl k,n-1 , 0 ) ( l =1 , 2 , … ,δ; k = 1 , 2 , … , r ;r≤n-1 ) one after
another. Coordinate of the point can be found out by using the following method:
In all r+n-1 inequalities of system of linear equations
gi (X)≤ai ( i = 1 , 2 , … , r )
164
X≥0,
take those inequalities which satisfy the given tm ( m =0 , 1, 2, …, δ;δis positive integer) and
change them into equalities, every n-2 of them can be a group and separately be simultaneous
with
( ) ( )
=
==
0
,,1,0
t
mtXf m δL
(the other r+1 inequalities keep changeless,but bracket them and put them under all equalities)and form feasible group(can form n-1 feasible groups at most).
( )
( ) ( )
( )
−≤≤−=≥
≤≤−=≤
=
==
−≤≤−−==
≤≤−−==
11,,2,1 0
1,,2,1
0
,,2,1
112,,2,1,0 0
12,,2,1,0
group
feasible
row
th-
the
222)(
222)()(
111)(
111)()(
22
2222
1 1
1111
njpnqx
riqnpaXg
t
mtXf
njpnqx
riqnpaXg
k
qj
pipi
m
qj
pipi
l
;
;
;
;
:
L
L
L
L
L
δ
(in the formula,r is the rank of coefficient matrix of system of linear equations gi (X)≤ai) and then
separately solve them .
Assume xημ and xξμ are the same variable belonging to different feasible group,then xημ and
xξμ are called corresponding solutions of πS′ ,and 0≥− ξµηµ xx is called absolute difference of
the corresponding solution. When πS′ approximates to optimum point,it has 0→− ξµηµ xx .
But,when 0=− ξµηµ xx , S will not always has optimum solution. Near some
non-optimum points of convex polyhedron,it also has the case 0→− ξµηµ xx . Assume when
0=− ξµηµ xx ,correspondingly it has t = tμ,if we want to check whether tμ has maximum
value or not,if only choose a minimum value Δ(Δ>0,Δ should correspondent with the exact
value required.For example,if we demand it accurate to 0.0001, then we can order Δ=0.0001) and
substitute t = tμ+Δ into objective function S and make it simultaneous with other equations of
( )( ) ( )! 1! 2
!1
+−
−+
rn
nr equations,find out whether feasible solutions exist or not . If all solutions are
unfeasible solutions ,then t = µt is the maximum value;If feasible solutions still exist ,then
t = µt is not the maximum value,computation should be carried on.
Considering equation sets which have unfeasible solutions ( be called unfeasible sets) can also
be determined by computation actually,then we will compute ( )
( ) ( )! 1! 2
!1
+−
−+
rn
nr equation sets
each time,the workload is very great. So,we must manage to simplify computation steps . In the
following we will introduce two methods of simplifying computation steps:graded search
methodand unconstrained search method.
First,graded search method. The concrete practice is,divide constraint condition into two
sections:
The first section is called preferred constraint equation ,it is composed of principal part
g (X)≥0 of constraint condition. According to what we introduced in §1,constraint condition can
be classified by magnitude of constraint force. So this section of constraint conditions all have great
constraint force;
165
The second section is called prevailing(or be called recognized) constraint equation,it is
composed of subsidiary section X≥0 of constraint condition . Constraint force of this section of
constraint condition is usually very weak.
Because in the model we created, πS′ usually translates from near the origin along search
direction ,its trace line moves to the first quartile of plane x1ox2 gradually. So in feasible groups(or
unfeasible groups),constraint conditions all composed of prevailing constraint equations are called
initial group;Constraint conditions all composed of preferred constraint equations are called
terminative group;Constraint conditions composed of both preferred and prevailing equations are
called intermediate group. Of which,according to the amount of prevailing equations included in
intermediate group, it can also be divided into first grade intermediate group,second grade
intermediate group,…,etc.
In example 3,the11-th, 21-th, 31-th feasible groups are all initial groups,the 12-th, the
22-th feasible groups are intermediate groups(because in the example there is only one prevailing
constraint equation in the intermediate groups,so they are called intermediate group a joint
name),the 32-th feasible group is terminative group.
Considering optimum solution is usually near terminative group,so in the computation of the
11-th , the 21-th , the 31-th three feasible groups,the value of t increases very fast,computational
accuracy is also very low(in fact, the value of t may increase much faster,accuracy may be even
lower. Even,we can consider giving up the computation of initial group and directly carry the
computation of intermediate group or terminative group);And in the 12-th , the 22-th , the 32-th
feasible groups,the increase speed of the value of t slow down,computational accuracy increased.
Second,unconstrained search method.
In spite it is feasible group or unfeasible group,we choose one among each of them and
compute together. For example in the equation set
( )
( ) ( )
=
==
−≤≤−−==
≤≤−−==
0
,,2,1
112,,2,1,0 0
12,,2,1,0
111)(
111)()(
1 1
1111
t
mtXf
njpnqx
riqnpaXg
m
qj
pipi
δL
L
L
;
;
( )
−≤≤−=≥
≤≤−=≤
11,,2,1 0
1,,2,1
group
positive
-contra
222)(
222)()(
22
2222
njpnqx
riqnpaXg
qj
pipi
;
;
L
L
inequalities bracketed are deleted,it is equivalent to being deleted constraint condition,so it is
called unconstrained search method. But constraint condition can't be wholly ignored,or else we
can't determine solutions found out are feasible solutions or not. So we put constraint condition
behind equation set as contrapositive group.
This equation set represents an edge and its extension line of convex polyhedron.
In concrete compution,we usually choose two adjacent edges' equations of the same vertex
of convex polyhedron.
Determine whether two equation sets represent two adjacent edges of the same vertex of
convex polyhedron or not,the method is very simple:according to interleaving theorem (refer it in
theorem 1——theorem 4 of the sixth chapter) between two linear figures of the sixth chapter,each
equation set has n-1 equalities,of which, n-2 equalities are the same,only one is different. When
the two equation sets are simultaneous,simultaneous group has the unique solution ,and the
166
unique solution is feasible solution.
But,solving the unique solution of simultaneous group of two eqution sets need increase
amount of calculation invisibly. So,in the two equation sets,we choose one having feasible
solution,the other has no feasible solution. Of which,equation set having feasible solutions (can be
called the original equation set for convenience) represents an edge of convex polyhedron,equation
set having no feasible solution represents extension line of the edge's adjacent edge. When
πS′ translates along search direction,absolute difference of corresponding solutions of two equation
sets should decrease gradually.If absolute difference of corresponding solutions of two equation
sets doesn't decrease but increase gradually,it represents that vertex of intersection position of two
edges has fallen behind πS′ ,now we should give up the adjacent edge's equation set(be called old
equation set for the convenience of description)having unfeasible solution,and choose the other
equation set(be called new equation set) having unfeasible solution. In order to ensure that absolute
difference of corresponding solutions between new equation set and the original equation set
decreases gradually,adjacent edge(its extension line) denoted by the new
equation set should intersect edge denoted by the original equation set at another vertex. This
demand that new equation is secondary intermediate group of the original equation set, To be
specific,the new equation set should have one prevailing constraint equation less than that of the
original equation set but one preferred constraint equation more than that of the original equation
set.
When absolute difference is zero, πS′ translates to intersection point of two edges of convex
polyhedron denoted by two equation sets. Further translate it,then absolute difference again
increases gradually, πS′ changes from intersecting extension line of one of two edges to intersecting
this edge,and originally it changes from intersecting the other edge to intersecting its extension
line .So,as long as absolute difference gradually decreases at the beginning,and suddenly increases
gradually in the sequel,it shows that πS′ has translated intersection point of the two edges.Now we
can choose another extension line of an edge having common point with the new edge and make
itcontinuously intersect the translating πS′ .……,until the optimum point is found. For example,in example 3,
≥
≤++
≤++
≤++
=
=
=
0
66223
4622
4622
ositive
-contrap
0
0
0
2
321
321
321
3
1
x
xxx
xxx
xxx
t
x
x
and
≥
≥≤++
≤++
=
=
=++
0
0
66223
4622
ositive
-contrap
0
0
4622
2
1
321
321
3
321
x
x
xxx
xxx
t
x
xxx
167
are separately equations of OQ and CQ . Take 2.418, 2.937 and 3.47 of t successively,and order
x3=0,substitute it into S ,it gets
3.472.937,2.418,
1414000339104452672
25
1273664
25
563824
25
936976
1157
1
1157
160
5785
92
5785
131
21
21
2
2
2
121
===
=
+++
++++−−−
xx
xxxxxx
and make them separately simultaneous with equations of OQ and CQ,the results are listed in table
4:
table 4( accurate to 0.001)
Value
of t
the orre-
sponding
solution
OQ
CQ
maximum absolute
difference
x1 0 -23.4 23.4
x2 13.71 34.7 20.99
2.418
x3 0 0 0
x1 0 -8 8
x2 18.3 27 8.7
2.937
x3 0 0 0
x1 0 0 0
x2 27 27 0
3.47
x3 0 0 0
It is observed from data of the table,when t<3.47 of objective function S,solution of
equation of CQ is unfeasible solution .So now πS′ intersects CQ at extension line. When t≥3.47,πS′ intersects CQ in feasible region,but intersects extension line of OQ in the outside of feasible
region. Now we can change equation of OQ into equation of CR and carry on the computation.……
In this method, πS′ continuously intersects another and yet another edge of convex
polyhedron,so,we call this method “one edge method”. Traditional method(such as interiorpoint method) using extreme value theory to solve
non-linear programming tries to include objective function and constraint condition these very
complex equations all in an equation. Although this idea is very new,it is unavoidable to be too
naive. Because such a result will always change property of equation ,its accuracy and reliability
are doubtable. And because the method is too complex,it is not only very inconvenient in
application,but also greatly limits people's study course in nonlinear programming field.
Preferred n-dimensional system method is an example of successfully applying High
Dimensional Euclidean Geometry .
Because of the appearance of preferred n-dimensional system method,we may observe
problems and consider problems at a higher initial point,so,many concepts which are very
ambiguous in the past become transparent. For example constant distance theorem and optimum
point theorem of §1 in this chaper,only on the basis of oblique axes transform and oblique axes
draughting ,it can be found and successfully solved.
Preferred n-dimensional system method regard objective function t as a variable,it is one
dimension more than traditional sloution in spacial thinking manner. The change of thinking
manner brought by preferred n-dimensional system graphical method help us realize the transform
from planar thinking manner to spatial thinking manner. What does this imply? It implies that we
put objects to be observed and studied all in a very spacious planar region (viz. generic coordinate
168
plane),but we stand in in the air to observe and study them i. Apparently,it has a more wide-field
than we standing in the original planar region to observe and study them,and more
easily to grasp the overall situation.
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2.Hongfan Basis of Discrete MathematicsWuchang:Huazhong Institute of Science and
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3.Edited by School of Mathematics of Nankai University Introduction of Analytical Geometry
of Three Dimensions Beijing:People's Education Press,1978
4.NI Guoxi Common Matrix Theory and Method Shanghai:Shanghai Scientific and
Technical Publishers,1984
5.Zhang Rijin,Li Shenglin Graphics Set of Stereoscopic Projection Beijing:National
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8.[USSR]Felibf wrote Xie Shenjian,Zhou Jiyi translated Hyperspace Descriptive
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12.《Mathematics Handbook》Compile GroupMathematics Handbook Beijing:People's
Education Press,1979
13.Jian Zhaoquan Solving Included Angle Between Two Planes of E n By Unit Circle
Method Engineering Graphics Archieves,1986(1)
14.Edited by Beijing Institute of Education Higher Algebra Beijing:Beijing Press,1979
15.Qian Songdi et al. wrote Operations Research Beijing:Tsinghua University Press,1990
169
Postscript
Owing to help and guidance of many experts and scholars and warm-hearted people of all
circles,then the author can dedicate its self taught and studied achievement of 17 years to readers
of wide scope. In initial stages of study,many scholars and experts gave the author illuminative guidance. Of
which,the most outstanding ones are Mr Li Shenglin of BeiHang University、Mr You Chengye of
Peking University、Mr Mei Xiangming of Capital Normal University ,Mr Fu Ruonan of Beijing
Normal University and Lu Daxiong of Beijing Institute of Education. When the study made some progress,Mr Zhou Jiyi of Tsinghua University、Mr Zhang Yunhe of
Beijing Institute of technology 、Mr Zhang Rijin of College of Building Materials and Light
Industry of Beijing Union University、Mr Liu Guangxu of Tianjin Nankai University and Mr Zhu
Xinmin of 《Guangming Daily》Press gave acknowledge and supporting in time,and made some
basic ideas and methods be published Beijing《Engineering Graphics Archieves》and《Potential
Science Magazine》,and be lucky to be published in annual meeting of 1987 in Beijing Engineering
Graphics Society.
Section of making a point of this book (the first chaper)is finished with the cooperation of
professor Hou Bingtao of professor Hou Bingtao of Beijing Institute of armored forces of Chinese
People's Liberation Army. Mr Hou thoroughly corrected the contents of some theorems and gave
strict and scientific proof to most theorems in person after reading over the manuscript of the first
chapter for many times,then section of making a point of this book kept its legs steady in science at
last. He also called section of making a point and methods extended by it "relation " method.
At the beginning this book is given to Professor Shen Yidan of Beijing Institute of Technology
in the form of series of manuscripts .Mr Shen carefully checked and approved series of manuscripts
and gave high appraise for them. He suggested that the author arrange series of manuscripts into
book manuscripts and directly publish it. After the book manuscripts are finished,Mr Shen
reviewed and embellished it for many times patiently,and worked out name for this book in person.
He also rushed about in many ways,appealed to concerned departments and experts to give
attention to ideas and methods of this book. And contacted issuing office in person using his
reputation and relation, did his best to make this book be published.Herein the author gave cordial
thanks and heartfelt admiration to him .
Besides,researcher Mr Pan Jianzhong of Geometry and Topology Research Office of
Graduate School of Chinese Academy of Sciences gave enthusiastic guidence to revision and
modification of this book,and strictly checked it on,hereon I especially give thanks to him.
During publication of this book,Mr Li Yingan of Atomic Energy Press 、Mr Ma Bohua of
Population Press、Mr Liu Wenchao of Military Medica Science Chinese、Mr Lan Chaowen of
Institute of Automation of Chinese Academy of Sciences and Mr Chen Linfeng of Beijing No.606
post box also gave great support,hereon I also give thanks to them!
Auto CAD making of iconographs in this book are finished under the guidance of Mr Bu
Yanwu,I also especially give thanks to him!
Limited to qualification of the author,there must be much error in this book,readers are
expected to point out them so that they can be corrected.
Contents of the ninth chapter aren't agreed with or acknowledged by any experts,the author is
responsible for the responsibility .
170
Compiler
July,1996 Beijing
Solutions to part exercises The first chapter
1.1 be changed into 15e1-14e2.
1.2 be changed into 7
33
7
1321 ee ′−′ .
1.3 -5e1+e2.
1.4 59e1-34e2.
1.5 ( ) ( ) 21 2472812 ee −−− .
The third chapter
3.4 Because generic points are divided into sufficient-order、difficient-order and zero-order,
then accordingly,there are three graphic methods of general graphic method、indirect graphic
method、and direct method.
3.8 36x1+126x2-198x3-132x4-88x5-99x6-72=0.
3.10 x1=8, x2= 4, x3=6, x4=10, x5=12, x6=18, x7=4 .
The sixth chapter
6.1 L1、L2 are mutually interleaving planes,the distance between them is about 5.5149.
The seven chapter
7.6 φ≈71°45′.
7.7 φ≈47°48′.
7.8 φ≈89°53′.
7.9 φ≈73°34′.
The eighth chapter
8.4 φ3≈74°12′(the others are all zero angles).
8.5 φ1≈25°31′, φ2≈64°30′.
8.6 φ1≈64°04′, φ2≈33°46′.
8.7 The characteristic root found out is λ1=1,
, 6764.1328659 , 3236.116328659 32 ≈−=≈+= λλ
so the nonzero included angle between two figures are φ2≈84°41′, φ3≈39°26′.
171
Indexes of special terms or symbols
vocable or symbol the ubiety of pages
"relation" method-----------------------------------------------------------------------------------------------1
has relation σ:σ ------------------------------------------------------------------------------------------2
has no relation σ: --------------------------------------------------------------------------------------2
overlapping direction-----------------------------------------------------------------------------------------14
oblique axes draughting--------------------------------------------------------------------------------preface
preferred n-dimensional system-----------------------------------------------------------------------------17
image relation sign“□”------------------------------------------------------------------------------------19
shadow relation sign“○”-----------------------------------------------------------------------------------19
oblique axes transform-----------------------------------------------------------------------------------------19
principal overlapping direction-------------------------------------------------------------------------------22
principal overlapping space-----------------------------------------------------------------------------------22
generic point---------------------------------------------------------------------------------------------------23
sufficient-order generic point--------------------------------------------------------------------------------30
difficient-order generic point--------------------------------------------------------------------------------23
zero-order generic point--------------------------------------------------------------------------------------23
projective trace------------------------------------------------------------------------------------------------24
opposite trace--------------------------------------------------------------------------------------------------24
generic curved surface----------------------------------------------------------------------------------------28
generic curve---------------------------------------------------------------------------------------------------28
generic plane---------------------------------------------------------------------------------------------------30
generic straight line-------------------------------------------------------------------------------------------30
generic coordinate plane--------------------------------------------------------------------------------------30
generic sphere--------------------------------------------------------------------------------------------------32
generic circle---------------------------------------------------------------------------------------------------42
generic cylinder surface---------------------------------------------------------------------------------------42
solid generic curved surface----------------------------------------------------------------------------------56
hollow generic curved surface-------------------------------------------------------------------------------56
generic circular conical surface------------------------------------------------------------------------------59
generic paraboloid---------------------------------------------------------------------------------------------61
projective generic cylinder surface-------------------------------------------------------------------------113
dimension theorem--------------------------------------------------------------------------------------------29
direct graphic method-----------------------------------------------------------------------------------------30
simplex principal overlapping direction---------------------------------------------------------------------36
oblique number“S(xs)” ----------------------------------------------------------------------------------36-37
oblique sign“A(xb)”-----------------------------------------------------------------------------------------37
indirect graphic method--------------------------------------------------------------------------------------30
general graphic method--------------------------------------------------------------------------------------30
points having common generic------------------------------------------------------------------------------33
points having common generic sign“ ”--------------------------------------------------------------33
points having common generic theory-----------------------------------------------------------------33-34
leading-axis method-------------------------------------------------------------------------------------------59
cutting-trace method------------------------------------------------------------------------------------------59
172
synthetical graphic method-----------------------------------------------------------------------------------61
singly through--------------------------------------------------------------------------------------------------62
mutually through-------------------------------------------- -------------------------------------------------73
through point--------------------------------------------------------------------------------------------------73
mutually interleaving-----------------------------------------------------------------------------------------74
orthodromic vector-------------------------------------------------------------------------------------------77
orthodromic space----------------------------------------------------------------------------------------76-77
common orthodromic space---------------------------------------------------------------------------------78
normal space--------------------------------------------------------------------------------------------------78
common-orthodromic normal space-----------------------------------------------------------------------78
common normal space---------------------------------------------------------------------------------------79
exterior direct sum -------------------------------------------------------------------------------------------81
exterior sum---------------------------------------------------------------------------------------------------81
passive set------------------------------------------------------------------------------------------------------87
path set----------------------------------------------------------------------------------------------------------87
accepting set---------------------------------------------------------------------------------------------------87
Jianshi solution theorem-------------------------------------------------------------------------------------90
angle vector----------------------------------------------------------------------------------------------------92
orthogonal-unit vector----------------------------------------------------------------------------------------92
common vector------------------------------------------------------------------------------------------------93
uncommon vector---------------------------------------------------------------------------------------------93
Jianshi solution solution-------------------------------------------------------------------------------------94
Jianshi method------------------------------------------------------------------------------------------------94
exterior product method and inner product method----------------------------------------------------106
exterior product----------------------------------------------------------------------------------------------106
Jianshisolution principle------------------------------------------------------------------------------------118
coefficient theorem------------------------------------------------------------------------------------------120
exterior product method and projective method---------------------------------------------------------132
objective trace line-------------------------------------------------------------------------------------------141
constant distance theorem----------------------------------------------------------------------------------142
optimum point theorem-------------------------------------------------------------------------------------144
translation method--------------------------------------------------------------------------------------------154
inflated translation method----------------------------------------------------------------------------------154
axis trace------------------------------------------------------------------------------------------------------156
expansivity----------------------------------------------------------------------------------------------------157
comparison method-------------------------------------------------------------------------------------------158
objective implicit function-----------------------------------------------------------------------------------158
feasible group--------------------------------------------------------------------------------------------------159
absolute difference--------------------------------------------------------------------------------------------160
absolute difference method----------------------------------------------------------------------------------163
unfeasible group-----------------------------------------------------------------------------------------------165
graded search method----------------------------------------------------------------------------------------164
initial group----------------------------------------------------------------------------------------------------165
terminative group---------------------------------------------------------------------------------------------165
intermediate group--------------------------------------------------------------------------------------------165
unconstrained search method--------------------------------------------------------------------------------164
173
Appendix::::solution process of part exercises The First Chapter
1.1 solution:
under this relaton expression,arrange and transpose the right end of vector
3e1+5e2-4e3σ 3 (3e1-5e2)+5 (-2e1+e2)-4 (-4e1+e2 ) ,
it has 15e1-14e2
viz. the vector become 15e1-14e2 after the transform.
1.2 solution:
because submatrix composed of the front rows of coefficients matrix is
−
−
12
53
we found out inverse matrix of the submatrix is
−−
−−
7
3
7
27
5
7
1
so
=
−−
−−•
−
−
−
=
7
17
7
2
1
1
7
3
7
27
5
7
1
14
12
53
M
now,the relation of exercise 1.1 becomes
′+′
′
′
213
22
11
7
17
7
2eee
ee
ee
σ
σσ
substitute the new relation expression into vector
3e1+5e2-4e3,
it has 2121217
33
7
13
7
17
7
2453 eeeeee ′−′=
′+′−′+′
viz.the vector becomes 217
33
7
13ee ′−′ .
1.3 Solution:
substitute the transform into vector 4e1-5e2+3e4,the vector becomes
4e1-5e2+3 (-3e1+2e2 )= -5e1+e2 . 1.4 solution: 5e1-4e2+6 (9e1-5e2 )=59e1-34e2 ,
the vector becomes 59e1-34e2. 1.5 Solution:
174
( ) ( ) , 24728122
1
2
28712 212121 eeeeee −−−=
+−+−
viz. the vector becomes ( ) ( ) 2472812 21 ee −−− .
The Second Chapter
2.7 Solution:
abbreviated formula of transform in exercise 2.6 is
−
−
+−
+−
214
213
214
213
456
23
456
23
eee
eee
eee
eee ○○□□
2.11 Solution:
the corresponding oblique axis transform is
.
489
485
483
489
485
483
213
213
213
213
213
213
○○○□□□
+
+
+
−−
−−
−−
eee
eee
eee
eee
eee
eee
The sixth chapter
6.1 solution:
from theorem 4,L1 and L2 are two mutually interleaving planes,and they have a
one-dimensional common-orthodromic space. The common-orthodromic space can be denoted by
vector {11,-13,-4,1}. Separately find out another orthodromic vector
{0,10,9,-16}, {-6,0,1,16}
on L1 and L2 ,then the common normal space is
=−+
=++−
=+−−
016910
0166
041311
432
431
4321
xxx
xxx
xxxx
T:
one solution vector of T is {715,715,-286,286}.
through one point
−− 0,0
5
3,
5
4A
of L1 , and make generic plane
S:715x1+715x2-286x3+286x4+1001=0
taking this vector as normal vector This is the equation of exterior sum
120 LLS +=
175
(L20 is a new plane gainded by translating L2 and making it intersect L1 at point A).order
,22
⊥′LL I
it has
=++−
=+−−
=−+++
=−+−+
0166
041311
05223
030432
131
1321
1321
1321
xxx
xxxx
xxxx
xxxx
M:
it gets intersection point M1 (2,3,-4,1) .
translate T and make it through point M1 and intersect S at
=+−+
=++−
=+−−
=++−+
02216910
0166
041311
01001286286715715
432
431
4321
4321
2
xxx
xxx
xxxx
xxxx
M :
it gets the intersection point
−−−−
4147
1859,
4147
10582,
4147
2574,
4147
67212M .
Then ,distance
5149.54147
1666.22870
6006600615015150154147
1 2222
21
≈≈
+++=MM
between the two points is just the distance between L1 and L2 .
The Seventh Chapter
7.6 solution:
from formula(1),
,3131.09999
31
942536916144251649
3410301228cos ≈=
++++++++++
−−−−−=ϕ
so it gets φ≈71°45′.
7.7 solution:
from formula(2),direction vectors of the two straight lines separately are
H1={2,-44,-87,-74}, H2={6,-16,-8,-2},
from formula(3) ,6717.036014985
1560cos
21
21 ≈=•
•=
HH
HHϕ
so φ≈47°48′.
7.8 solution:
two planes intersect at the straight line
176
=++−−
=−+++
=+−−+
054353
0233
035432
4321
4321
4321
xxxx
xxxx
xxxx
H: ,
then the orthodromic vetor of H is H={79,-82,93,-92}.
taking the vector as normal vector and make generic plane separately intersecting I1 , I2 ,it gets
straight line
=−+−
=−+++
=+−−+
092938279
0233
035432
4321
4321
4321
1
xxxx
xxxx
xxxx
H :
=−+−
=++−−
=++−−
092938279
06246
054353
4321
4321
4321
2
xxxx
xxxx
xxxx
H :
separately found out H1={1372,-853,-1050,877}, H2={913,956,2217,2173}.
so, , 0021.0113845234481622
15039cos ≈=ϕ φ≈89°53′.
7.9 solution:
because it found out H1={2,-44,-87,-74},
then it is known that ⊥′
1H is a generic plane
F:2x1-44x2-87x3-74x4= 0
use F∩I1 ,it gets the straight line
=−−−
=−+++
=+−−+
07487442
0233
035432
4321
4321
4321
xxxx
xxxx
xxxx
H:
H={124,-223,586,-553}
H⊥′:124x1-223x2+586x3-553x4= 0
=−+−
=−+++
=+−−+
0553586223124
0233
035432
4321
4321
4321
2
xxxx
xxxx
xxxx
H :
H2={8031,-5802,-2371,1628}
is found out,so , 2828.010643219014985
357155cos ≈=ϕ φ≈73°34′.
The Eighth Chapter
8.4 solution:
arrange the equation of generic plane into
177
,2
4
2
5
2
3 3214
xxxx −+=
substitute it into the equation of unit generic sphere
, 12
4
2
3
2
2
2
1 =+++ xxxx
it gets ,4402430202913 323121
2
3
2
2
2
1 =−−+++ xxxxxxxxx
its characteristic matrix is
−
−−
−−
202012
202915
121513
λ
λ
λ
: G
make elementary transformation
( ) ( )( )
⇒
+−−−
−−−
−−
⇒
−
−−
−−
116334150
45430
131512
122020
152920
131512
2 λλλ
λλ
λ
λ
λ
λ
( ) ( )( )( )
−−
−⇒
+−
−−−
54400
040
001
2165800
45430
001
2 λλ
λ
λλ
λλ
for G then the equation of the generic sphere becomes
,14
54 2
3
2
2
2
1 =++ xxx
so, cosφ1=cosφ2=1,
,2722.054
2cos 3 ≈=ϕ
becauseφ1=φ2=0, then choose φ3≈74°12′to define the included angle between the two
generic planes.
8.5 solution:
determinant of coefficient
, 1145
14−=
−−=D
of the two posterior terms of the plane,
( )( )
( )( )
, 21935
234
, 51343
123 but
21
21
21
2
21
21
21
1
xxxx
xxD
xxxx
xxD
−=+−
−−−=
−−=+−
−−−=
so it gets , 11
2
11
19 ,
11
5
11
13214213 xxxxxx +−=+=
substitute it into the equation of unit generic sphere,it gets
178
, 121154150651 2
2
2
2
1 =++ xxxx
from coefficient theorem,it has
( )
( ) ,3922.5651150541506511212
1
,2277.1651150541506511212
1
22
22
≈ −+++×
=
≈ −+−+×
=
B
A
so, cosφ1≈0.9025, cosφ2≈0.4306, φ1≈25°31′,φ2≈64°30′.
8.6 solution:
determinant of coefficient , 1625
42=
−−=D
( )( )
( )( ) 21
21
21
2
21
21
21
1
1924325
542
216232
454
xxxx
xxD
xxxx
xxD
+=+−
−−−=
+−=+−
−−−=
of the two posterior terms of I1 it gets
. 16
19
16
24 ,
16
2
16
16214213 xxxxxx +=+−=
substitute it into the equation of unit generic sphere ,it gets
, 2568486211088 21
2
2
2
1 =++ xxxx
from coefficient theorem,it gets
( ) ( )
( )
, 8313.04564.370
16cos , 4373.0
5436.1338
16cos
, 4564.37093719317092
1
, 5436.133893719317092
110886218486211088
2
1
21
22
≈=≈=
≈−=
≈+= −+++=
ϕϕ
B
A
φ1≈64°04′,φ2≈33°46′。
8.7 solution:
determinant of coefficient , 4
212
423
652
−=
−
−−
−−
=D
( )( )
( ), 12820
21322
4232
65326
321
321
321
321
1 xxx
xxx
xxx
xxx
D +−=
−−+−
−+−−
−+−−
=
( )( )
( )
( )( )
( ), 12816
22322
4332
62326
23222
4323
63262
321
321
321
321
321
321
321
2
xxx
xxx
xxx
xxx
xxx
xxx
xxx
D
−+−=
=
−−+−
−+−−
−+−−
=
−+−
−+−−−
−+−−−
=
179
( )( )
( ). 12824
32212
3223
32652
321
321
321
321
3 xxx
xxx
xxx
xxx
D −+−=
−+−−
+−−−
+−−−
=
of the three posterior terms of G1.then it gets
+−=
+−=
−+−=
,
326
324
325
3216
3215
3214
xxxx
xxxx
xxxx
substitute it into the equation of unit generic sphere,it gets
. 1369060281378 323121
2
3
2
2
2
1 =−+−++ xxxxxxxxx
make elementary transformation
( )( )
+−−−
−⇒
+−
−
⇒
+−−
−−⇒
+−−
−−
−−
⇒
−−
−
−−
⇒
−−
−
−−
32865932865900
010
001
19511800
010
001
159106660
6210
001
15910630300
62550
783045
451828
301318
783045
281845
181330
453078
2
22
λλ
λ
λλ
λ
λλλ
λλ
λλλ
λλ
λ
λ
λ
λ
λ
λ
λ
for characteristic matrixso,the characteristic root of the characteristic matrix is λ1=1,
, 328659 , 328659 32 −=+= λλ
then, cosφ1=1,
, 7724.0328659
1cos , 0927.0
328659
1cos 32 ≈
−=≈
+= ϕϕ φ1=0°,φ2≈84°41′, φ3≈39°26′.
choose nonzero angleφ2, φ3 to define the included angle between G1, G2.
8.10 solution:
first found out two solution vectors
{ }
+
−
−−
−
−
−
−−
=
−
−
−−=
−=−−=−
−−=
322
252
424
320
253
425
3202
2532
4254
3,2,0,2
252
4248
1
0010
2532
4254
8
1
21
4321
4
431
4321
3
ee
eeee
eeeeeee
β
β
{ }82,70,136,53
202
532
254
312
232
454
43 −−−=
−
−
−
−−
−
+ ee
180
of I1 and find out two new normal vectors
{ }, 0,8,1,8
7013653
2234
1
1000
827013653
3202
34
1421
4321
1 −=
−−
−−=−−−
−−=
eeeeeee
β
{ }, 258,173,328,214
7013653
818
202
178213653
018
302
17
827053
088
322
178270136
081
320
17
827013653
0818
3202
17
1
43
21
4321
2
−−=
−−
−
−
−
−−
−+
+
−−
−
−
−−−
−
−
=
−−−
−
−=
ee
ee
eeee
β
perpendicular to each other of I1.assume
{ }
{ }
{ }
{ },258,173,328,214249873
1
0,8,1,8129
1
82,70,136,5332929
1
3,2,0,217
1
2
24
1
13
4
42
3
31
−−==′
−==′
−−−==′
−==′
β
β
β
β
β
β
β
β
e
,e
,e
,e
then it gets orthogonal matrix
−−
−
−−−
−
=
249873
258
249873
173
249873
328
249873
214
0129
8
129
1
129
832929
82
32929
70
32929
136
32929
5317
3
17
20
17
2
Q
from formual (3) X=Q′Y
of the eighth chapter it gets
−−
−−−
−−
−
4
3
2
1
4
3
2
1
1
249873
2580
32929
82
17
3
249873
173
129
8
32929
70
17
2
249873
328
129
1
32929
1360
249873
214
129
8
32929
53
17
2
y
y
y
y
x
x
x
x
=: σ
choose β1 ,β2 as new normal vector of I1,then it gets
181
=−−+
=+−
0258173328214
088
4321
321
xxxx
xxx
substitute σ-1 into the new equation of I1 , it gets
=
=′
0
0
4
3
1y
yI :
and substituteσ-1 into the equation of I2,then it gets
=−−+
=++−
0249873
166
129
5
32929
683
17
20
0249873
992
129
19
32929
931
17
4
4321
4321
2
yyyy
yyyy
I :
determinant of coefficient
,193712917
8322
12917
400
193717
683
17
20
129
5
193717
931
17
4
129
19
,12917
270
129193717
20504
1937129
166
193717
683
17
20
1937129
992
193717
931
17
4
,1937
14
1937129
166
129
51937129
992
129
19
21
21
21
2
21
21
21
1
yy
yy
yy
D
yy
yy
yy
D
D
+=
+−
−
=
−−=
−
+
−
=
=−−
=
of the two posterior terms of I2′, so
,21937
4161
21937
1937200
,21937
1937135
21937
10252
214
213
yyy
yyy
+=
−−=
substitute the equation of y3 , y4 into the equation of unit generic sphere
12
4
2
3
2
2
2
1 =+++ yyyy
it gets ,1074575272320319374432440182690961 2
221
2
1 =++ yyyy
( )
, 4621.03.503268
21937cos , 0214.0
7.234910895
21937cos
, 3.5032687.117203813117707082
, 7.2349108957.117203813117707082
106891618.1108055317.3235414164
21
1616
≈≈≈≈
≈−≈
≈+≈
≈×+×+≈
ϕϕ
B
A
φ1≈88°46′, φ2≈62°28′.