Vector Calculus COPYRIGHTED MATERIAL

34
1 Vector Calculus The vector is the basic tool in the formalism of mechanics because it brings together in one concept two fundamental ideas, that is the size of the used parameter or the studied phenomenon and the direction in which it must be considered or in which it applies. The calculus rules that describe it are continuously exploited in the mathematical expression of the motion of bodies. This chapter lists them and develops them for the ease of use. 1.1. Vector space 1.1.1. Definition The vector space E is a set with two operating laws: an internal law, from E E , which confers an Abelian group structure (commutative), and an external law, the multiplication by a scalar. The elements of a vector space are called vectors and, in the formalism of mechanics, are generally represented by an alphabetical symbol topped with an arrow: u G . 1.1.1.1. Properties of the internal composition law The internal composition law in the formalism of mechanics is the vector addition, denoted as +, and that has the following properties: – if , uv E GG , so, u v E + G G ; COPYRIGHTED MATERIAL

Transcript of Vector Calculus COPYRIGHTED MATERIAL

Page 1: Vector Calculus COPYRIGHTED MATERIAL

1

Vector Calculus

The vector is the basic tool in the formalism of mechanics because it brings together in one concept two fundamental ideas, that is the size of the used parameter or the studied phenomenon and the direction in which it must be considered or in which it applies. The calculus rules that describe it are continuously exploited in the mathematical expression of the motion of bodies. This chapter lists them and develops them for the ease of use.

1.1. Vector space

1.1.1. Definition

The vector space E is a set with two operating laws: an internal law, from E E→ , which confers an Abelian group structure (commutative), and an external law, the multiplication by a scalar. The elements of a vector space are called vectors and, in the formalism of mechanics, are generally represented by an alphabetical symbol topped with an arrow: u .

1.1.1.1. Properties of the internal composition law

The internal composition law in the formalism of mechanics is the vector addition, denoted as +, and that has the following properties:

– if ,u v E∈ , so, u v E+ ∈ ;

COPYRIG

HTED M

ATERIAL

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2 Movement Equations 2

– it is commutative: , ,u v E u v v u∀ ∈ + = + ;

– it is associative: ( ) ( ), , ,u v w E u v w u v w∀ ∈ + + = + + ;

– it has a neutral element denoted as 0 so that:, 0 0u E u u u∀ ∈ + = + = ;

– any element has an inverse (or opposite), that is to say: , ,∀ ∈ ∃ − ∈u E u E such as: ( ) 0u u u u+ − = − = .

1.1.1.2. Properties of the external composition law

The external composition law is identified in the mechanical formalism as the multiplication by a scalar λ ∈ of a vector ,u E∈ such that:

u u Eλ λ× = ∈ .

This law has the following properties:

– there is a neutral element, the scalar 1 as: 1 u u× = ;

– the law is distributive with respect to addition and in relation to the addition and multiplication in , as:

( )( )

( ) ( )

, , :

, , :

, , :

u v E u v u v

u E u u u

u E u u

λ λ λ λ

λ μ λ μ λ μ

λ μ λ μ λμ

⎧∀ ∈ ∀ ∈ × + = +⎪⎪∀ ∈ ∀ ∈ + × = +⎨⎪

∀ ∈ ∀ ∈ × × =⎪⎩

1.1.2. Vector space – dimension – basis

We say that n vectors 1 2, , , nu u u E∈… are linearly independent if the relation:

1 1 2 2 0n nu u u uα αλ λ λ λ+ + + = =…

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Vector Calculus 3

has only the following solution:

0, 1, ,nαλ α= ∀ = … .

We say that a vector space E is of dimension n when it holds at most n linearly independent vectors; any other vector element of this space is then expressed as a linear combination with coefficients

αλ ∈ of these n vectors. Any other vector of the vector space E of dimension n can be expressed from only these n linearly independent vectors, which constitute the basis of what we call space.

Thus, if this basis consists of n linearly independent vectors

1 2, , , nu u u E∈… , any vector V of E can be written as:

1 1 2 2 n nV u u uμ μ μ= + + +… ,

where the coefficients 1 2, , , nμ μ μ ∈… are the components of V in the considered basis.

1.1.3. Affine space

The first area to be considered as the frame of the movement of physical bodies is terrestrial space. This can be represented by geometrical parameters and its properties fit well with the concept of vector space, a mathematical entity to which it is advised to give a reality; hence the notion of affine space of vector space.

This notion follows a precise mathematical definition that is not necessary to repeat here; it merely gives the physical space the mathematical specificity, which is necessary to make it the mechanical workplace, and thus the characteristics of the vector space rules applied to it.

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6 Movement Equations 2

The sign of the scalar product that is an algebraic quantity depends on whether θ is acute ( )cos 0θ > or obtuse ( )cos 0θ < .

1.3.1. Properties of the scalar product

Vis-à-vis the vector operations, specific to vector spaces, scalar product has the following properties:

– it is commutative:

a b b a⋅ = ⋅ ,

– it is distributive right and left with respect to the vector addition:

( )1 2 1 2a b b a b a b⋅ + = ⋅ + ⋅ and ( )1 2 1 2a a b a b a b+ ⋅ = ⋅ + ⋅ ,

– its multiplication by a scalar gives:

( ) ( ) ( )a b a b a bλ λ λ⋅ = ⋅ = ⋅ ,

– the scalar product of two linear combinations of the vectors unfolds as follows:

( )1 1 1 1

m n m n

i i j j i j i ji j i j

a b a bλ μ λ μ= = = =

⎛ ⎞ ⎛ ⎞⋅ = ⋅⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠∑ ∑ ∑ ∑ .

1.3.2. Scalar square – unit vector

As ( )cos , 1a a = , we obtain the scalar square of a by the

operation:

2 2a a a a⋅ = = so

2a a= .

If we consider a group of three linearly independent vectors ( )1 2 3, ,U U U of the affine space, we can define the three vectors:

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8

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Vector Calculus 9

1.3.4. Solving the equation a x⋅ = 0

Consider a given vector a and the equation 0a x⋅ = . It accepts two types of solution:

so and

0

, cos 02 2

x

x a a x π π

⎧ =⎪⎨⎪ ⊥ = =⎩

.

Therefore, the scalar product of two vectors is zero if one of the two vectors, at least, is zero, or if the two vectors are orthogonal.

1.4. Vector product ∧a b

1.4.1. Definition

The vector product of the two vectors a and b of affine space 3E of dimension 3 is represented by the operation a b∧ . Its result is the vector c a b= ∧ , which has the following properties:

– it is orthogonal to the plane formed by the vectors a and b : ( ),c a b⊥ Π ;

– it is oriented so that the trihedron ( ), ,a b c is direct (see below);

– sinc a b θ= × × with π θ π− ≤ ≤ .

NOTE.– According to the rule of the corkscrew by Maxwell, a corkscrew planted perpendicular to the plane ( ),a bΠ progresses,

when rotated from a toward b , in the direction of the vector c .

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10

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Vector Calculus 11

1.4.3. Properties of vector product

The vector product has properties that are either its own or combined with the other operations of vectors:

– the vector product is anticommutative, that is to say:

b a a b∧ = − ∧ .

This property results from the fact that the trihedrons ( ), ,a b a b∧

and ( ), ,b a b a∧ must be both direct;

– multiplication by a scalar obeys the rule:

( ) ( ) ( )a b a b a bλ λ λ∧ = ∧ = ∧

– the vector product is distributive with respect to the addition of vectors:

( )( )

1 2 1 2

1 2 1 2

a b b a b a b

a a b a b a b

∧ + = ∧ + ∧

+ ∧ = ∧ + ∧

– the vector product of two linear combinations of vectors is developed:

( )1 1 1 1

m n m n

i i j j i j i ji j i j

a b a bλ μ λ μ= = = =

⎛ ⎞ ⎛ ⎞∧ = ∧⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠∑ ∑ ∑ ∑ .

1.4.4. Solving the equation a x∧ = 0

Consider a given vector a and equation 0a x∧ = . It accepts two types of solution:

and, as a result, and

0

, 0 sin 0 0

x

x a a xλ

⎧ =⎪⎨⎪ = = =⎩

.

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12

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Vector Calculus 13

The segment OH, orthogonal projection of a on the vector b c∧which is orthogonal to the plane ( ),b cΠ , is the height of the

parallelepiped constructed on the three vectors a , b , c . The volume of the parallelepiped is equal to:

( )OBDC OH×Aire , with OH cosa θ= × .

We deduce that

( ) ( )[ ]

( ) ( )parallelepiped , , 6 tetrahedron OABC

" " , ,

a b c

a b c a b c

⎡ ⎤ = ×⎣ ⎦

= ⋅ ∧ =

Vol Vol.

If we express this result as:

( ) ( )parallelepiped , ,a b c a b c⎡ ⎤ =⎣ ⎦Vol ,

this algebraic volume is positive when the trihedron ( ), ,a b c is direct,

negative when it is indirect.

1.5.3. Properties of the mixed product

1.5.3.1. Cases of nullity of the mixed product

The relationship ( ), , 0a b c = corresponds to the following cases:

– 0a = ;

– 0b c∧ = , which means one of these 0

0

such as

b

c

b cλ λ

⎧ =⎪⎪ =⎨⎪∃ ∈ =⎪⎩

;

– a b c⊥ ∧ , which means , such as a b cλ μ λ μ∃ ∈ = + .

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14 Movement Equations 2

The mixed product of three vectors is zero when one of the vectors is zero or when two of them are at least linked.

1.5.3.2. Circular permutation of terms

If we consider the parallelepiped constructed on the three vectors a , b , c , the calculation of its volume does not depend on the order in which one considers them. We can thus calculate at first the vector product of a and b , then project c onto the vector a b∧ , or the vector product c a∧ and project b onto this vector; the main thing is to conserve the order of vectors, since the algebraic sign of the volume depends on the direct or indirect order of the three vectors. We can either write the equalities:

( ) ( ) ( )a b c b c a c a b⋅ ∧ = ⋅ ∧ = ⋅ ∧ ,

that is to say we can invert the operations, such that:

( ) ( )a b c a b c⋅ ∧ = ∧ ⋅ .

Symbolically therefore we write:

( ) ( ) ( ), , , , , ,a b c b c a c a b= = .

1.5.3.3. Permutation of two terms

For the same reasons related to the direct or indirect nature of the trihedron formed by the three vectors of a mixed product, in the order they are listed, the permutation of two vectors changes the sign of the result of the operation. So

( ) ( ), , , ,a c b a b c= − .

1.5.3.4. Multiplication by a scalar

Multiplying a mixed product by a scalar amounts to multiplying one of the vectors by this scalar:

( ) ( ) ( ) ( ), , , , , , , ,a b c a b c a b c a b cλ λ λ λ× = = = .

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Vector Calculus 15

1.5.3.5. Distributivity

The operation is distributive with respect to the addition of vectors on each member of the mixed product, as, for example:

( ) ( ) ( )1 2 1 2, , , , , ,a a b c a b c a b c+ = + .

1.5.3.6. Mixed product of a combination of vectors

( )1 1 1 1 1 1

, , , ,l m n l m n

i i j j k k i j k i j ki j k i j k

a b c a b cλ μ ν λ μ ν= = = = = =

⎛ ⎞ =⎜ ⎟⎝ ⎠∑ ∑ ∑ ∑ ∑ ∑ .

1.6. Vector calculus in the affine space of dimension 3

1.6.1. Orthonormal basis

The orthogonal projection of vectors applies practically trigonometric functions. A trirectangular trihedron vector corresponds to linearly independent vectors since the orthogonal projection of one of them, on the plane formed by the two others, is always zero. The mechanical formalism is thereby greatly simplified. So it is worthwhile to resort, if another need does not justify it, to the orthogonal bases and, moreover, normed.

The vectors that constitute the orthonormal basis ( ) ( )1 2 3u u u u=

therefore verify all the relations:

2 2 21 2 3

1 2 2 3 3 1

1 2 3 2 3 1 3 1 2

1 1 1

0 0 0

u u u

u u u u u u

u u u u u u u u u

= = =

⋅ = ⋅ = ⋅ =

∧ = ∧ = ∧ =

.

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16 Movement Equations 2

1.6.2. Analytical expression of the scalar product

Consider the orthogonal basis ( ) ( )1 2 3u u u u= , the two vectors

a and b , and their respective components ( )1 2 3, ,a a a and ( )1 2 3, ,b b bon it. These vectors are thus written:

1 1 2 2 3 3a a u a u a u= + + and 1 1 2 2 3 3b b u b u b u= + + .

The scalar product:

( ) ( )1 1 2 2 3 3 1 1 2 2 3 3a b a u a u a u b u b u b u⋅ = + + ⋅ + + ,

thus is developed, taking into account the properties of the unit vectors of the orthonormal basis:

1 1 2 2 3 3a b a b a b a b⋅ = + + .

If the basis was not orthonormal, we should take it into account, in the development, the expression of the different scalar products i ju u⋅ that are involved.

In the analytical form, the scalar square of the vector a is written as:

2 2 2 21 2 3a a a a a a= ⋅ = + + .

1.6.3. Analytical expression of the vector product

Similarly, the vector product of the two vectors:

1 1 2 2 3 3a a u a u a u= + + and 1 1 2 2 3 3b b u b u b u= + +

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Vector Calculus 17

is written, taking into account the properties of the orthonormal basis:

( ) ( )( ) ( ) ( )

1 1 2 2 3 3 1 1 2 2 3 3

2 3 3 2 1 3 1 1 3 2 1 2 2 1 3"

a b a u a u a u b u b u b u

a b a b u a b a b u a b a b u

∧ = + + ∧ + +

= − + − + −.

We can calculate, using a practical method, this vector product by using the following determinant that is effectively developed with respect to the first line:

( ) ( ) ( )1 2 3

1 2 3 2 3 3 2 1 1 3 3 1 2 1 2 2 1 3

1 2 3

u u u

a a a a b a b u a b a b u a b a b u

b b b

= − − − + −

in order to obtain the expression.

1.6.4. Analytical expression of the mixed product

Consider the three vectors:

1 1 2 2 3 3 1 1 2 2 3 3, ,a a u a u a u b b u b u b u= + + = + +

1 1 2 2 3 3,c c u c u c u= + +

and their mixed product ( ), ,a b c :

,

( ) ( ) ( )( ) ( ) ( )

( ) ( ) ( )

1 1 2 2 3 3

2 3 3 2 1 3 1 1 3 2 1 2 2 1 3

1 2 3 3 2 2 3 1 1 3 3 1 2 2 1

, ,

"

a b c a b c a u a u a u

b c b c u b c b c u b c b c u

a b c b c a b c b c a b c b c

= ⋅ ∧ = + +

⎡ ⎤⋅ − + − + −⎣ ⎦= − + − + −

( ) ( ) ( ) ( )1 2 3 3 2 2 3 1 1 3 3 1 2 2 1, ,a b c a b c b c a b c b c a b c b c⇒ = − + − + −

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18

re

1.

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Th

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8 Movement E

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.7. Applicat

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Vector Calculus 19

1 1 2 2 3 3

1 1 2 2 3 3

1 1 2 2 3 3

a a u a u a u

b b u b u b u

c c u c u c u

⎧ = + +⎪⎪ = + +⎨⎪ = + +⎪⎩

,

where ( ) ( ) ( )2 3 3 2 1 3 1 1 3 2 1 2 2 1 3b c b c b c u b c b c u b c b c u∧ = − + − + − .

If we consider the component 1u of the vector ( )d a b c= ∧ ∧ ,

which is of the form d b cλ μ= + , we can write:

( ) ( )( ) ( )( ) ( )

1 1 2 1 2 2 1 3 3 1 1 3

1 1 2 2 3 3 1 1 1 2 2 3 3 1

1 1

"

"

b c a b c b c a b c b ca c a c a c b a b a b a b c

a c b a b c

λ μ+ = − − −= + + − + +

= ⋅ − ⋅

.

In the same way, we obtain:

( ) ( ) ( ) ( )2 2 2 2 3 3 3 3,b c a c b a b c b c a c b a b cλ μ λ μ+ = ⋅ − ⋅ + = ⋅ − ⋅ ,

and we can identify:

,a c a bλ μ= ⋅ = ⋅ .

This gives the development of the double vector product

( ) ( ) ( )a b c a c b a b c∧ ∧ = ⋅ − ⋅ .

1.7.1.2. Particular characteristics of the double vector product

( ) ( ) ( ) ( )a b c c b a c a b c b a∧ ∧ = ∧ ∧ = ⋅ − ⋅ .

The difference, memberwise, of this relationship and the previous yields:

( ) ( ) ( ) ( )a b c a b c c b a a b c∧ ∧ = ∧ ∧ + ⋅ − ⋅ .

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20 Movement Equations 2

In particular, when c a= , we have:

( ) ( )a b a a b a∧ ∧ = ∧ ∧ ,

and if λ∃ ∈ such as c aλ= :

( ) ( ) ( ) ( ) 0a b a a b a a b a a b aλ λ λ λ∧ ∧ − ∧ ∧ = ⋅ − ⋅ = ,

with the result that if a and c are collinear:

( ) ( )a b c a b c∧ ∧ = ∧ ∧ .

1.7.1.3. The case for nullity with the double vector product

Consider the relationship ( ) 0a b c∧ ∧ = . It is checked if:

– at least one of the three vectors, a , b or 0c = ;

– the vector 0b c∧ = λ⇒ ∃ ∈ such as b cλ= ;

– the vector a is collinear with b c∧ ;

λ⇒ ∃ ∈ such as ( )a b cλ= ∧ or else ( ),a b c⊥ Π .

1.7.1.4. Projection of a vector on a plane

Consider the plane ( )aΠ whose normal vector is a , and the unit

vector aua

= ; consider the vector b and its orthogonal projection on

the plane (parallel to a ).

The vector b is expressed in terms of its two projections:

( )a ab proj b proj bΠ= + with ( ) ( )2a

a bproj b u b u a

a

⋅= ⋅ = .

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We

p

From

u

we obtvector axis:

b

The

the dir

given b

p

Figu

deduce that:

( )aproj bΠ =

m the formul

( )u b u∧ ∧ =

tain the expon an axis o

( )b u b u= ⋅ +

erefore, the p

ection defin

by:

( )aproj bΠ =

ure 1.9. Projec

:

ab proj b−

la of the dou

(2u b u b= − ⋅

pression of tof unit vector

( )u b u+ ∧ ∧

projection of

ned by vecto

( )u b u∧ ∧

ction of a vect

( )2

a bb

a

⋅= −

uble vector pr

)b u ,

the three-dimr u and on t

) .

f a vector b

or a , and th

(2

a b a

a

∧ ∧=

Ve

or on a plane

a .

roduct:

mensional pthe plane orth

on an orthog

he unit vect

)a.

ctor Calculus

projection ofhogonal to th

gonal plane,

or aua

= ,

21

f a his

in

is

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22 Movement Equations 2

1.7.2. Resolving the equation a x b⋅ =

Consider two data, a vector 1 1 2 2 3 3a a u a u a u= + + and a scalar b .

This is about determining the vector 1 1 2 2 3 3x x u x u x u= + + ,

satisfying the equation a x b⋅ = , that is to say the relationship 1 1 2 2 3 3a x a x a x b+ + = that admits, as an equation, a double infinity of

solutions.

Indeed, if there are infinitely many solutions to choose 1x , for example, this value taken, there are still an infinite number of values to choose for 2x ; 3x then depends on these two choices.

Suppose that we know a particular solution 0x that satisfies the

equation 0a x b⋅ = ; it also satisfies the relationship:

( )0 0a x x⋅ − = ,

that is to say that the vector 0x x− is orthogonal to a .

We only needs to know one of these particular solutions in order to then express the general form of the equation’s solution.

Choosing one, among all the possible solutions, which is collinear with a , so the form 0x aλ= .

2

0 2ba x a a a b

aλ λ λ⋅ = ⋅ = = ⇒ = .

By stating:

OA a= , 0 0OM x= and OM x= ,

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the ext( 0MΠ

1.7.3.

Suchthat is t

a

If wwrite:

a

that is t

x

Amorthogo

a

treme M of)a , passin

Figur

Resolving

h an equatioto say if:

0a b⋅ = so a

we know a pa

( )0a x x∧ −

to say that th

0x x aλ− =

ong all the ponal to a , th

0 0a x⋅ = w

f the vector g through M

re 1.10. Resol

the equati

on only has

1 1 2 2a b a b+ +

articular solu

0= ,

he vector x −

so 0x x=

possible vecthat is to say:

with 0a x∧ =

position OM

0M and orthog

lving the equa

ion a x b∧ =

a solution if

3 3 0a b = .

ution 0x sati

0x− is colline

aλ+ .

tor solutions,

b= .

Ve

M is situategonal to OA

ation a x b⋅ =

b

f a and b a

isfying a x∧

ear to a , and

, we choose,

ctor Calculus

ed in the plaA .

are orthogon

0x b= , we c

d is:

if there is, x

23

ane

nal,

can

0x

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24

ob

a

thpa

4 Movement E

Multiplyingbtain:

( )0a a x∧ ∧

The genera

bxa

∧=

By stating,

OA a=

he extremityassing throug

F

Equations 2

g vectorially

( )0a x a= ⋅ −

al solution is

2a aλ∧ + .

as per the di

a , OB b= ,

M of the vegh 0M and p

Figure 1.11. R

y left by a

20a x a− = −

thus in the f

iagram below

0 0OM x=

ector OMparallel to O

Resolving the

the vector

20x a b= ∧

form of:

w:

and OM

is located oA .

equation a ∧

product abo

0x⇒ =

x= ,

on the straig

x b=

ove, we

2b a

a

∧= .

ght line

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Vector Calculus 25

1.7.4. Equality of Lagrange

Consider two vectors a and b , and perform the scalar square of the two following terms, the one, the scalar a b⋅ , the other, the vector a b∧ .

( ) ( )

( ) ( ) ( ) ( )

2 21 1 2 2 3 3

2 2 2 22 3 3 2 3 1 1 3 1 2 2 1

a b a b a b a b

a b a b a b a b a b a b a b

⋅ = + +

∧ = − + − + −.

The sum memberwise of these two expressions yields, after development and simplification:

( ) ( ) ( )( ) ( )( )2 2 2 22 2 2 2 2 21 2 3 1 2 3a b a b a a a b b b a b∧ + ⋅ = + + + + = .

This equality of Lagrange is also written as:

2 2

1a b a ba b a b

⎛ ⎞ ⎛ ⎞∧ ⋅+ =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

,

and reduces, in fact, to the trigonometric expression 2 2sin cos 1θ θ+ = .

1.7.5. Equations of planes

1.7.5.1. Plane normal to a vector and passing through a point

In the affine space that is associated with the frame 1 2 3O u u u ,

we consider a vector 1 1 2 2 3 3a a u a u a u= + + represented by the vector

position OA and a unit vector 1 1 2 2 3 3v v u v u v u= + + .

The question is to express the equation of the plane ( )A νΠ

passing through the point A and orthogonal to v .

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26

eq

an

1.po

path

6 Movement E

Figure 1.

Consider th

OM x=

This point

AM v⋅ =

The plane

quation:

OM v⋅ =

nd has the Ca

1 1 2v x v+

.7.5.2. Planoint

Consider tassing throughe plane ( AΠ

AM ∈

Equations 2

12. Plane norm

he point M i

1 1 2 2x u x u+ +

M is such th

( OM OA= −

( )A vΠ i

OA v= ⋅ ,

artesian equa

2 2 3 3x v x+ =

ne defined

the two vectgh the point

)A ,b c is su

( )A ,b cΠ

rmal to a vecto

n this plane:

3 3x u+ .

hat:

)A 0v⋅ = .

is the locus

ation:

1 1 2 2v a v a+ +

by two vec

tors b and A of vector

uch that:

,λ μ⇒ ∃ ∈

or and passing

of points

3 3v a .

ctors and p

c , the planposition OA

such tha

g through a po

M that sati

passing thro

ne containingA . Any poin

at AM bλ=

oint

isfy the

rough a

g them, nt M of

b cμ+ .

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Acc

O

In consdefinesthe prev

The

(

1.7.6.

Not

Figure 1.13.

cording to Fig

OM OA= +

sidering the s the orthogovious proble

e equation of

( ) OMb c∧ ⋅

Relations w

e the triangle

Fi

Plane defined

gure 1.13, w

AM OA+ = +

scalar produonal directioem.

f the plane is

( ) Ob c= ∧ ⋅

within the t

e represented

igure 1.14. Co

d by two vecto

we have:

b cλ μ+ + .

uct of this exn to the plan

given by the

OA .

triangle

d in the figur

onfiguration of

Ve

rs and through

xpression byne, one is br

e equation:

re below.

f a triangle

ctor Calculus

h a point

y b c∧ , whirought back

27

ich to

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28 Movement Equations 2

Consider the relation of the Chasles BC BA AC= + and multiply its two members, scalarly, by BC . We obtain

2

2

BC BC BA BC AC BC BA CB CAcos cosa ac abβ γ

= ⋅ + ⋅ = ⋅ + ⋅⇒ = +

,

therefore cos cosa c bβ γ= + .

Considering now the scalar square of the Chasles relation:

( )2 2 2 2

2 2

BC BA AC BA AC 2 BA AC

AC AB 2 AC AB,

= + = + + × ⋅

= + − × ⋅

so 2 2 2 2 cosa b c bc α= + − .

According to the definition of the vector product, the area A of the triangle is given by:

2 sin sin sinbc ca abα β γ= = =A ,

and, by dividing each term of this expression by the product abc , we obtain a fundamental relationship of the triangle:

sin sin sina b c

α β γ= = .

1.8. Vectors and basis changes

1.8.1. Einstein’s convention

Given the size of mathematical expressions that will be developed in this work and formulas to be used therein, it quickly becomes essential to have a condensed notation system that facilitates writing and exploitation. The Einstein notation convention is a response to

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Vector Calculus 29

that expectation. To illustrate the principle, consider the example of a system of linear equations of n variables ix for 1, ,i n= …

11 1 12 2 1 1

21 1 22 2 2 2

1 1 2 2

n n

n n

n n nn n n

a x a x a x ba x a x a x b

a x a x a x b

+ + + =⎧⎪ + + + =⎪⎨⎪⎪ + + + =⎩

……

……

.

Because of the Einstein notation convention, this system can be written in the following condensed form

, , 1, ,ij j ia x b i j n= ∀ = … ,

and is interpreted as follows:

– the index i that appears only once in each member of the equality indicates that there are as many relations as values of this index, so here n ; it is identified as free index. These are the n equations of the above system mixed into one relationship;

– the index j that appears only twice in the left section of the equality means that in each of the n equations of the given system, this member consists of the sum of n monomials obtained by giving to j its n values. This index is called summation index. The summation index only has meaning if the index appears only twice in a monomial convention.

NOTE.– Widely used in digital programming, this agreement is particularly interesting if the various indices used have the same range of variation; as will often be the case in the situations that will be discussed in this book.

For complete information on using the Einstein convention, take the example of the expression index:

ij i ja b c dα αβ β=

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30 Movement Equations 2

– indices i and j appear only once in each of the two members. They are free indices. When their values are chosen, the expression of the relevant term ija outcomes;

– the indices α and β appear, however, twice in the right side of the index relationship; so these are the summation indices.

Consider, for the four indices, the values 1,2,3 . We obtain, for example, for 1, 2i j= = :

12 11 11 12 11 12 22 11 13 32

12 21 12 12 22 22 12 23 32

13 31 12 13 32 22 13 33 32

a b c d b c d b c db c d b c b b c db c d b c d b c d

= + ++ + ++ + +

…… ……

.

1.8.2. Transition table from basis ( )e to basis ( )E

Consider two direct orthonormal bases:

( ) ( ) ( ) ( )1 2 3 1 2 3,e x x x E X X X= = ,

and a vector f whose projection is expressed in the two bases.

( ) ( ) ( ) ( )( ) ( ) ( ) ( )

1 1 2 2 3 3

1 1 2 2 3 3

f f x x f x x f x x f x x f x

f f X X f X X f X X f X X F X

α α α α

β β β β

⎧ = ⋅ + ⋅ + ⋅ = ⋅ =⎪⎨

= ⋅ + ⋅ + ⋅ = ⋅ =⎪⎩

The relationship between the two bases is expressed by the following expressions:

; ; ,p x X x p X X p xαβ α β α αβ β β αβ α= ⋅ = =

which can be summarized by the transition table ( ),p e E from the basis ( )e to basis ( )E where the reading is done by line to express the

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Vector Calculus 31

vectors of basis ( )e according to those of basis ( )E or by column to determine those ( )E based on those ( )e .

( ) ( ) 1 2 3

1 11 12 13

2 21 22 23

3 31 32 33

e E X X X

x p p p

x p p p

x p p p

.

This table is also used to express the components of a vector of a basis in the other.

( ) ( )( ) ( )

f f x f p X p f X p F

F f X f p x p f x p f

α α αβ β αβ β αβ β

β β αβ α αβ α αβ α

= ⋅ = ⋅ = ⋅ =

= ⋅ = ⋅ = ⋅ =.

If we use a matrix representation of vectors and basis change operation, we can write:

1 11 12 13 1 1 11 21 31 1

2 21 22 23 2 2 12 22 32 2

3 31 32 33 3 3 13 23 33 3

orf p p p F F p p p ff p p p F F p p p ff p p p F F p p p f

⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥= =⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦

.

From ( )[ ], ,p e E the matrix directly transcribed from the transition table from basis ( )e to basis ( )E , we have:

( )[ ]11 12 13

21 22 23

31 32 33

,p p p

p e E p p pp p p

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

,

and we see that the matrix that corresponds to a vector from the basis ( )E to a vector of basis ( )e is the transpose of the previous one, which can be expressed by:

( )[ ] ( )[ ], ,p E e p e E= .

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32 Movement Equations 2

The basis change of the vector f between ( )e and ( )E is therefore written as a matrix:

[ ] ( ) ( )[ ][ ] ( )

[ ] ( ) ( )[ ][ ] ( ) ( )[ ][ ] ( )

,

, ,e E

E e e

f p e E f

f p E e f p e E f

=

= =.

1.8.3. Characterization of the transition table

The fact that the transition table takes place between two direct orthonormal bases means that its nine elements pαβ are not independent of each other. First note that the bases are orthonormal:

x xα β αβδ⋅ = and X Xμ ν μνδ⋅ = ,

,i i

x x p X p X p p X Xp p p p

α β αμ μ βν ν αμ βν μ ν

αμ βν μν α β αβδ δ⇒ ⋅ = ⋅ = ⋅

= = =

which gives six relations that should verify the elements of the transition table between two bases.

We get the same six relations between the elements of the transition table by writing:

i iX X p x p x p p x x p p p pμ ν αμ α βν β αμ βν α β αμ βν αβ μ ν μνδ δ⋅ = ⋅ = ⋅ = = = .

In addition, the orientation of the bases introduces new relationships, whether they are direct or indirect. If the basis ( )e is direct, as it is orthonormal, the relationship below is enough to express its orientation.

1 2 3x x x= ∧ ,

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Vector Calculus 33

since the three vectors are orthogonal and this relationship enlightens unequivocally on the basis orientation. In the change of basis, the relation is written as:

1 2 3 2 3p X p X p X p p X Xα α β β γ γ β γ β γ= ∧ = ∧ .

The terms of this expression exist only if the values of the three indices , ,α β γ come within a circular permutation 1,2,3 or 1,3,2 , which gives another three relationships that should cross-check with the terms of the table. Since these are six distinct terms, it is clear that these relations are redundant.

When a transition table or a matrix change in basis has been established, we should check that:

– non-diagonal terms are anti-symmetric;

– vectors are unitary, that is to say, for each row or each column, the sum of squares of the three elements is equal to 1.

– vectors are orthogonal, that is to say the sum of the products of the relevant terms of two lines or two columns is zero, in checking it in the three combinations of lines or columns (1+2, 1+3, 2+3).

This control is required when drawing up transition table.

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