14.1Vectors in Three-dimensional Rectangular Coordinate System 14.2Vector Product and Scalar Triple...

14.1 Vectors in Three-dimensional Rectangular Coordinate

System14.2 Vector Product and Scalar Triple

ProductChapter Summary

Case Study

Vectors in Three-dimensional Space14

P. 2

A plane is approaching Hong Kong International Airport. The flight-control operator of the control toweris trying to give instructions to thepilot to provide a safe route for landing.

Case StudyCase Study

To describe the position and the route ofthe plane, we can introduce the three-dimensional rectangular coordinate system as shown in the figure.

Let the airport be the origin of the coordinate system. Then we use the triplet (x, y, z) to describe the horizontal position (x, y as in the two-dimensional case) and the height of the plane (z).

The plane is located at the point A(2, 1, 4),B(−1, −2, 3), C(2, 0, 2), D(1, 1, 1) are points in space, such that the plane follows the route A B C D O.

We are now ready for landing. Please indicate a flight route.

Captain, please follow the following flight route.

P. 3

1144 .1 .1 Vector in Three-dimensional Vector in Three-dimensional Rectangular Coordinate Rectangular Coordinate SystemSystem

A. A. Vectors in Three-dimensional SpaceVectors in Three-dimensional Space

The addition, subtraction, scalar multiplication, negative, parallelism of vectors, and the rules of operations of vectors are also defined in the same way as in the case of plane vectors.

For any two points A and B in space, the directed line segment from A to B is called the vector from A to B, and is denoted by . The magnitude of is denoted by , which is the same as the vectors on the plane defined before.

AB ABAB

For example, for the cube ABCDEFGH, we have1. (equal vectors) HGDCEFAB 2. (negative vectors) BAAB 3. (parallel vectors) HGDCEFAB //////

4. (addition of vectors ) DFADBFABAF 5. (subtraction of vectors) DBDEABAEBE

P. 4

Example 14.1T

Solution:(a)

(b)

(c)



AEDADE AEAD

cbbc

CGBCABAG AEADAB

cba

EBBD )()( ABEAADBA ABAEADAB

AEAD bc

The figure shows a cube ABCDEFGH. Let = a, = b and = c. Express the following in terms of a, b and c. (a) (b) (c) EBBD AGDE

AE ADAB

P. 5

Example 14.2T

Solution:



The figure shows a cube ABCDEFGH. Prove that .2GCEGDBFDEC

EGDBFDEC DBFDEGEC

FBGC GCGC

GC2

P. 6


BB. . Representation of Representation of Vectors in Three-dimensional Vectors in Three-dimensional Rectangular Coordinate SystemRectangular Coordinate System

The directions of these axes are aligned in such a way that they obey the right-hand rule.

The three-dimensional Cartesian coordinate system R3 consists of three mutually perpendicular axes: x, y and z.

If the x- and y-axes are represented by the index finger and the middle finger respectively, then the thumb represents the z-axis.

P. 7



These three values are called x-, y- and z-coordinates of the point respectively.

Every point in space can be represented in the three-dimensional coordinate system by the triplet (x, y, z), where x, y and z represent the directed distances from the yz-, zx- and xy-planes respectively.

The point of intersection of the three axes is called the origin O and its coordinates are (0, 0, 0). In the figure, i, j and k are the unit vectors in the positive directions of x-, y- and z-axes respectively. They have a common starting point at the origin, and their terminal points are (1, 0, 0), (0, 1, 0) and (0, 0, 1) respectively. P(x, y, z) is a point in R3, so we can express the position vector asOP

By Pythagoras’ theorem, we have

. kji zyxOP

222 zyxOP

P. 8



kji )()()( 121212 zzyyxxAB 2

122

122

12 )()()( zzyyxxAB

Now if two points A(x1, y1, z1) and B(x2, y2, z2) in R3 are given, the vector from A to B can be found by subtracting the position vector from which is the same as we did in the case of R2, then we have

OA OB

P. 9

Example 14.3T

Solution:



Given two points P(–3, –2, 8) and Q(0, –5, 4). Find the unit vector in the direction of . PQ

kji 823 OPkj 45 OQ

OPOQPQ

kjikjikj

433)823()45(

222 )4()3(3 PQ

34

Unit vector 34

433 kji

PQ

kji34

4

34

3

34

3

P. 10



Property 14.1 (a) pi + qj + rk = si + tj + uk if and only if p = s, q = t and

r = u, and (b) pi + qj + rk = 0 if and only if p = q = r = 0.

Consider two vectors pi + qj + rk and si + tj + uk in R3. As i, j and k are non parallel vectors, we have

P. 11

Example 14.4T

Solution:

Given three points A(–3, 1, 5), B(2, 5, –1) and C(–6, 4, 3). Find the coordinates of a point D if (a) ABCD forms a parallelogram, (b) ABDC forms a parallelogram.



Let the coordinates of D be (x, y, z). (a) If ABCD forms a parallelogram, .DCAB

kji )51()15()]3(2[ AB kji 645 kji )3()4()6( zyxDC

kjikji 645)3()4()6( zyx

We have

963

044

1156

zz

yy

xx

The coordinates of D are (11, 0, 9).

P. 12

Example 14.4T

Solution:

Given three points A(–3, 1, 5), B(2, 5, –1) and C(–6, 4, 3). Find the coordinates of a point D if (a) ABCD forms a parallelogram. (b) ABDC forms a parallelogram.



(b) If ABDC forms a parallelogram, .CDAB kji 645 AB

kji )3()4()6( zyxCD

kjikji 645)3()4()6( zyx

We have

363

844

156

zz

yy

xx

The coordinates of D are (1, 8, 3).

P. 13

Example 14.5T

Solution:

Consider the three vectors a = 3i – 4j + 2k, b = i – 3j – k and c = 5i + 2j + k. If m = –16i – 8j – 7k, express m in terms of a, b and c.



Let m = a + b + c. )25()3()243( kjikjikji kji 7816

kji )2()234()53( kji 7816

728234

1653

Consider the determinant of the coefficient matrix: 112234513

55

By Cramer’s rule, 165 and 110,55

355

165,2

55

110,1

55

55 cbam 32

P. 14

Example 14.6T

Solution:



(a)

2

OBOA

OC

)]726()583[(2

1kjikji

kji 32

9

(b)

21

21

OBOA

OC

)]726(2)583[(3

1kjikji

kji 33

45

Given two points A and B with = 3i – 8j + 5k and = 6i + 2j – 7k. C is a point on the line segment AB.Find if (a) C is the mid-point of AB, (b) C divides AB in the ratio 2 : 1.

OBOA

OC

P. 15


CC. . Scalar ProductScalar Product

In the three-dimensional rectangular coordinate system R3, as the unit base vectors i, j and k are mutually perpendicular, we have i i = j j = k k = 1

i j = j i = 0j k = k j = 0i k = k i = 0

If a = x1i + y1j + z1k and b = x2i + y2j + z2k are two non-zero vectors, then a b = x1x2 + y1y2 + z1z2,

where is the angle between a and b.

,cos2

22

22

22

12

12

1

212121

zyxzyx

zzyyxx

We also have the following properties of scalar product:a b = b a a (b + c) = a b + a c (ka) b = k(a b) = a (kb)

P. 16

Example 14.7T

Solution:

Two vectors r = 2i + 3j – k and s = i + 2k are given. (a) Find the value of r s. (b) Hence find the angle between r and s.



)2()32( kikjisr (a)

0)2)(1()0)(3()1)(2(

(b) 0srsr

The angle between r and s is 90.

P. 17

Example 14.8T

Solution:

If the vectors ci + 5j – 3k and 2ci + cj + k are perpendicular to each other, find the value(s) of c.

ci + 5j – 3k and 2ci + cj + k are perpendicular to each other. 0)2()35( kjikji ccc0)1)(3())(5()2)(( ccc

0352 2 cc0)12)(3( cc

2

1or 3c



P. 18

Example 14.9T

Solution:

Given three points A(2, 1, 6), B(– 5, 3, 5) and C(0, –6, 5) are vertices of ABC. Solve ABC. (Give the answers in surd form or correct to the nearest degree.)



kji )65()13()25( AB kji 27kji )55()36()5(0[ BC ji 95

kji )65()16()20( AC kji 72222 )1(2)7( AB 63

222 0)9(5 BC 106222 )1()7()2( AC

)63)(63(

)1)(1()7)(2()2)(7(cos

BAC54

1

9389.88BAC89 (cor. to the nearest degree)

63

P. 19

Example 14.9T

Solution:

Given three points A(2, 1, 6), B(– 5, 3, 5) and C(0, –6, 5) are vertices of ABC. Solve ABC. (Give the answers in surd form or correct to the nearest degree.)



)106)(63(

)0)(1()9)(2()5)(7(cos

ABC6363

53

5305.45ABC46 (cor. to the nearest degree)

ABCBACACB 180 5305.459389.88180

46 (cor. to the nearest degree)

P. 20

Example 14.10T



Solution:

Given two vectors x = 6j – 5k and y = 3i – 4j. (a) Find the angle between x and y, correct to the

nearest degree. (b) Find the length of the projection of y on x.

)43()56( jikjyx 24)5)(0()4)(6()3)(0( Let be the angle between x and y.

yx

yxcos

(a)

222222 0)4(3)5(60

24

615

24

128 (cor. to the nearest degree)

The angle between x and y is 128.

(b) The length of the projection of y on x

cosyx

yx

222 )5(60

24

61

24

P. 21

Suppose we have two non-zero vectors a and b in the three-dimensional space. The vector product of a and b, denoted by a b, is the vector which is perpendicular to both a and b, with the magnitude equal to

|a b| = |a||b|sin,where is the angle between a and b (with 0° ≤ ≤ 180°).

1144 ..22 Vector Product and Scalar Vector Product and Scalar Triple ProductTriple Product

AA. . Definition of Vector ProductDefinition of Vector Product

Note:1. a b is read as ‘a cross b’. Therefore the vector product is also called the cross product. 2. The vector product is only defined in the three-dimensional space.3. In contrast to the scalar product of two vectors, the vector product

is a vector while the scalar product is a scalar.

a b = |a||b|sin , where is the angle between a and b, and is a unit vector whose direction is defined by the right-hand rule.

n̂n̂

Particularly, the direction of a b is defined in such a way that a, b and a b always obey the right-hand rule. In conclusion,

P. 22


AA. . Definition of Vector ProductDefinition of Vector Product

In particular, if b = a, we have

For the unit vectors i, j and k:

For and two non-zero vectors a and b, a b = 0 if and only if a and b are parallel to each other.

a a = 0.

i i = j j = k k = 0 i j = k j i = k j k = i k j = i k i = j i k = j

P. 23


BB. . Properties of Vector ProductProperties of Vector Product

Proof of (a):

b a = (a b)

Property 14.2 Properties of Vector Product(a) b a = (a b)(b) (a + b) c = a c + b c(c) a (b + c) = a b + a c(d) (ka) b = a (kb) = k(a b)(e) |a b|2 = |a|2|b|2 – (a b)2

n baba ˆsin)ˆ(sin nabab n ba ˆsin

P. 24



Proof of (d):If k = 0 or a = 0 or b = 0, then

(ka) × b = a × (kb) = (k a × b) = 0.

Assume that k 0 and a and b are non-zero.

When k < 0, n baba ˆsin kk

n ba ˆsink bak

n ba ˆsin k

When k > 0, n baba ˆsinkk

bakn ba ˆsink

Similarly, it can be proved that a × (kb) = k(a × b).

(ka) × b = a × (kb) = k(a × b)

Let be the angle between a and b, and be the unit vector in the direction of a × b.

n̂

P. 25



Proof of (e):

|a × b|2 = (|a||b|sin )2

Since |a × b| = |a||b|sin

= |a|2|b|2sin2 = |a|2|b|2 − |a|2|b|2cos2 = |a|2|b|2 – (a b)2

sin2 = 1 – cos2a – b = |a||b|cos

P. 26


CC. . Calculation of Vector ProductCalculation of Vector Product

We can use the determinant to represent the vector product:

If a = x1i + y1j + z1k and b = x2i + y2j + z2k,

then .

222

111

zyxzyxkji

ba

P. 27

Example 14.11T

Solution:

For the following pairs of vectors m and n, find the vector products m × n. (a) m = 3i + 8j, n = 6k(b) m = –4i + 2j + 6k,

(a)

(b)



kjin4

3

4

1

2

1

600

083

kji

nm kji00

83

60

03

60

08 ji 1848

4

3

4

1

2

1624

kji

nm kji4

1

2

124

4

3

2

164

4

3

4

162

kji 000 0

P. 28

Example 14.12T

Solution:



P, Q and R are three points with position vectors i + j + k, –2j and –i + 3j – k respectively. Find the unit vectors which are perpendicular to and . PRPQ

kjikjij 3)(2PQ

kjikjikji 222)()3( PR

PRPQ222

131

kji

kji22

31

22

11

22

13

ki 88

28)8(08 222 PRPQ

Unit vectors which are perpendicular to and PQ PR

PRPQ

PRPQ

28

88 ki

ki

2

1

2

1

P. 29


DD. . Applications of Vector ProductApplications of Vector Product

Consider a parallelogram ABCD.

Area of the parallelogram ABCD = ADAB

Since the area of ABD is half that of parallelogram ABCD, we can obtain a formula for the area of triangle:

Area of ABD = ADAB2

1

The above formula can be further rewritten as

Area of ABD = 222

)(2

1ADABADAB

P. 30

Example 14.13T

Solution:

Find the area of the triangle formed by vertices X(2, 1, 1), Y(0, –1, 0) and Z(–2, 1, –1).


DD. . Applications of Vector ProductApplications of Vector Product

kjikjij 22)2(XY

kikjikji 24)2()2( XZ

204

122

kji

XZXY kji04

22

24

12

20

12

ki 84

Area of XYZ XZXY 2

1

52

)8(042

1 222

P. 31

Since the scalar product of two vectors is a scalar, thus a (b c) is a scalar as the name suggests.

The volume of a parallelepiped (a prism with all faces are parallelograms) with sides a, b and c is given by |a (b × c)|.


EE. . Scalar Scalar TripleTriple Product Product

The expression a (b c) is called the scalar triple product of a, b and c.

In the three-dimensional rectangular coordinate system, suppose a = x1i + y1j + z1k, b = x2i + y2j + z2k and c = x3i + y3j + z3k, then

.)(

333

222

111

zyxzyxzyx

cba

Note:If a (b c) = 0, the volume of the parallelepiped with sides a, b and c equals zero. This only when a, b and c are coplanar.

P. 32



Property 14.3 Properties of Scalar Triple Product(a) (a b) c = a (b c)(b) a (b c) = b (c a) = c (a b)

Since a determinant is unchanged when interchanging the rows twice and for any non-zero vectors x and y,x y = y x, we have the following properties of the scalar triple product:

P. 33

Example 14.14T

Solution:

If p = 2i + j + 3k, q = 3i – j – 2k and r = –i + 2j – k, find (a) r × p, and (b) q (r × p).

(a)

(b)



312

121 kji

pr

kji12

21

32

11

31

12

kji 57

)( prq )57( kjiq )57()23( kjikji )5)(2()1)(1()7)(3(

30

P. 34

Example 14.15T

Solution:



Consider A(2, 1, 0), B(–3, 4, 5), C(0, –2, 4) and D(1, 2, 5). Find the volume of the parallelepiped with sides , and . AD

ACAB

kjikji

535)05()14()23(

AB

kjikji

432)04()12()20(

AC

kjikji

5)05()12()21(

AD

Volume of the parallelepiped

511

432

535

88

P. 35

14.1 Vectors in Three-dimensional Rectangular Coordinate System

Chapter Chapter SummarySummary

1. Every point in the space can be represented in the three-dimensional coordinate system by the triplet (x, y, z), where x, y and z represent the directed

distances from the yz-, zx- and xy-planes respectively.

2. The distance between two points A(x1, y1, z1) and B(x2, y2, z2) is given by

.)()()( 221

221

221 zzyyxxAB

P. 36

Chapter Chapter SummarySummary14.1 Vectors in Three-dimensional Rectangular Coordinate System

1. The rules of operations and properties of vectors in the space are the same as vectors on a plane.2. In R3, we define three mutually perpendicular unit vectors i, j and

k, which point in the positive direction of x-, y- and z-axes respectively.

3. For a point P(x, y, z) in R3, the position vector can be expressed

as , where .222 zyxOP kji zyxOP

P. 37

Chapter Chapter SummarySummary14.1 Vectors in Three-dimensional Rectangular Coordinate System

Scalar ProductIf a = x1i + y1j + z1k and b = x2i + y2j + z2k, are two non-zero vectors, then

where is the angle between a and b.

,212121 zzyyxx ba

,cos2

22

22

22

12

12

1

212121

zyxzyx

zzyyxx

P. 38

14.2 Vector Product and Scalar Triple Product


Vector Product1. If a = x1i + y1j + z1k and b = x2i + y2j + z2k, are non-zero vectors and is the angle between them, then

n bab a ˆsin||||

222

111

zyx

zyx

kji

2. Area of ABC

ACAB2

1

222)(

2

1ACABACAB

P. 39

Scalar Triple Product

14.2 Vector Product and Scalar Triple Product


2. Volume of the parallelepiped with sides a, b and c = |a (b c)|.

1. If a = x1i + y1j + z1k and b = x2i + y2j + z2k and c = x3i + y3j + z3k are non-zero vectors, then

.)(

333

222

111

zyxzyxzyx

cba

14.1Vectors in Three-dimensional Rectangular Coordinate System 14.2Vector Product and Scalar Triple...

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