CHAPTER
1
VECTOR ALGEBRA
Books are the quietest and most constant friends; they are the most accessible and
wisest of counselors, and most patient of teachers.
—CHARLES W. ELLIOT
1.1 INTRODUCTION
Electromagnetics (EM) may be regarded as the study of the interactions between electric
charges at rest and in motion. It entails the analysis, synthesis, physical interpretation, and
application of electric and magnetic fields.
Electromagnetics (EM) is a branch of physics or electrical engineering in which
electric and magnetic phenomena are studied.
EM principles find applications in various allied disciplines such as microwaves, antennas,
electric machines, satellite communications, bioelectromagnetics, plasmas, nuclear research,
fiber optics, electromagnetic interference and compatibility, electromechanical energy conversion, radar meteorology, and remote sensing.1,2 In physical medicine, for example, EM power,
in the form either of shortwaves or microwaves, is used to heat deep tissues and to stimulate
certain physiological responses in order to relieve certain pathological conditions. EM fields
are used in induction heaters for melting, forging, annealing, surface hardening, and soldering
operations. Dielectric heating equipment uses shortwaves to join or seal thin sheets of plastic
materials. EM energy offers many new and exciting possibilities in agriculture. It is used, for
example, to change vegetable taste by reducing acidity.
EM devices include transformers, electric relays, radio/TV, telephones, electric motors,
transmission lines, waveguides, antennas, optical fibers, radars, and lasers. The design of
these devices requires thorough knowledge of the laws and principles of EM.
1
For numerous applications of electrostatics, see J. M. Crowley, Fundamentals of Applied Electrostatics. New
York: John Wiley & Sons, 1986.
2
For other areas of applications of EM, see, for example, D. Teplitz, ed., Electromagnetism: Paths to Research.
New York: Plenum Press, 1982.
3
4
CHAPTER 1 VECTOR ALGEBRA
†
1.2 A PREVIEW OF THE BOOK
The subject of electromagnetic phenomena in this book can be summarized in Maxwell’s
equations:
= # D 5 rv
(1.1)
=#B50
(1.2)
'B
't
'D
=3H5J1
't
=3E52
(1.3)
(1.4)
where = 5 the vector differential operator
D 5 the electric flux density
B 5 the magnetic flux density
E 5 the electric field intensity
H 5 the magnetic field intensity
v 5 the volume charge density
J 5 the current density
Maxwell based these equations on previously known results, both experimental and theore­
tical. A quick look at these equations shows that we shall be dealing with vector quantities. It
is consequently logical that we spend some time in Part 1 examining the mathematical tools
required for this course. The derivation of eqs. (1.1) to (1.4) for time-invariant conditions
and the physical significance of the quantities D, B, E, H, J, and v will be our aim in Parts 2
and 3. In Part 4, we shall reexamine the equations for time-varying situations and apply
them in our study of practical EM devices such as transmission lines, waveguides, antennas,
fiber optics, and radar systems.
1.3 SCALARS AND VECTORS
Vector analysis is a mathematical tool with which electromagnetic concepts are most conveniently expressed and best comprehended. We must learn its rules and techniques before
we can confidently apply it. Since most students taking this course have little exposure to
vector analysis, considerable attention is given to it in this and the next two chapters.3 This
chapter introduces the basic concepts of vector algebra in Cartesian coordinates only. The
next chapter builds on this and extends to other coordinate systems.
A quantity can be either a scalar or a vector. A scalar is a quantity that is completely
specified by its magnitude.
†
Indicates sections that may be skipped, explained briefly, or assigned as homework if the text is covered in one
semester.
3
The reader who feels no need for review of vector algebra can skip to the next chapter.
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1.4 Unit Vector 5
A scalar is a quantity that has only magnitude.
Quantities such as time, mass, distance, temperature, entropy, electric potential, and population are scalars. A vector has not only magnitude, but direction in space.
A vector is a quantity that is described by both magnitude and direction.
Vector quantities include velocity, force, momentum, acceleration displacement, and electric
field intensity. Another class of physical quantities is called tensors, of which scalars and vectors
are special cases. For most of the time, we shall be concerned with scalars and vectors.4
To distinguish between a scalar and a vector
it> is customary to represent a vector by
>
a letter with an arrow on top of it, such as A and B , or by a letter in boldface type such as
A and B. A scalar is represented simply by a letter—for example, A, B, U, and V.
EM theory is essentially a study of some particular fields.
A field is a function that specifies a particular quantity everywhere in a region.
A field may indicate variation of a quantity throughout space and perhaps with time.
If the quantity is scalar (or vector), the field is said to be a scalar (or vector) field. Examples
of scalar fields are temperature distribution in a building, sound intensity in a theater, electric
potential in a region, and refractive index of a stratified medium. The gravitational force on
a body in space and the velocity of raindrops in the atmosphere are examples of vector fields.
1.4 UNIT VECTOR
A vector A has both magnitude and direction. The magnitude of A is a scalar written as
A or 0 A 0 . A unit vector aA along A is defined as a vector whose magnitude is unity (i.e., 1)
and its direction is along A; that is,
aA 5
A
A
5
0A0
A
Note that 0 aA 0 5 1. Thus we may write A as
A 5 AaA
(1.5)
(1.6)
which completely specifies A in terms of its magnitude A and its direction aA.
A vector A in Cartesian (or rectangular) coordinates may be represented as
1 Ax, Ay, Az 2 or Axax 1 Ayay 1 Azaz
(1.7)
4
For an elementary treatment of tensors, see, for example, A. I. Borisenko and I. E. Tarapor, Vector and Tensor
Analysis with Applications. New York: Dover, 1979.
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6
CHAPTER 1 VECTOR ALGEBRA
FIGURE 1.1 (a) Unit vectors ax, ay, and az, (b) components of A
along ax, ay, and az.
where Ax, Ay, and Az are called the components of A in the x-, y-, and z-directions, respectively; ax, ay, and az are unit vectors in the x-, y-, and z-directions, respectively. For example,
ax is a dimensionless vector of magnitude one in the direction of the increase of the x-axis.
The unit vectors ax, ay, and az are illustrated in Figure 1.1(a), and the components of A along
the coordinate axes are shown in Figure 1.1(b). The magnitude of vector A is given by
A 5 "A2x 1 A2y 1 A2z
(1.8)
and the unit vector along A is given by
aA 5
Axax 1 Ayay 1 Azaz
"A2x 1 A2y 1 A2z
(1.9)
1.5 VECTOR ADDITION AND SUBTRACTION
Two vectors A and B can be added together to give another vector C; that is,
C5A1B
(1.10)
The vector addition is carried out component by component. Thus, if A 5 1 Ax, Ay, Az)
and B 5 1 Bx, By, Bz).
C 5 1 Ax 1 Bx 2 ax 1 1 Ay 1 By 2 ay 1 1 Az 1 Bz 2 az
(1.11)
Vector subtraction is similarly carried out as
D 5 A 2 B 5 A 1 1 2B 2
5 1 Ax 2 Bx 2 ax 1 1 Ay 2 By 2 ay 1 1 Az 2 Bz 2 az
01_Sadiku_Ch01.indd 6
(1.12)
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1.6 Position and Distance Vectors 7
FIGURE 1.2 Vector addition C 5 A 1 B: (a) parallelogram rule,
(b) head-to-tail rule.
FIGURE 1.3 Vector subtraction
D 5 A 2 B: (a) parallelogram rule,
(b) h
­ ead-to-tail rule.
Graphically, vector addition and subtraction are obtained by either the parallelogram rule
or the head-to-tail rule as portrayed in Figures 1.2 and 1.3, respectively.
The three basic laws of algebra obeyed by any given vectors A, B, and C are summarized as follows:
Law
Commutative
Associative
Distributive
Addition
Multiplication
A1B5B1A
A 1 1B 1 C2 5 1A 1 B2 1 C
k 1 A 1 B 2 5 kA 1 kB
kA 5 Ak
k(,A) 5 (k,)A
where k and , are scalars. Multiplication of a vector with another vector will be discussed
in Section 1.7.
1.6 POSITION AND DISTANCE VECTORS
A point P in Cartesian coordinates may be represented by (x, y, z).
The position vector rP (or radius vector) of point P is defined as the directed distance from the origin O to P; that is,
rP 5 OP 5 xax 1 yay 1 zaz
01_Sadiku_Ch01.indd 7
(1.13)
01/11/17 1:14 PM
8
CHAPTER 1 VECTOR ALGEBRA
FIGURE 1.4 Illustration of position vector
rP 5 3ax 1 4ay 5 5az.
O
FIGURE 1.5 Distance vector rPQ.
The position vector of point P is useful in defining its position in space. Point (3, 4, 5), for
example, and its position vector 3ax 1 4ay 1 5az are shown in Figure 1.4.
The distance vector is the displacement from one point to another.
If two points P and Q are given by (xP, yP, zP) and (xQ, yQ, zQ), the distance vector (or
separation vector) is the displacement from P to Q as shown in Figure 1.5; that is,
rPQ 5 rQ 2 rP
5 1 xQ 2 xP 2 ax 1 1 yQ 2 yP 2 ay 1 1 zQ 2 zP 2 az
(1.14)
The difference between a point P and a vector A should be noted. Though both P
and A may be represented in the same manner as (x, y, z) and (Ax, Ay, Az), respectively,
the point P is not a vector; only its position vector rP is a vector. Vector A may depend on
point P, however. For example, if A 5 2xyax 1 y2ay 2 xz2az and P is 1 2, 21, 4 2 , then A at
P would be 24ax 1 ay 2 32az. A vector field is said to be constant or uniform if it does
not depend on space variables x, y, and z. For example, vector B 5 3ax 2 2ay 1 10az is a
uniform vector while vector A 5 2xyax 1 y2ay 2 xz2az is not uniform because B is the
same everywhere, whereas A varies from point to point.
EXAMPLE 1.1
01_Sadiku_Ch01.indd 8
If A 5 10ax 2 4ay 1 6az and B 5 2ax 1 ay, find (a) the component of A along ay, (b) the
magnitude of 3A 2 B, (c) a unit vector along A 1 2B.
01/11/17 1:14 PM
1.6 Position and Distance Vectors 9
Solution:
(a) The component of A along ay is Ay 5 24.
(b) 3A 2 B 5 3 1 10, 24, 6 2 2 1 2, 1, 0 2
Hence,
5 1 30, 212, 18 2 2 1 2, 1, 0 2
5 1 28, 213, 18 2
0 3A 2 B 0 5 "282 1 1 213 2 2 1 1 18 2 2 5 "1277
5 35.74
(c) Let C 5 A 1 2B 5 1 10, 24, 6 2 1 1 4, 2, 0 2 5 1 14, 22, 6 2 .
A unit vector along C is
ac 5
or
1 14, 22, 6 2
C
5
2
0C0
"14 1 1 22 2 2 1 62
ac 5 0.9113ax 2 0.1302ay 1 0.3906az
Note that 0 ac 0 5 1 as expected.
PRACTICE EXERCISE 1.1
Given vectors A 5 ax 1 3az and B 5 5ax 1 2ay 2 6az, determine
(a) uA 1 Bu
(b) 5A 2 B
(c) The component of A along ay
(d) A unit vector parallel to 3A 1 B
Answer: (a) 7, (b) (0, 22, 21), (c) 0, (d) 6(0.9117, 0.2279, 0.3419).
EXAMPLE 1.2
01_Sadiku_Ch01.indd 9
Points P and Q are located at (0, 2, 4) and 1 23, 1, 5 2 . Calculate
(a) The position of vector rP
(b) The distance vector from P to Q
(c) The distance between P and Q
(d) A vector parallel to PQ with magnitude of 10
01/11/17 1:14 PM
10 CHAPTER 1 VECTOR ALGEBRA
Solution:
(a) rP 5 0ax 1 2ay 1 4az 5 2ay 1 4az
(b) rPQ 5 rQ 2 rP 5 1 23, 1, 5 2 2 1 0, 2, 4 2 5 1 23, 21, 1 2
or rPQ 5 23ax 2 ay 1 az
(c) Since rPQ is the distance vector from P to Q, the distance between P and Q is the magnitude of this vector; that is,
d 5 0 rPQ 0 5 "9 1 1 1 1 5 3.317
Alternatively:
d 5 " 1 xQ 2 xP 2 2 1 1 yQ 2 yP 2 2 1 1 zQ 2 zP 2 2
5 "9 1 1 1 1 5 3.317
(d) Let the required vector be A, then
A 5 AaA
where A 5 10 is the magnitude of A. Since A is parallel to PQ, it must have the same unit
vector as rPQ or rQP. Hence,
aA 5 6
and
A56
rPQ
0 rPQ 0
56
1 23, 21, 1 2
3.317
10 1 23, 21, 1 2
5 6 1 29.045ax 2 3.015ay 1 3.015az 2
3.317
PRACTICE EXERCISE 1.2
Given points P(1, 23, 5), Q(2, 4, 6), and R(0, 3, 8), find (a) the position vectors of P and
R, (b) the distance vector rQR, (c) the distance between Q and R.
Answer: (a) ax 2 3ay 1 5az, 3ax 1 8az, (b) 22ax 2 ay 1 2az, (c) 3.
EXAMPLE 1.3
A river flows southeast at 10 km/hr and a boat floats upon it with its bow pointed in the
direction of travel. A man walks upon the deck at 2 km/hr in a direction to the right and
perpendicular to the direction of the boat’s movement. Find the velocity of the man with
respect to the earth.
Solution:
Consider Figure 1.6 as illustrating the problem. The velocity of the boat is
ub 5 10 1 cos 45° ax 2 sin 45° ay 2
5 7.071ax 2 7.071ay km/hr
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1.7 Vector Multiplication 11
FIGURE 1.6 For Example 1.3.
The velocity of the man with respect to the boat (relative velocity) is
um 5 2 1 2 cos 45° ax 2 sin 45° ay 2
5 21.414ax 2 1.414ay km/hr
Thus the absolute velocity of the man is
uab 5 um 1 ub 5 5.657ax 2 8.485ay
256.3°
0 uab 0 5 10.2 li
ii
t
that
is, 10.2 km/hr at 56.3° south of east.
PRACTICE EXERCISE 1.3
An airplane has a ground speed of 350 km/hr in the direction due west. If there is a wind
blowing northwest at 40 km/hr, calculate the true air speed and heading of the airplane.
Answer: 379.3 km/hr, 4.275° north of west.
1.7 VECTOR MULTIPLICATION
When two vectors A and B are multiplied, the result is either a scalar or a vector depending
on how they are multiplied. Thus there are two types of vector multiplication:
1. Scalar (or dot) product: A # B
2. Vector (or cross) product: A 3 B
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12 CHAPTER 1 VECTOR ALGEBRA
Multiplication of three vectors A, B, and C can result in either:
or
3. Scalar triple product: A # 1 B 3 C 2
4. Vector triple product: A 3 1 B 3 C 2
A. Dot Product
The dot product of two vectors A and B, written as A ? B, is defined geometrically
as the product of the magnitudes of A and B and the cosine of the smaller angle
between them when they are drawn tail to tail.
Thus,
A # B 5 AB cos u AB
(1.15)
where uAB is the smaller angle between A and B. The result of A # B is called either the scalar
product because it is scalar or the dot product due to the dot sign. If A 5 1 Ax, Ay, Az 2 and
B 5 1 Bx, By, Bz), then
A # B 5 AxBx 1 AyBy 1 AzBz
(1.16)
which is obtained by multiplying A and B component by component. Two vectors A and B
are said to be orthogonal (or perpendicular) with each other if A # B 5 0.
Note that dot product obeys the following:
(i) Commutative law:
A#B5B#A
(1.17)
A # 1B 1 C2 5 A # B 1 A # C
(1.18)
A # A 5 0 A 0 2 5 A2
(1.19)
ax # ay 5 ay # az 5 az # ax 5 0
ax # ax 5 ay # ay 5 az # az 5 1
(1.20a)
(1.20b)
(ii) Distributive law:
(iii)
Also note that
It is easy to prove the identities in eqs. (1.17) to (1.20) by applying eq. (1.15) or (1.16).
If A # B 5 0, the two vectors A and B are orthogonal or perpendicular.
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1.7 Vector Multiplication 13
B. Cross Product
The cross product of two vectors A and B, written as A 3 B, is a vector quantity
whose magnitude is the area of the parallelogram formed by A and B (see Figure 1.7)
and is in the direction of advance of a right-handed screw as A is turned into B.
Thus,
A 3 B 5 AB sin u ABan
(1.21)
where an is a unit vector normal to the plane containing A and B. The direction of an is
taken as the direction of the right thumb when the fingers of the right hand rotate from
A to B as shown in Figure 1.8(a). Alternatively, the direction of an is taken as that of the
advance of a right-handed screw as A is turned into B as shown in Figure 1.8(b).
The vector multiplication of eq. (1.21) is called cross product owing to the cross
sign; it is also called vector product because the result is a vector. If A 5 1 Ax, Ay, Az) and
B 5 1 Bx, By, Bz), then
ay
az
A 3 B 5 3 Ax Ay
Az 3
ax
Bx
By
(1.22a)
Bz
5 1 AyBz 2 AzBy 2 ax 1 1 AzBx 2 AxBz 2 ay 1 1 AxBy 2 AyBx 2 az
(1.22b)
which is obtained by “crossing” terms in cyclic permutation, hence the name “cross
product.”
AB
B
A
FIGURE 1.7 The cross product of A and B is a vector with magnitude equal
to the area of the parallelogram and direction as indicated.
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14 CHAPTER 1 VECTOR ALGEBRA
FIGURE 1.8 Direction of A 3 B and an using (a) the right-hand rule and (b) the
right-handed-screw rule.
Note that the cross product has the following basic properties:
(i) It is not commutative:
A3B2B3A
(1.23a)
A 3 B 5 2B 3 A
(1.23b)
(1.24)
(iii) It is distributive:
A 3 1B 3 C2 2 1A 3 B2 3 C
(1.25)
(iv) Scaling:
A 3 1B 1 C2 5 A 3 B 1 A 3 C
(1.26)
(v)
kA 3 B 5 A 3 kB 5 k 1 A 3 B 2
A3A50
(1.27)
ax 3 ay 5 az
ay 3 az 5 ax
az 3 ax 5 ay
(1.28)
It is anticommutative:
(ii) It is not associative:
Also note that
which are obtained in cyclic permutation and illustrated in Figure 1.9. The identities in eqs.
(1.23) to (1.28) are easily verified by using eq. (1.21) or (1.22). It should be noted that in
obtaining an, we have used the right-hand or right-handed-screw rule because we want to
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01/11/17 1:14 PM
1.7 Vector Multiplication 15
FIGURE 1.9 Cross product using cyclic permutation. (a) Moving
clockwise leads to positive results. (b) Moving counterclockwise
leads to negative results.
be consistent with our coordinate system illustrated in Figure 1.1, which is right-handed.
A right-handed coordinate system is one in which the right-hand rule is satisfied: that is,
ax 3 ay 5 az is obeyed. In a left-handed system, we follow the left-hand or left-handed
screw rule and ax 3 ay 5 2az is satisfied. Throughout this book, we shall stick to righthanded coordinate systems.
Just as multiplication of two vectors gives a scalar or vector result, multiplication of
three vectors A, B, and C gives a scalar or vector result, depending on how the vectors are
multiplied. Thus we have a scalar or vector triple product.
C. Scalar Triple Product
Given three vectors A, B, and C, we define the scalar triple product as
A # 1B 3 C2 5 B # 1C 3 A2 5 C # 1A 3 B2
(1.29)
obtained in cyclic permutation. If A 5 1 Ax, Ay, Az), B 5 1 Bx, By, Bz), and C 5 1 Cx, Cy, Cz),
then A # 1 B 3 C 2 is the volume of a parallelepiped having A, B, and C as edges and is easily
obtained by finding the determinant of the 3 3 3 matrix formed by A, B, and C; that is,
Ax A y
Az
A # 1 B 3 C 2 5 3 Bx By Bz 3
Cx
Cy
(1.30)
Cz
Since the result of this vector multiplication is scalar, eq. (1.29) or (1.30) is called the scalar
triple product.
D. Vector Triple Product
For vectors A, B, and C, we define the vector triple product as
A 3 1B 3 C2 5 B1A # C2 2 C1A # B2
01_Sadiku_Ch01.indd 15
(1.31)
01/11/17 1:14 PM
16 CHAPTER 1 VECTOR ALGEBRA
which may be remembered as the “bac-cab” rule. It should be noted that
but
1A # B2C 2 A1B # C2
(1.32)
1A # B2C 5 C1A # B2
(1.33)
1.8 COMPONENTS OF A VECTOR
A direct application of scalar product is its use in determining the projection (or component) of a vector in a given direction. The projection can be scalar or vector. Given a vector
A, we define the scalar component AB of A along vector B as [see Figure 1.10(a)]
or
AB 5 A cos u AB 5 0 A 0 0 aB 0 cos u AB
A B 5 A # aB
(1.34)
The vector component AB of A along B is simply the scalar component in eq. (1.34) multiplied by a unit vector along B; that is,
AB 5 ABaB 5 1 A # aB 2 aB
(1.35)
Both the scalar and vector components of A are illustrated in Figure 1.10. Notice from Figure
1.10(b) that the vector can be resolved into two orthogonal components: one c­ omponent AB parallel to B, another 1 A 2 AB 2 perpendicular to B. In fact, our Cartesian representation of a vector
is essentially resolving the vector into three mutually orthogonal components as in Figure 1.1(b).
We have considered addition, subtraction, and multiplication of vectors. However, division of vectors A/B has not been considered because it is undefined except when A and B are
parallel so that A 5 kB, where k is a constant. Differentiation and integration of vectors will be
considered in Chapter 3.
FIGURE 1.10 Components of A along B: (a) scalar component AB,
(b) vector component AB.
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1.8 Components of a Vector 17
EXAMPLE 1.4
Given vectors A 5 3ax 1 4ay 1 az and B 5 2ay 2 5az, find the angle between A and B.
Solution:
The angle uAB can be found by using either dot product or cross product.
A # B 5 1 3, 4, 1 2 # 1 0, 2, 25 2
50182553
0 A 0 5 "32 1 42 1 12 5 "26
cos u AB 5
0 B 0 5 "02 1 22 1 1 25 2 2 5 "29
3
A#B
5
5 0.1092
0A0 0B0
" 1 26 2 1 29 2
u AB 5 cos21 0.1092 5 83.73°
Alternatively:
ax
ay
az
A 3 B 5 33
4
1 3
0 2 25
1
5 220 2 2 2 ax 1 1 0 1 15 2 ay 1 1 6 2 0 2 az
5 1 222, 15, 6 2
0 A 3 B 0 5 " 1 222 2 2 1 152 1 62 5 "745
sin u AB 5
0A 3 B0
"745
5
5 0.994
0A0 0B0
" 1 26 2 1 29 2
u AB 5 sin21 0.994 5 83.73°
PRACTICE EXERCISE 1.4
If A 5 ax 1 3az and B 5 5ax 1 2ay 2 6az, find uAB.
Answer: 120.6°.
EXAMPLE 1.5
Three field quantities are given by
P 5 2ax 2 az
Q 5 2ax 2 ay 1 2az
R 5 2ax 2 3ay 1 az
Determine
(a) 1 P 1 Q 2 3 1 P 2 Q 2
(b) Q # R 3 P
(c) P # Q 3 R
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18 CHAPTER 1 VECTOR ALGEBRA
(d) sin uQR
(e) P 3 1 Q 3 R 2
(f) A unit vector perpendicular to both Q and R
(g) The component of P along Q
Solution:
(a)
1P 1 Q2 3 1P 2 Q2 5 P 3 1P 2 Q2 1 Q 3 1P 2 Q2
5P3P2P3Q1Q3P2Q3Q
501Q3P1Q3P20
5 2Q 3 P
ax
ay
az
5 232
2 3
21
2
0 21
5 2 1 1 2 0 2 ax 1 2 1 4 1 2 2 ay 1 2 1 0 1 2 2 az
5 2ax 1 12ay 1 4az
(b) The only way Q # R 3 P makes sense is
ax
Q # 1 R 3 P 2 5 1 2, 21, 2 2 # 3 2
ay
az
23
1 3
0
21
2
Q # 1R 3 P2 5 3 2
2
2
13
21
2
5 1 2, 21, 2 2 # 1 3, 4, 6 2
5 6 2 4 1 12 5 14
Alternatively:
21
23
0
To find the determinant of a 3 3 3 matrix, we repeat the first two rows and cross multiply;
when the cross multiplication is from right to left, the result should be negated as shown
diagrammatically here. This technique of finding a determinant applies only to a 3 3 3
matrix. Hence,
2
2
Q # 1R 3 P2 5 5 2
2
2
21
23
0
21
23
2
1
21 5 2 1 5 16 1 0 2 2 1 12 2 0 2 2
5 14
as obtained before.
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1.8 Components of a Vector 19
(c) From eq. (1.29)
P # 1 Q 3 R 2 5 Q # 1 R 3 P 2 5 14
or
P # 1 Q 3 R 2 5 1 2, 0, 21 2 # 1 5, 2, 24 2
5 10 1 0 1 4
5 14
(d)
sin u QR 5
5
1 5, 2, 24 2 0
0Q 3 R0
5
0Q0 0R0
0 1 2, 21, 2 2 0 0 1 2, 23, 1 2 0
"45
3"14
5
"5
"14
5 0.5976
(e)
P 3 1 Q 3 R 2 5 1 2, 0, 21 2 3 1 5, 2, 24 2
5 1 2, 3, 4 2
Alternatively, using the bac-cab rule,
P 3 1Q 3 R2 5 Q1P # R2 2 R1P # Q2
5 1 2, 21, 2 2 1 4 1 0 2 1 2 2 1 2, 23, 1 2 1 4 1 0 2 2 2
5 1 2, 3, 4 2
(f ) A unit vector perpendicular to both Q and R is given by
a5
6 1 5, 2, 24 2
6Q 3 R
5
0Q 3 R0
"45
5 6 1 0.745, 0.298, 20.596 2
Note that 0 a 0 5 1, a # Q 5 0 5 a # R. Any of these can be used to check a.
(g) The component of P along Q is
PQ 5 0 P 0 cos u PQaQ
5 1 P # aQ 2 aQ 5
1P # Q2Q
0Q02
1 4 1 0 2 2 2 1 2, 21, 2 2
2
5 1 2, 21, 2 2
14 1 1 1 42
9
5 0.4444ax 2 0.2222ay 1 0.4444a
5
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20 CHAPTER 1 VECTOR ALGEBRA
PRACTICE EXERCISE 1.5
Let E 5 3ay 1 4az and F 5 4ax 2 10ay 1 5az.
(a) Find the component of E along F.
(b) Determine a unit vector perpendicular to both E and F.
Answer: (a) (20.2837, 0.7092, 20.3546), (b) 6(0.9398, 0.2734, 20.205).
EXAMPLE 1.6
Derive the cosine formula
a2 5 b2 1 c2 2 2bc cos A
and the sine formula
sin B
sin C
sin A
5
5
a
c
b
using dot product and cross product, respectively.
Solution:
Consider a triangle as shown in Figure 1.11. From the figure, we notice that
a1b1c50
that is,
b 1 c 5 2a
Hence,
a2 5 a # a 5 1 b 1 c 2 # 1 b 1 c 2
5 b # b 1 c # c 1 2b # c
2
a 5 b2 1 c2 2 2bc cos A
where (p 2 A) is the angle between b and c.
The area of a triangle is half of the product of its height and base. Hence,
1
1
1
` a 3 b` 5 ` b 3 c` 5 ` c 3 a `
2
2
2
ab sin C 5 bc sin A 5 ca sin B
Dividing through by abc gives
sin B
sin C
sin A
5
5
a
c
b
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1.8 Components of a Vector 21
FIGURE 1.11 For Example 1.6.
PRACTICE EXERCISE 1.6
Show that vectors a 5 (4, 0, 21), b 5 (1, 3, 4), and c 5 (25, 23, 23) form the sides
of a triangle. Is this a right angle triangle? Calculate the area of the triangle.
Answer: Yes, 10.5.
EXAMPLE 1.7
Show that points P1 1 5, 2, 24 2 , P2 1 1, 1, 2 2 , and P3 1 23, 0, 8 2 all lie on a straight line.
Determine the shortest distance between the line and point P4 1 3, 21, 0 2 .
Solution:
The distance vector rP1P2 is given by
Similarly,
rP1P2 5 rP2 2 rP1 5 1 1, 1, 2 2 2 1 5, 2, 24 2
5 1 24, 21, 6 2
rP1P3 5 rP3 2 rP1 5 1 23, 0, 8 2 2 1 5, 2, 24 2
5 1 28, 22, 12 2
rP1P4 5 rP4 2 rP1 5 1 3, 21, 0 2 2 1 5, 2, 24 2
5 1 22, 23, 4 2
ax
ay
az
rP1P2 3 rP1P3 5 3 24
21
63
28 22
5 1 0, 0, 0 2
12
showing that the angle between rP1P2 and rP1P3 is zero 1 sin u 5 0 2 . This implies that P1, P2,
and P3 lie on a straight line.
Alternatively, the vector equation of the straight line is easily determined from Figure
1.12(a). For any point P on the line joining P1 and P2
rP1P 5 lrP1P2
where λ is a constant. Hence the position vector rP of the point P must satisfy
rP 2 rP1 5 l 1 rP2 2 rP1 2
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22 CHAPTER 1 VECTOR ALGEBRA
FIGURE 1.12 For Example 1.7.
that is,
rP 5 rP1 1 l 1 rP2 2 rP1 2
5 1 5, 2, 24 2 2 l 1 4, 1, 26 2
rP 5 1 5 2 4l, 2 2 l, 24 1 6l 2
This is the vector equation of the straight line joining P1 and P2. If P3 is on this line, the
position vector of P3 must satisfy the equation; r3 does satisfy the equation when l 5 2.
The shortest distance between the line and point P4 1 3, 21, 0 2 is the perpendicular
distance from the point to the line. From Figure 1.12(b), it is clear that
d 5 rP1P4 sin u 5 0 rP1P4 3 aP1P2 0
0 1 22, 23, 4 2 3 1 24, 21, 6 2 0
5
0 1 24, 21, 6 2 0
5
"312
"53
5 2.426
Any point on the line may be used as a reference point. Thus, instead of using P1 as a reference
point, we could use P3. If jP4P3 P2 5 , then
d 5 0 rP3P4 0 sin u r 5 0 rP3P4 3 aP3P2 0
PRACTICE EXERCISE 1.7
If P1 is (1, 2, 23) and P2 is (24, 0, 5), find
(a) The distance P1P2
(b) The vector equation of the line P1P2
(c) The shortest distance between the line P1P2 and point P3 (7, 21, 2)
Answer: (a) 9.644, (b) (1 2 5l)ax 1 2(1 2 l)ay 1 (8l 2 3)az, (c) 8.2.
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