elements of electromagnetics third edition e book chapter 01

7/22/2019 Elements of Electromagnetics Third Edition E Book Chapter 01

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PART 1V E C T O R A N A L Y S I S


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h pter 7 V .S- f

VECT OR A LGEBR AOne thing I have learned in a long life: that all our science, measured againstreality, is primitive and childlike and yet is the m ost precious thing w e have.

ALBERT EINSTEIN

1.1 INTRODUCTIONElectromagnetics (EM) may be regarded as the study of the interactions between electriccharges at rest and in motion. It entails the analysis, synthesis, physical interpretation, andapplication of electric and magnetic fields.

Kkctioniiiniutics k.Yli is a branch of physics or electrical engineering in whichelectric and magnetic phenomena are studied.

EM principles find applications in various allied disciplines such as microwaves, an-tennas, electric machines, satellite communications, bioelectromagnetics, plasmas, nuclearresearch, fiber optics, electromagnetic interference and compatibility, electromechanicalenergy conversion, radar meteorology, and remo te sensing.1'2 In physical medicine, forexample, EM power, either in the form of shortwaves or microwaves, is used to heat deeptissues and to stimulate certain physiological responses in order to relieve certain patho-logical cond itions. EM fields are used in induction hea ters for melting, forging, anne aling,surface hardening, and soldering operations. Dielectric heating equipment uses shortwavesto join or seal thin sheets of plastic materials. EM energy offers many new and excitingpossibilities in agriculture. It is used, for example, to change vegetable taste by reducingacidity.

EM devices include transform ers, electric relays, radio/TV, telephon e, electric mo tors,transmission lines, waveguides, antennas, optical fibers, radars, and lasers. The design ofthese devices requires thorough know ledge of the laws and principles of EM .

For numerous applications of electrostatics, see J. M. Crowley,FundamentalsofAppliedElectro-statics.NewYork:John Wiley Sons, 1986.2For otherareasof applications of EM ,see,for example,D.Teplitz,ed.,Electromagnetism:PathstoResearch.NewYork:Plenum Press, 1982.


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4 Vector A lgebra+1.2 A PREVIEW OF THE B OO K

The subject of electromagnetic phenom ena in this book can be summ arized in M axw ell'sequations:V - D = p v (1.1)V B = 0 (1.2)

* * V X E = - 1 . 3 )dtV X H = J + (1.4)dt

where V = the vector differential operatorD = the electric flux densityB = the magnetic flux densityE = the electric field intensityH = the magnetic field intensityp v = the volume charge densityand J = the current density.

Maxw ell based these equations on previously known results, both experimental and theo-retical. A quick look at these equations shows that we shall be dealing with vector quanti-ties.It is consequently logical that we spend some time in Part I examining the mathemat-ical tools required for this course. The derivation of eqs. (1.1) to (1.4) for time-invariantconditions and the physical significance of the quantities D, B, E, H, J and p vwill be ouraim in Parts II andIII.In Part IV, we shall reexam ine the equations for time-v arying situa-tions and apply them in our study of practical EM devices.

1.3 SCALARS A N D VECTORSVector analysis is a mathem atical tool with which electromagn etic (EM ) concepts are mostconveniently expressed and best comprehended. We must first learn its rules and tech-niques before we can confidently apply it. Since mo st students taking this course ha ve littleexposure to vector analysis, considerable attention is given to it in this and the next twochapters.3This chapter introduces the basic concepts of vector algebra in Cartesian coordi-nates only. The next chapter builds on this and extends to other coordinate systems.A quantity can be either a scalar or a vector.

Indicates sections that may be skipped, explained briefly, or assigned as homework if the text iscovered in one semester.

3The reade r who feels n o need for review of vector algebra can skip to the next ch apter.


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1.4 U N I T VECTOR

A scalaris a quantity that has only magnitude.Quantities such as time, mass, distance, temperature, entropy, electric potential, and popu-lation are scalars.

Avectoris a quantity that has both magnitude and direction.Vector quantities includ e velocity, force, displacemen t, and electric field intensity. Ano therclass of physical quantities is calledtensors,of which scalars and vectors are special cases.For most of the time, we shall be concerned with scalars and vectors.4To distinguish between a scalar and a vector it is customary to represent a vector by aletter with an arrow on top of it, such as A and B , or by a letter in boldface type such as AandB.A scalar is represented simply by a lettere.g.,A, B, U ,andV.

EM theory is essentially a study of some particular fields.Afieldis a function that specifies a particular quantity every whe re in a region.

If the quantity is scalar (or vector), the field is said to be a scalar (or vector) field. Exam-ples 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 gravitationalforce on a body in space and the velocity of raindrops in the atmosphere are examples ofvector fields.

1.4 U N IT VECTORA vector A has both magnitude and direction. Themagnitudeof A is a scalar written asAor |A |. Aunit vector aAalong A is defined as a vector whose mag nitude is unity (i.e., 1) andits direction is along A, that is,

(1-5)

(1.6)

(1.7)

Note that |aA | = 1. Thus we may write A asA = AaA

which completely specifies A in terms of its magnitudeAand its direction aA.A vector A in Cartesian (or rectangular) coordinates may be represented as(A x, Ay, A z) or Ayay + Azaz

4For an elementary treatment of tensors, see, for example, A. I. Borisenko and I. E. Tarapor,Vectorand T ensor Analysis with Application.Englewood Cliffs, NJ: Prentice-Hall, 1968.


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Vector A lgebra

H 1 \-y

a) b)Figure 1.1 (a) Unit vectorsax, ay,and az, (b) components ofAalongax, a ,,and az.

where Ax, A r andAz are called the components of A in thex, y,and z directions respec-tively;ax, aT andazare unit vectors in thex, y, andzdirections, respectively. For example,axis a dimensionless vector of magnitude one in the direction of the increase of the x-axis.The unit vectorsax,a,,, andazare illustrated in Figure 1.1(a), and the components ofAalongthe coordinate ax es are shown in Figure 1.1(b). The magnitude of vectorA is given byA = VA2X Al A\

and the unit vector along A is given byAxa x Aza z

VAT+AT+AI

(1-8)

(1.9)

1.5 VECT O R A D D I T I O N A N D S UB T R A C T IO NTwo vectors A and B can be added together to give another vector C; that is,

C = A + B (1.10)The vector addition is carried out com ponent by com ponent. Thus, if A = (Ax, A y, A z) andB = (Bx,B y,B z).

C = (A x + Bx)a x + {Ay + By)a y + (Az + Bz)a zVector subtraction is similarly carried out as

D = A - B = A + ( - B )= (Ax - Bx)a x + (Ay - By)a y + (Az - Bz)a z

( l . H )

(1.12)


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Ba)

1.6 POSITION A N D DISTANCE VECTORS

b)Figure 1.2 Vector addition C = A + B: (a) parallelogram rule,(b) head-to-tail rule.

Figure 1.3 Vector subtraction D = A -B:(a) parallelogram rule, (b) head-to-tailfA rule.

a) b)

Graphically, vector addition and subtraction are obtained by either the parallelogram ruleor the head-to-tail rule as portrayed in Figures 1.2 and 1.3, respectively.

The three basic laws of algebra obeyed by any giveny vectors A, B, and C, are sum-marized as follows:

Law A ddition MultiplicationCommutative A + B = B + A kA = AkAssociative A + (B + C) = (A + B) + C k( (A) = (k()ADistributive k( A + B) = kA +ZfcB

wherekand are scalars. Multiplication of a vector with another vector will be discussedin Section 1.7.

1.6 PO SITIO N A N D DISTANCE VECTORSA pointPin Cartesian coordinates may be represented by(x, y, z).

The position vector r,. or radius vector) of point P is as (he directed silancc fromthe origin ) lo P:i.e..r P = OP = xax + yay (1.13)


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8 Vector Algebra

4,5)/ I- - / I

111 I A I

Figure 1.4 Illustration of position vector rP3a, + 4a., + 5az.

Figure 1.5 Distance vector rPG .

The position vector of pointPis useful in defining its position in space . Point (3 , 4, 5), forexample, and its position vector 3a x + 4a>( + 5azare shown in Figure 1.4.The distance vector isihcdisplacement from one point to another.

If two pointsPandQare given by(xP, yP, zp) and(xe, yQ, Z Q , thedistance vector(orseparation vector)is the displacement fromPtoQas shown in Figure 1.5; that is,rPQ ~ rQ rP= (x Q - xP)a x + (yQ - yP) y + (zQ - zP)a z (1.14)

The difference between a pointPand a vector A should be noted. Though bothPandA m ay be represented in the same manner as(x,y, z)an d( Ax,Ay, A z),respectively, the pointP is not a vector; only its position vector i> is a vector. Vector A may depend on point P,however. For example, if A = 2xya,t + y2ay - xz2az and Pis (2, - 1 , 4 ) , then A at Pwould be 4a^ + ay 32a;,. A vector field is said to beconstantoruniform if it does notdepend on space variables x, y, and z. For example, vector B = 3a^ 2a^, + 10az is auniform vector while vector A = 2xya x + y2ay xz2a z is not uniform because B is thesame everywhe re whereas A varies from point to point.

EXAMPLE 1 .1 If A = 10ax - 4ay + 6 az a n d B = 2 x+ av, find: (a) the com ponent ofAalongay,(b) themagnitude of 3A - B, (c) a unit vector along A + 2B.


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1 6 POSIT ION ANDDISTANCE VECTORS

Solution:a) The component ofAalongayisAy= -4.b) 3A - B = 3 10, -4, 6) - 2,1,0)

= 30,-12,18) - 2, 1,0)= 28,-13,18)

Hence|3A - B| = V282 + -13)2 + 18)2= VT277

= 35.74c) Let C = A + 2B = 10, -4, 6) + 4, 2, 0) = 14, -2, 6).Aunit vector along C is

14,-2,6)

orVl42 + -2)2 + 62

ac = 0.91\3ax - 0.1302a,, + 0.3906azNote that |ac| = 1asexpected.

PRACTICE EXERCISE 1.1Given vectorsA =ax+ 3a. andB =5ax+ 2av- 6a,, determine(a) |A + B(b) 5A - B(c) The componentofA alongav(d) A unit vector parallelto3A 4-BAnswer: (a)7, b)(0,- 2 ,21), (c)0, d) (0.9117, 0.2279, 0.3419).

PointsPandQare locatedat(0, 2,4) and( - 3 , 1, 5). Calculate(a) The position vectorP(b) The distance vector fromPtoQ(c) The distance betweenPandQ(d) A vector parallel toPQwith magntudeof 10


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1 Vector AlgebraSolution:(a) i> = 0ax + 2av + 4az = 2a, + 4az(b) rPQ =rQ - i> = ( - 3 , 1, 5) - (0, 2, 4) = ( - 3 , - 1 , 1)or = - 3 ax - ay + az(c) SincerPQis the distance vector fromPtoQ,the distance betweenPandQis the mag-nitude of this vector; that is,

Alternatively:d = |i>e | = V 9 + 1 + 1 = 3.317

d= V xQ - xPf + yQ- yPf + zQ- zPf= V 9 + T + T = 3.317

(d) Let the required vector be A, thenA = AaA

whereA = 10 is the magnitude ofA.Since A is parallel toPQ,it must have the same unitvector asrPQorrQP.Hence,rPQ

and

(-3,-1,1)3.317

A = I 0 ( 3' * ' - = (-9.045a^ - 3.015a, + 3.015az)

PRACTICE EXERCISE 1.2Given points P(l , - 3 , 5),Q(2,4, 6), andR(0,3, 8), find: (a) the position vectors ofPandR,(b) the distance vectorrQR,(c) the distance betweenQandR,Answer: a)ax 3ay + 5az, 3a* + 33,, b) 2a* - ay + 2az.

EXAMPLE 1.3 A river flows southeast at 10 km/hr and a boat flows upon it with its bow pointed in the di-rection oftravel.A man walks upon the deck at 2 km/hr in a direction to the right and per-pendicular to the direction of the boat's movement. Find the velocity of the man withrespect to the earth.Solution:Consider Figure 1.6 as illustrating the problem. The velocity of the boat is

ub = 10(cos 45ax - sin 45 a,)= 7.071a^ - 7.071a, km/hr


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w-

1.7 VECTO R MULTIPLICATION 11Figure 1.6 For Example 1.3.

The velocity of the man with respect to the boat (relative velocity) isum = 2( -c os 45 ax - sin 45 a,,)= -1 .414a , - 1.414a,,km/hr

Thus the absolute velocity of the man isuab = um + uh = 5.657a., - 8.485ay| u j = 1 0 . 2 / -5 6 . 3

that is, 10.2 km/hr at 56.3 south of east.

PRACTICE EXERCISE 1.3An airplane has a ground speed of 350 km/hr in the direction due west. If there is awind blowing northwest at 40 km/hr, calculate the true air speed and heading of theairplane.Answer: 379.3 km/hr, 4.275 north of west.

1.7 VECT OR MU LT IP L ICA T IONWhen two vectors A and B are multiplied, the result is either a scalar or a vector depend-ing on how they are multiplied. Thus there are two types of vector multiplication:

1. Scalar (or dot) product: A B2. Vector (or cross) product: A X B


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12 HI Vector A lgebraMultiplication of three vectors A, B, and C can result in either:

3. Scalar triple product: A(B X C )or

4. Vector triple product: A X (B X C )

A. Dot ProductThedot productof two vectors A and B , wrilten as A B. is defined geom etricallyas the product of the magnitudes of A and B and the cosine of the angle betweenthem.

Thus:A B = ABco sI (1.15)

where 6AB is thesmallerangle between A and B. The result of A B is called either thescalar product because it is scalar, or the dot product due to the dot sign. If A =(A x, Ay, A z) and B = (Bx, By, B z), then

A B = AXBX + AyBy + AZBZ (1.16)which is obtained by mu ltiplying A and B com ponent by com ponent. Two vectors A and Bare said to beorthogonal(or perpendicular) with each other if A B = 0.

Note that dot product obeys the following:(i) Com mutative law:

A - B = B - A (1 .1 7)(ii) Distributive law:

A (B + C) = A B + A C (1.18)A - A = |A |2 = A2 (1.19)

(iii) Also note thatax ay = ayaz = az ax = 0 (1.20a)ax ax = ay ay = az az = 1 (1.20b)

It is easy to prove the identities in eqs. (1.17) to (1.20) by applying eq. (1.15) or (1.16).


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1.7 VECTOR MULT IP L ICA T ION H 13

B. Cross P roductThe cross product of two vectors A ;ind B. written as A X B. is a vector quantitywhose magnitude is ihe area of the parallclopiped formed by A and It (see Figure1.7) and is in the direction of advance ofaright-handed screw as A is turned into B.

ThusA X B = AB sin6ABan (1.21)

where an is a unit vector normal to the plane containing A and B. The direction of an istaken as the direction of the right thum b when the fingers of the right hand rotate from A toB as shown in Figure 1.8 a). Alternatively, the direction of an is taken as that of theadvance of a right-handed screw as A is turned into B as shown in Figure1.8 b).

The vector multiplication of eq. (1.21) is calledcrossproductdue to the cross sign; itis also called vector product because the result is a vector. If A = (AxB = (Bx, B y, B z)then

A X B = axAxBxavAyBy

azKB z- A zBy)a x + (AZBX - AxB z)a y + (AxBy - AyBx)a z

Ay, Az) and

(1.22a)

(1.22b)which is obtained by crossing terms in cyclic permutation, hence the name crossproduct.

Figure 1.7 Thecross product ofA and Bisavectorwithmagnitude equaltothearea oftheparallelogram and direction as indicated.


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14 H Vector A lgebraA X B A X B

* - A

(a) (b)Fig ure 1.8 Direction ofA X B and anusing (a) right-hand rule, (b) right-handedscrew rule.

Note that the cross product has the fol lowing basic proper t ies:( i) I t i s not com mu tat ive:

I t i s ant icommutat ive:A X B ^ B X A

A X B = - B X A(ii) I t is not ass ocia tive:

A X (B X C) =h (A X B) X C(ii i) I t is distr ibutiv e:

(iv)A X B + C) = A X B + A X C

A X A = 0Also note that

ax X ay = aza, X a z = axaz X ax = ay

(1.23a)

(1.23b)

(1.24)

(1.25)

(1.26)

(1.27)

which are obtained in cyclic perm utation and illustrated in Figure 1.9. The identities in eqs.(1.25) to (1.27) are easily verified using eq. (1.21) or (1.22). It should be noted that in ob-tainingan,we have used the right-hand or right-handed screw rule because we w ant to beconsistent with our coordinate system illustrated in Figure 1.1, which is right-handed. Aright-handed coordinate system is one in which the right-hand rule is satisfied: that is,ax Xay = az is obeyed. In a left-handed system, we follow the left-hand or left-handed


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1.7 VECTOR MULT IP L ICAT ION 15

a) b)Figure 1.9 Cross product using cyclic permutation: (a) movingclockwise leads to positive results: (b) moving counterclockwiseleads to negative results.

screw rule and ax Xay = -az is satisfied. Throug hout this book, we shall stick to right-handed coordinate systems.Just as multiplication of two vectors gives a scalar or vector result, multiplication ofthree vectors A, B, and C gives a scalar or vector result depending on how the vectors aremultiplied. Thus we have scalar or vector triple product.

C. Scalar T riple P roductGiven three vectors A, B, and C , we define the scalar triple product as

A (B X C) = B (C X A) = C (A X B) (1.28)obtained in cyclic permu tation. IfA = (Ax, A y, A z),B = (Bx, B y, B z), and C = (Cx, C y,Cz),then A (B X C ) is the volum e of a parallelepiped h aving A, B , and C as edges and iseasily obtained by finding the determinant of the 3 X 3 matrix formed by A, B, and C;that is,

A (B X C) = Bx By BzCy C,

(1.29)

Since the result of this vector multiplication is scalar, eq. (1.28) or (1.29) is called thescalar triple product.

D. Vector T riple P roductFor vectors A, B, and C, we define the vector tiple product as

A X (B X C) = B(A C) - C(A B) (1.30)


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16 Vector A lgebraobtained using the bac -cab rule. It should be noted that

bu t(A B)C # A(B C)

(A B)C = C(A B).

(1.31)

(1.32)

1.8 CO MP ON ENT S OF A VECTORA direct application of vector product is its use in determining the projection (or compo-nent) of a vector in a given direction. T he projection ca n be scalar or vector. Given a vectorA, we define thescalar compone nt ABof A along vector B as [see Figure1.10 a)]

AB = A cos6AB = |A| |aB | cos6ABor

AR =Aafl (1.33)The vector component AB ofAalong B is simply the scalar component in eq. (1.33) multi-plied by a unit vector along B; that is,

AB = ABaB = (A (1-34)Both the scalar and vector components of A are illustrated in Figure 1.10. Notice fromFigure1.10 b)that the vector can be resolved into two orthogonal compo nents: one com-ponent A Bparallel to B, another (A - As) perpendicular to B. In fact, our Cartesian repre-sentation ofavector is essentially resolving the vector into three mutually orthogonal com-ponents as in Figure l.l(b).

We have considered addition, subtraction, and multiplication of vectors. However, di-vision of vectors A/B has not been considered because it is undefined except when A andB are parallel so that A = kB , where k is a constant. D ifferentiation and integration ofvectors will be considered in Chapter 3.

- - B -B

a)Fig ure 1.10 Com ponents of A along B: (a) scalar comp onent AB, (b) vectorcomponent A B.


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EXAMPLE 1.4

1.8 COMPONEN TS OF A VECTOR 17

Given vectors A = 3a x + 4ay + az and B = 2ay - 5az, find the angle between AandB .Solution:The angledAB can be found by using either dot product or cross product.

A B = 3 , 4, 1) (0, 2, - 5 )= 0 + 8 - 5 = 3

Alternatively:

A| = V3 2 + 42 + I2 = V 2 6

CO S BA R =

B | = VO2 + 22 + ( -5)2 = V29A B 3I A I I B I V 2 6 ) 2 9 ) = 0.1092

9AR = cos 1 0.1092 = 83.73

A X B = 3 4 az10 2 - 5= ( -20 - 2 )ax + (0 + 15)ay + (6 - 0)az= -22,15,6)

|A X B + 152 + 62 = V 7 4 5sin 6AB = A X Bj V 74 5/(26X29)= 0.994

dAB = cos 10.994 = 83.73

PRACTICE EXERC ISE 1.4If = ax + 3azandB = 5a* + 2ay - 6a.,find6AB .Answer: 120.6.

EXAMPLE 1.5 Three field quantities are given byP = 2ax - a,Q = 2a^ - ay + 2azR = 2ax - 33^, + az

Determine(a) (P + Q) X (P - Q)(b) Q R X P


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18 Vector A lgebra(c)P Q XR(d) sin0 e R(e)P X (Q X R)(f) A unit vector perpen dicular to both Q and R(g) The component of P alongQSolution:a) P Q) X P- Q)= PX P- Q) Q X P- Q)= P X P - P X Q + Q X P - Q X Q

= O + Q X P + Q X P - O= 2Q XP= ay a2 - 1 2

2 0 - 1= 2(1- 0)ax 2(4 2)ay 2(0 2)az= 2ar 12av 4a,

(b) The only way Q RX P m akes sense is

Q R X P)= ( 2 , - 1 , 2 ) 22ay- 30

a1- 1= ( 2, - 1 , 2 ) - (3 ,4, 6)=6 - 4 12= 14.

Alternatively:

Q (R X P)=2 - 1 22 - 3 12 0 - 1

To find the determinan t of a 3 X 3 matrix, we repeat the first two rows and cross m ultiply;when the cross multiplication is from right to left, the result should be negated as shownbelow. This technique of finding a determinant applies only to a 3 X 3 matrix. Hence

Q RXP)=_

= 6 0-2 12-0-2= 14

as obtained before.


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1.8 COMP ONENT S OF A VECTOR 19(c) From eq. (1.28)

or

(d)

P (Q X R) = Q (R X P) = 14

P (Q X R) = (2, 0, - 1 ) (5, 2 , - 4 )= 10 + 0 + 4= 14

| Q X RIQIIRI 1(2,/45 V5

3V14 V14 = 0.5976(e) P X (Q X R) = (2, 0, - 1 ) X (5, 2, - 4 )= (2, 3, 4)Alternatively, using the bac-cab rule,

P X (Q X R) = Q(P R) - R(P Q)= (2, - 1 , 2)(4 + 0 - 1) - (2, - 3 , 1)(4 + 0 - 2 )= (2, 3, 4)

(f) A unit vector perpendicular to both Q and R is given by Q X R ( 5 , 2 , - 4 )

3 |QXR|= ( 0 . 745 , 0 . 298 , - 0 . 5 96 )

No te that |a | = l , a - Q = 0 = a - R . An y of these can be used to check a .(g) The comp onent of P a long Q i s

cos6PQaQQ == (P aG)a e =

(4(P Q)Q

IQI2= 294 + 1 + 4 )

= 0.4444a r - 0.2222av + 0.4444a7.

PRACTICE EXERC ISE 1.5LetE = 3av + 4a,andF = 4a^ - 10av + 5 ar(a) Find the compon ent of E along F.(b) Determ ine a unit vector perpen dicular to bothEandF.Answer: (a) (-0 .28 37 , 0.7092, -0 .35 46 ), (b) (0.9398, 0.2734, -0 .2 05 ).


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20 Vector AlgebraFXAMPIF1f. Derive the cosine formula

and the sine formulaa2 = b2 + c2 - 2bccosA

sinA sinB sinCa b c

using dot product and cross product, respectively.Solution:Consider a triangle as show n in Figure 1.11. From the figure, w e notice that

a + b + c = 0that is,

b + c = - aHence,

a2 = a a = (b + c) (b + c)= b b + c c + 2 b c

a2 = b2 + c2 - 2bccosAwhereAis the angle between b and c.

The area of a triangle is half of the product of its height and base. Hence,l-a X b | = l-b X c| = l-c X alabsin C = besinA = casinB

Dividing through byabcgivessinA sinB sin C

Figure 1.11 For Example 1.6.


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1.8 COMP ONENT S OF A VECTOR 21

PRAC TICE EXERC ISE 1.6Show that vectors a = (4, 0, - 1 ) , b = (1 ,3 , 4) , and c = ( - 5 , - 3 , - 3 ) form thesides of a triangle. Is this a right angle triangle? Calculate the area of the triangle.Answer: Yes, 10.5.

EXAMPLE 1.7 Show that pointsPtf, 2, - 4 ) ,P2{\ , 1, 2), andP3 ( - 3 , 0, 8) all lie on a straight line. De ter-mine the shortest distance between the line and pointP4(3 , - 1 , 0).Solution:The distance vectorfptp2is given by

Similarly,

rPJ P2 rp2

Tp,P3 = Tp3-1

rP tP4 = rP4 -

rP P X rP

rP,= (1,1= ( -4

>, = ( 3,= ( -8 ,>, = (3, -= ( -2 ,

p ==

a*- 4- 8(0,0,

,2 ), - 1

0,8)- 2 ,1,0)- 3 ,a,- 12

0)

- ( 56)

, 2, -4 )

- (5, 2, - 4 )12)- (5, 2, - 4 )

4)az612

showing that the angle between r> iP 2andrPiPiis zero (sin6= 0). This implies thatPh P2 ,andP3lie on a straight line.Alternatively, the vector equation of the straight line is easily determined from Figure1.12 a).For any pointPon the line joining P , andP2

whereXis a constant. Hence the position vector r> of the pointP must satisfyi> - i>, = M*p2~ rP)that is,

i> = i>, + \ i> 2 - i>,)= (5, 2, - 4 ) - X(4, 1, - 6 )i> = (5 - 4X, 2 - X, - 4 + 6X)

This is the vector equation ofthestraightlinejoiningPxandP2-IfP3is on this line, the po-sition v ector of F 3must satisfy the equation; r3does satisfy the equation whenX= 2.


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22 Vector Algebra

a)Figure 1.12 For Example 1.7.

The shortest distance between the line and pointP4(3, - 1 , 0) is the perpen dicular dis-tance from the point to the line. From Figure 1.12 b),it is clear thatd = rPiPt si n 6 = | rP |p 4 X aP]p2

312 = 2.42653Any point on the line may be used as a reference point. Thus, instead of usingP\ as a ref-erence point, we could useP3so that

d= sin

PRACTICE EXERC ISE 1.7If P, is (1,2, -3) and P2is (- 4 , 0,5), f ind(a) The distance P]P2(b) The vector equation of the line P]P2(c) The shortest distance between the lineP\P2and point P3( 7 , - 1 , 2 )Answer: (a) 9.644, (b) (1 - 5X)ax + 2(1 - X)av + (8X - 3)a ,(c) 8.2.

S U M M A R Y 1 . A field is a function that specifies a quantity in space. For examp le,A(x, y, z)is a vectorfield whereasV(x, y, z)is a scalar field.

2. A vector A is uniquely specified by its magnitude and a unit vector along it, that is,A = AaA.


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R EVIEW QUEST IONS M 23

3. Multiplying two ve ctors A and B results in either a scalar A B = AB cos 6AB or avector A X B = AB sin 9AB an. Multiplying three vectors A, B, and C yields a scalarA (B X C ) or a vector A X (B X C ).4. The scalar projection (or comp onent) of vector A onto B isAB = A aBwhereas vectorprojection of A onto B isAB = ABaB.

1.1 Identify which ofth efollowing quantities is not a vector: (a) force, (b) momentum, (c) ac-celeration, (d) work, (e) weight.1.2 Which of the following is not a scalar field?

(a) Displacement of a mosquito in space(b) Light intensity in a drawing room(c) Temperature distribution in your classroom(d) Atmospheric pressure in a given region(e) Hum idity of a city

1.3 The rectangular coordinate systems shown in Figure 1.13 are right-handed except:1.4 Which of these is correct?

(a) A X A = |A |2( b ) A X B + B X A = 0(c) A B C = B C A(d) axay = az(e ) ak = ax - aywhereakis a unit vector.

a)-* y

d) e)Figure 1.13 For Review Question 1.3.

c )y

f)


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24 H Vector Algebra1.5 Whichofthefollowing identitiesis notvalid?

(a) a b + c) = ab + be(b) a X b + c) = a X b + a X c(c) ab = ba(d) c a X b) = - b a X c)(e) aAaB = cosdAB

1.6 Whichofthe following statements are meaningless?(a) AB + 2A = 0(b) AB + 5 = 2A(c) A A + B) + 2 = 0(d) AA + BB = 0

1.7 Let F = 2ax- 63 + 10a2and G =ax Gyay 5az. If F and Ghavethesame unitvector,Gyis(a) 6 d) 0(b) - 3 e) 6

1.8 Given that A = ax+aay azand B = (is(a) -12ax - 9ay- 3az(b) 30a, - 40a^(c) 10/7(d) 2(e) 10

1.10 Given A = 6ax 3ay+ 2az, theprojectionofAalongayis(a) -12(b) - 4(c) 3(d) 7(e) 12

Answers: Lid, 1.2a, 1.3b,e, 1.4b, 1.5a, 1.6b,c, 1.7b, 1.8b, 1.9d, 1.10c.


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PRO L MS

PROBLEMS

1.1 Find the unit vector along the l ine joining point (2, 4, 4) to point ( - 3 , 2, 2).25

1.2 Let A = 2a^ + 53^, - 3a z , B = 3a^ - 4a y , and C = ax + ay + az . (a) DetermineA + 2B . (b) Calculate |A - 5 C |. (c) For what values of k is |kB| = 2? (d) Find(A X B)/(A B ) .1.3 If

A = ay - 3az

C = 3ax 5 a v 7a zdetermine:

(a) A - 2B + C(b) C - 4(A + B)

2A - 3B(d) A C - |B |2( e) |B X ( |A + |C )

1.4 If the position vectors of points T an d S are 3a^ 23^, + az an d Aax 4- 6ay + 2ax , re-spectively, find: (a) the coordinate s of Tand S,(b) the distance vector from TtoS, (c) thedistance between Ta nd S.

1.5 If

Aay + 6azA = 5a xB = -*xC = 8a x + 2 a ,

find the values of a an d/3such that a A + 0B + Cis parallel to the y-axis.1.6 Give n vectors

A = aax + ay + Aazo ^ a x ~T~ p3y O3ZC = 5a x - 2ay + 7 a ,

determine a, /3, and 7 such that the vectors are mutually orthogonal.1.7 (a) Show that

(A B )2 + (A X B)2 = (AB)2

a z X axa , aY X a . ' a, =

a^ X a ya* ay X az


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26 Vector A lgebra1.8 Given that

P = 2ax - Ay - 2 azQ = 4a_, + 3a y + 2a2C = ~ax + ay + 2azfind: (a) |P + Q - R |, (b) P Q X R, (c) Q XP R, (d) (P X Q) (Q X R) ,

(e) (P X Q) X (Q X R), (f) cos6PR,(g) sin6PQ .1.9 Given vectors T = 2ax 6ay + 3azand 8 = 3^-4- 2ay + az, find: (a) the scalar projec-tion of T on S, (b) the vector projection ofSon T, (c) the smaller angle between T and S.1.10 IfA = ax + 6ay + 5az andB = ax + 2ay + 3ax,find: (a) the scalar projections ofAon B, (b) the vector projection of B on A, (c) the unit vector perpendicular to the planecontaining A and B.

1.11 Calculate the angles that vector H = 3a x + 5ay - 8azmakes with thex-,y-, and z-axes.1.12 Find the triple scalar product ofP ,Q , and R given that

P = 2ax - ay + azQ = a^ + a y + az

andR = 2a, + 3az

1.13 Simplify the following expressions:(a) A X (A X B)(b) A X [A X (A X B)]

1.14 Show that the dot and cross in the triple scalar product may be interchanged, i.e.,A (B X C) = (A X B) C.1.15 PointsPi(l, 2, 3),P2(~5, 2, 0), and P3(2 , 7, - 3 ) form a triangle in space. Calculate thearea ofthetriangle.1.16 The vertices ofatriangle are located at(4, 1 , - 3 ) , ( - 2 , 5,4), and (0,1 ,6) . Find the threeangles ofthetriangle.1.17 Points P, Q,andRare located at ( - 1 , 4, 8), ( 2 , - 1 , 3), and ( - 1 , 2, 3), respectively.Determine: (a) the distance betweenPandQ,(b) the distance vector fromPtoR,(c) theangle betweenQPandQR ,(d)the area of trianglePQR,(e)theperimeter of trianglePQR.

*1.18 If r is the position vector of the point(x , y, z)and A is a constant vector, show that:(a) (r - A)A = 0 is the equation ofaconstant plane(b) (r A) r = 0 is the equation ofasphere

*Single asterisks indicate problems ofintermediatedifficulty.


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PROBLEMS

Figure 1.14 For Proble m 1.20.27

(c) Also show that the result of part (a) is of the form Ax + By + Cz + D = 0 whereD = -(A 2 + B2 + C2), and that of part (b) is of the form x2 + y2 + z2 = r2.

*1.19 (a) P rove that P = cos 0i x + sin6xay and Q = cos82ax + sin02ayare unit vectors inthe xy-plane respectively making angles i an d 82with the x-axis.(b) By mean s of dot produc t, obtain the formula for cos( 0 2 #i)- By similarly formulat-ing P and Q, obtain the formula for cos(0 2 + #i) .(c) If 6 is the angle between P and Q, find |P Q | in terms of 6.

1.20 Con sider a rigid body rotating with a constan t angular velocityw radians per second abouta fixed axis through O as in Figure 1.14. Let r be the distance vector from O to P, th eposition of a particle in the body. The velocity u of the body at P is |u| = dw =r sin 6 |co or u =

elements of electromagnetics third edition e book chapter 01

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