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VECTOR CALCULUSANDGEOMETRYPrepared ByDr. Sanjay JainAsstt. Prof. (Maths)Govt. Holkar Science College, Indore1

UNIT : 1PRODUCT OF VECTORS ANDVECTOR DIFFERENTIATION2

UNIT- 1BLOCK INTRODUCTIONIn previous courses students have already been studied the concepts ofdot and cross product of two vectors. In the present block initially a brief reviewof basic concepts and some useful formulae are given to recall the matter. In thesecond phase of this block we introduce the concepts of product of three andfour vectors in a simple manner followed by illustrated examples.Object :- At the end of this block the reader would be able to understandthe said concept.3

UNIT : 1Unit-I : Product of Three & Four Vectors And Vector DifferentiationSTRUCTURE1.0Introduction1.1Brief Review1.2Scalar Triple Product1.3Vector Triple Product1.4Scalar Product of Four Vectors1.5Vector Product of Four Vectors1.6Reciprocal System of Vectors1.7Vector Differentiation1.8Gradient, Divergence & Curl4

1.0IntroductionIn Physics and Geometry we generally deal with the different physicalquantities e.g. length, mass, volume, density, weight, velocity, force etc.According to the basic properties of these quantities, they are classified into twotypes viz. scalar and vector quantities.Physical quantities which have only magnitude but have no definitedirection are defined as scalars while those having both magnitude and directiondefined as vectors.1.1 Brief ReviewWe know : a . b ab cos , a scalar quantity .(1) a b ab sin . n̂ , a vector quantity where n̂ is a unit vector to the(2) plane of vectors a and b .(3) If a 1 ,then a is called unit vector. a . b . b . a i.e. dot product is commutative(4) (7)b .( b c ) a . b a . c i .i j .j k .k 1 i .j j .k k .i 0(8)Two vectors are said to be mutually perpendicular if a . b 0(9)Two vectors are said to be parallel or collinear if a b where is a(5)(6) scalar. (10) If a b 0 then vectors a and b are parallel. Also a a 0i.e. [every vector is parallel to itself]5

(11)(12)(13)(14) i i j j k k 0 i j k, j k i , k i j a b b a but b a – ( a b ) a ( b c ) a b a c (15) If a a1i a2 j a3 k and b b1i b2j b3k then a . b a1b1 a2b2 a3b3andjka b a1 a2a3 ib11.2b2b3SCALAR TRIPLE PRODUCT: The product of two vectors one ofwhich is itself the vector product of two vectors is a scalar quantity called scalar triple product. In short if a, b , c are three vectors, then a. (b c) is calledscalar triple product. Symbolically, it is denoted by [abc]THINGS TO REMEMBERNOTE-1 (a.b ) c is meaningless since ( a.b ) is not a vector quantity.NOTE-2 a. (b c) ( b c ) . aNOTE-3 a. (b c ) (a b ).cNOTE-4Volume with coterminous edges a , b , c of a parallelopiped a.( b c ) [abc] NOTE-5Scalar triple product of vectors a , b , c also represent as [abc] [bca] [cab]NOTE-6[abc] –[bac] – [acb] i.e. sign is changed provided the cycleorder be changed .NOTE-7 [ijk] i .(j k) i .i 16

NOTE-8Three vectors a , b , c are said to be coplanar if [ a , b , c ] 0 .NOTE-9Determinantform a a1 i a 2 j a 3k , ofscalartripleproductofvector b b1 i b 2 j b 3 k and c c1 i c 2 j c 3 k is given bya1a2a3 abc b1b2b3c1c2c3NOTE-10 Scalar triple product [abc] 0 when two of them are equal i.e. [aac] [abb] [aba] etc. 0SOLVED EXAMPLESEX.1 If a 2 i j 3k , b i 2 j k and c 3 i j 2k , find a. b c ij k Sol. Here b c 1 3 2 1 3i 5 j 7 k1 2therefore a. b c (2i j 3k ) .( 3i 5j-7k ) 2.3 1.5 3.( 7) 6 5-21 –10. i. j 0 etc. i 2 1 j 2 k 2Ans.Alternative method :7

2 a (b c ) 1 2 1 3Ex.21 31 2 2(4 – 1) – 1 (–2–3) 3(–)–6) 6 5 – 21 –10Ans. Show that the vectors a i 3 j k , b 2i j k and c 7 j 3k areparallel to the same plane.1Sol. abc 2031 1 1 1 3 7 2 9 7 4 4 07[See note 10]3so the vectors are coplanar i.e. parallel to the same plane .EX 3 Find the volume of a parallelopiped whose edges are represented bySol. a 2i 3 j 4k , b i 2 j k and c 3i j 2k . The required volume V [ a b c ]2 3 4 a. b c 1 2 1 2 4 1 3 2 3 4 1 6 6 15 28 7 7 3 12EX 4. Show that three points a 2b 3c, 2a 3b 4canda 3b 5c are coplanar . Sol. Let A a 2b 3c , B 2a 3b 4c , C a 3b 5c Then [ [ ABC ] A.( B C ) ( a 2b 3c ) .[ ( 2a 3b 4c )x( a 3b 5c )] ( a 2b 3c ) .[ 6( a b )-10 ( a c ) 3( b a ) 15( b c) 4(c a) 12(c b) ] ( a 2b 3c ) . [ 3( a b ) 6( c a ) 3( b c )]{see note 13(1.1)} 3[aab] 6[aca] 3[abc] 6[bab] 12[bca] 6[bbc] 9[cab] 18[cca] 9[cbc] 3[abc] 12[bca] 9[cab]{see note 10}8

The vectors A, B and C are coplanar .EX 5. Prove that [ a b, b c, c a] 2[abc]LHS. [ a b, b c, c a ]Soln. ( a b ) . [ (b c) (c a) (a b).[b c b a c a] c c 0 a.(b c) a.(b a) a.(c a) b.(b c) b.(b a) b.(c a) [abc] [aba] [aca] [bbc] [bba] [bca] [abc] [bca ]{note 10} [abc] [abc] 2 [abc]Hence Proved.CHECK YOUR PROGRESS : Q.1 If a -2i - 2j 4k,b -2i 4j - 2k and c 4i - 2j - 2k then evaluate a.(b c)and interpret the result [Ans 0] Q. 2 Find the value of so that the vectors 2 i j k , i 2 j 3k and 3 i j 5kare coplanar. [Ans. -4] Q .3 Show that if a, b, c are non coplanar than a b, b c, c a are also non coplanar. Is this true for a b, b c, c a ?Q.4 If a, b, c are non coplanar vectors then show that [axb bxc cxa] [abc]2.Q. 5 Showthatthefourpoints a, b, cand darecoplanarif[bcd] [cad] [abd] [abc].1.3 VECTOR TRIPLE PRODUCT :Definition:- The vector product of two vectors one of which is itself the vectorproduct of two vectors is a vector quantity called vector triple product. In short9

if a, b, c are three vectors , then a b c and a b c are called vector tripleproduct .THINGS TO REMEMBER :Note 1. a b c a.c b a.b c Way to remember : a b c (outer . remote) adjacent – (outer . adjacent)remoteWhere a outer, b adjacent , c remote.Note b c a a b c 2. -[(a.c)b-(a.b)c] (b.a)c–(c.a)b - [(a.c)b - (a.b)c] (b.a)c - (c.a)bSOLVED EXAMPLES Ex. 1 Verify that a (b c ) (a.c )b (a.b )c , given that a i 2j 3k , b 2i - j k and c 3i 2j - 5kSoln . Hereij k b c 2 1 1 3i 13i 7k therefore32i a b c 1 5j2k3 25i 2 j 7k.(1)3 13 7 a.c ( i 2j 3k ) . (3i 2j - 5k) - 8 similarlya.b (i 2j 3k). (2i - j k ) 3 (a.c ) b (a.b ) c - 8 (2i - j k ) - 3(3i 2j - 5k) Now - 25i 2j 7k .(2)from (1) and (2) a (b c) (a.c) b - (a.b) cverified10

Ex. 2 Prove that a (b c ) b (c a) c (a b ) 0 Soln. LHS. a (b c ) b (c a) c (a b ) (a.c )b - (b .a)c (b .a)c - (b .c )a (c .b )a - (c .a)b [since a.b b.a etc.] 0 Ex. 3 Prove that i (a i ) j (a j ) k (a k) 2aSol. We know that i (a i ) ( i . i ) a - ( i .a) i a - a 1 i [ i . i 1 also if a a1 i a 2 j a 3 k then a. i a1 ]similarly j (a j ) a - a2 j k (a k ) a - a3 k. (1) .(2) .(3)on adding (1 ),(2) & (3) we get 3a a 2a . Ex. 4 Prove that [a b b c c a] [abc] 2Soln. Let b c d Then (b c ) (c a) d (c a) (d.a)c - (d.c)a [(b c.a)]c - [(b c.c)]a [bca]c - [bcc]c [bca] csince [bcc] 0 .(2) Now [axb bxc cxa] (a b ) . [(b c ) (c a)] (a b). [bca] c [by (2) ]11

[bca] (a b ).c[ [ [bca] [abc] ] [bca ] [abc] [abc] 2Hence proved .1.4SCALAR PRODUCT OF FOUR VECTORS : Definition:- Let a,b , c and d are four vectors , then (a b ).(c d) is a scalarproduct of four vectors and in brief it is defined as a b . c d ba.cc a.d a.c b.d a.d b.c b.d1.5 VECTOR PRODUCT OF FOUR VECTORS : Definition :- Let a,b , c and d are four vectors , then (a b ) (c d) is a scalarproduct of four vectors and it defined in brief as (a b ) (c d) [abd ]c [abc]d [acd ]b [bcd ]away to remember : product c [remaining] – d [remaining].SOLVED EXAMPLES Ex.1 If a i 2j - k , b 3i - 4k , c - i j and d 2i - j 3k then find(i)Soln (i) (a b ) . (c d) (ii) (a b ) (c d) k 1 8i j 6k i j a b 1 23 0 4Similarly i c d 12 j1 k 0 3i 3 j k 1 312

now (a b ) . (c d) (-8i j - 6k). (3i 3j - k) - 24 3 6 - 15(ii) (a b ) (c d) [abd]c - [abc]d1nowAns. abd 32 10 4 4 34 3 352 132 10 4 4 8 3 91 adc 3and 1 10 (a b ) (c d) - 35 (-i j) - 9 (2i - j 3k) 17i - 26j - 27k Ans. Ex. 2 If the vectors a ,b, c, d are coplanar then show that (a b ) (c d) 0 . Solv. We know that (a b ) is a vector perpendicular to the plane of a and b .Similarly (c d)isperpendiculartotheplaneofcandd. But a, b , c , d are coplanar Hence (a b ) and (c d) are perpendicular to the same plane in which a, b , c , dare coplanar. Therefore (a b ) and (c d) are parallel to each other. Hence (a b ) (c d) 0 . Ex. 3 Prove that 2a2 a i 2 a j 2 a k 2 , where a a. Solv. a i 2 a i . a i a.a a.i2 a 2 a.i i.a i.i .(i)similarly a i 2 a 2 a. j 2 .(ii) a k 2 a 2 a.k 2 .(iii)On adding (i), (ii), (iii) ,we get13

2 2 2222a i a j a k 3a 2 a.i a. j a.k 3a2 – a2 2a2Hence proved.1.6 RECIPROCAL SYSTEM OF VECTORS :Definition:- The three vectors a', b', c' defined as a' b c /[abc] ,b' c a/[abc] and c' a b/[abc] are called the reciprocal to the vectors a, b , c which are non-coplanar so that [abc] 0 .THINGS TO REMEMBER :Note1 a.a' a.(b c)/[abc] [abc]/[abc ] 1 ,similarly b.b 1, c.c 1Remark : Due to this property the word reciprocal has been used.Note 2 '''a.b' a.(c a)/[abc] [aca]/[abc] 0 , similarly a.c 0 b.c b.a c.a' c.b'.Note 3If a', b', c' are three non-coplanar vectors then [abc] is reciprocal to[a' b' c'] where a',b',c' are the reciprocal vectors . (b c) (c a) (a b)Remark : [a' b' c' ] [,,][abc] [abc] [abc]Note 4 1.{ [b c, c a, a b] }3[abc] 12. abc 3[abc] 1 abc {by Ex.3 ,1.3}therefore [a' b' c'] [abc] 1, as required.The system of unit vectors i, j, k is its own reciprocal, because i' j k/[ijk] i/1 i {since j k i & [i j k] 1}, similarly j ' j and k ' k .14

SOLVED EXAMPLESEx.1 Findasetofvectorsreciprocaltotheset 2i 3j - k , i - j - 2k , - i 2j 2k .Soln Let a 2i 3j - k, b i - j - 2k, c - i 2j 2kWe know that by the system of reciprocal vectors as a' b c /[abc] ,b' c a/[abc] and c' a b /[abc]Thereforei b c 122ij c a 1 22i a b 2andk 1 2 2i k 1similarlyjk 2 8i 3 j 7k3 1jk 1 7i 3 j 5k31 1 22 abc 1also13 1 1 2 4 1 322so the required reciprocal vectors area' (2i k ) / 3, b' ( 8i 3 j 7k ) / 3, c' ( 7i 3 j 5k ) / 3 . Ex 2 Prove that (a a' ) (b b' ) (c c' ) 0 Solv.We know that a' (b c ) / [abc] ,therefore (a a' ) [a (b c )] / [abc] [(a.c)b- (a.b)c] / [abc]similarly (b b ' ) [(b.a)c- (b.c)a] / [abc] (ii)and (c c ' ) [(c.b)a - (c.a)b] / [abc] (iii). . (i)15

now on adding (i), (ii), (iii),we get (a a' ) (b b ' ) (c c ' ) 0. hence proved. Ex 3 If a' b c / [abc] ,b' c a / [abc] and c' a b / [abc] , then prove that a (b ' c ' )/[a' b' c' ],b ( c ' a' )/[a' b' c' ],c ( a' b ' ) / [a' b' c' ]. Solv. We have b' c' {(c a) (a b)} / [abc] 2 { d (a b)} / [abc] 2 [ where d c a ] {(d.b)a - (d.a)b} / [abc]2by the def. of vector triple product {[cab] a – [caa]b} / [abc]2since d c a [abc ]a / [abc]2since [caa] 0 .(1) a / [abc] now [a' b' c'] a'.(b ' c ' ) {(b c ) / [abc]}. {a / [abc]}, from{1} (b c).a / [abc]2 [abc] / [abc]2 1 / [abc] .(2)clearly from (1) & (2) we have ( b ' c ' )/[a' b' c' ] asimilarly we can prove the other parts.CHECK YOUR PROGRESS :Q.1 Find the value of a (b c ) if the vectors are a i 2j - 2k,b 2i - j k,c i 3j - k. Ans.20i - 3j 7k]Q.2Verify that a (b c) (a.c)b-(a.b)c whereQ.3 a i 2j 3k,b 2i - j k,c 3i 2j - 5k a. Prove that a (b a) (a b ) a . Show that i (j k) (i j ) k 0 .Q.416

Q.5Prove that (b c ).(a d) (c a).(b d) (a b ).(c d) 0.Q.6Prove that (a b) (c d) (a c) (d b) (a d) (b c) - 2[bcd]a .Q.7If a,b,c and a',b',c' are the vectors of reciprocal system, prove that (i) a.a' b.b' c.c' 3. (ii) a' b' b' c' c' a' (a b c) / [abc] , [abc] 0.1.7VECTOR DIFFERENTIATION :1.7.1 Vector Function : If a variable vector a depends on a scalar variable t in such a way that as t varies in some interval, the vector a also varies ,then a is called vector functionof the scalar t and we write a a(t) or f(t).Ex (1) f(t) at2i 2atj btk (2) a(t) cos ti sin tj tan tk1.7.2 Differentiation of Vector Function with respect to a scalar : Let a be a vector function of a scalar t . Let a be the small increment in acorresponding to the small increment t in t Then a a t dt a(t ) or a / t {a (t dt ) a(t )} / t d anow as t 0 the vector functiontends to a limit which is denoted by tdtand is called the derivative of the vector function a w.r.t. t .i.e. dlim a (t t ) a (t ) dt t 0 t1.7.3 Imp Formulae of Differentiation :(1) d d ka k , where k is constant .dtdt17

(2)(3)(4)(5) d da dba b k dtdt dt d dbdaa.b a. b.dtdtdt d db da a b a bdtdt dt If a be a vector function of some scalar t and we have a a x i a y j a z k referred to rectangular axes OX, OY, OZ then da da x da y da z i j kdtdtdtdt(6) da If a is a constant vector, then 0dtSOLVED EXAMPLES Ex.1 If a (cos nt)i (sin nt)j , where n is constant , then show that da(i)a n.kdt da(ii) a. 0dt Soln. Given a cos nt i sin nt j da - n sin nt i n cos nt jdt .(i) .(ii)ijk danow (i) a cos ntsin nt 0 k n cos2 nt n sin 2 nt n.k hence proved.dt n sin nt n cos nt 0 da(ii) a. {(cos nt i sin nt j)}.{ - n sin nt i n cos nt j }dt - n cos nt .sin nt n cos nt .sin nt 0Hence proved.18

da d 2 a da d 2 a 2 Ex.2 If a t i - t j (2t 1) k then find, 2,, 2 at t 0dt dtdt dtSoln. Given a t 2i - tj (2t 1)ktherefore da 2t i - j 2kdtnow d 2a 2i0j 0k 2idt 2 (i) .(ii)also from (i) & (ii)da dt 4t2 4t 1 4 2 5 d 2a (2) 2 22dt .(iii) (iv)Hence from (i) (ii) (iii) &(iv) at t 0 , we have d 2a dadad 2a j 2k , 2 2i 5 and 2.dtdtdtdt 2 r dr Ex.3 Show that rˆ drˆ 2 , where r r.rˆr rSoln. Since rˆ rAns. .(i)1 1 drˆ dr r d r r 1 1 dr r 2 dr r r 1 r rˆ drˆ rˆ dr 2 dr r r r 1 r dr 2 dr [by (i)]r rr 1 1 r dr 3 (r dr )drr2r 1 2 r dr [ r r 0]henceprovedr 19

Ex.4 If r cos it asin t j a tan t k, then find dr / dt d 2 r / dt 2 and dr / dt, d r / dt22, d 3 r / dt 3 Solv. Given r a costi a sin tj a tan tk,Thereforedr / dt a sin ti a costj a tan k (i)d 2 r / dt 2 a costi a sin tj 0k a costi a sin tj,andnow (ii) .(iii)d 3r / dt 3 a sin ti a cos tj i(dr / dt d 2 r / dt 2 ) a sin t ja cos t ka tan a cos t a sin t0a 2 sin t tan i a 2 cost tan j a 2 ktherefore dr / dt d 2 r / dt 2 a 4 sin 2 t tan 2 a 4 cos 2 t tan 2 a 4 a4 tan 2 a 4 a 2 sec Ans a sin t a cos t a tan dr d 2 r d 3r 0 dt dt 2 dt 3 a cos t a sin t a sin t a cos t0 a tan a 2 cos 2 t a 2 sin 2 t a3 tan Ans Ex.5 If a a(t) has a constant magnitude, then show that a . da dt 0. Soln. Given a has a constant magnitude ,therefore a 2 a.a constant (i)On differentiating (i) both side w.r.t. t , we get d (a.a )dt 0 a . da dt da dt . a 0 2 a . da dt 0 a . da dt 0 .20

CHECK YOUR PROGRESS : If r a cos t b sin t, then show that (i)Q.1 r dr /dt (a b ii d r dt - r If r a sinh nt bcosh nt , where a and b are constant vectors, then showQ.2that d2r/dt2 n2 r. If a t 2i tj (2t 1)k and b (2t - 3)i j - tk, then findQ.3 d(a.b)/dt , at t 1[Ans. 6](ii) d/dt [a, da/dt, d2a/dt2]Q.4Evaluate (i) d [abc]/dtQ.5If r be the position vector of P. Find the velocity and acceleration of P att /6 where r sec ti tan t j {Ans v (2/3)i (4/3)j , a 2(5i 4j)/3 3}[Hint: velocity dr/dt, acceleration d2r/dt2]1.8 Gradiant, Divergence & Curl :1.8.1 Partial Derivative of Vector : Let F(x,y) be a vector function of independent variable x and y . Then partial derivatives of F w.r.t. x and y are denoted by Fx and Fy resp. & defined as Fx F/ x Lt. x 0 {[F(x x, y)] - F(x,y)}/ x And Fy F/ y Lt. y 0 {[F(x,y y)] - F(x,y)}/ ySecond order partial derivatives are denoted by 2F/ x2, 2F/ x y , 2F/ y2. means F/ x ( F/ x)/ x, F/ x y ( F/ y)/ x, F/ y ( F/ y)/ y.22222THINGS TO REMEMBER :Note 1To find Fx , differentiate F w.r.t. x (treating other variable y asconstant). In the same way to find Fy treat x as constant.Note 2 (F F )/ x (F )/ x (F )/ x .21

Note 3 (k.F )/ x k. (F )/ x.Note 4 (F F )/ x F . (F )/ x (F )/ x . F .Note 5 F F )/ x F (F )/ x (F )/ x F Note 6δ(axi ayj azk)/δt (δax/δt)i (δay/δt)j (δaz/δt)k.Note 7In general vector r is taken as r xi yj zk. Then r [x 2 y 2 z 2 ] , r 2 x2 y2 z2 and δr/δx x/ (x2 y2 z2 ) x/retc.1.8.2 Partial Derivative of Vector:Operator is generally called delta and in brief read as del and is defined by i / x j / y k / z1.8.3 The Gradient :The gradient of a scalar function F(x,y,z) written by grad F or F is defined bygrad F F I F/ x j F/ y k F/ z1.8.4 The Divergence :let F(x,y,z) F1i F2j F3k then divergence of F written as divF or .F isdefined by divF .F (i / x j / y / z) . (F F j F3 F / x F / y F / zTHINGS TO REMEMBER :Note (1) Divergence of a vector F is a scalar quantity .Note (2) A vector F is solenoidal if divF 0.1.8.5The Curl : let F(x,y,z) F1i F2 j F3 k then the curl of F written as curlF or xF isdefined by curlF xF (i / x j / y k / z) x (F1i F2j F3k)22

i j k x y zF1F2F3 ( F3/ y - F2 / z) ( F1/ z - F3 / x) ( F2 / x - F1 / y )THINGS TO REMEMBER :Note (1)curl F is also known as rotation of F and written as rot F .Note (2)curl F is a vector quantity.Note (3)A vector F is called ir-rotational if curlF 0.Aid to memory :Gradient - G – general product .Divergence- D means dot product , curl – Cmeans cross product .SOLVED EXAMPLESEx.1 (a) If F x3 y3 z3-3xyz, find

VECTOR CALCULUS AND GEOMETRY Prepared By Dr. Sanjay Jain Asstt. Prof. (Maths) Govt. Holkar Science College, Indore . 2 UNIT : 1 PRODUCT OF VECTORS AND VECTOR DIFFERENTIATION. 3 UNIT- 1 BLOCK INTRODUCTION In previous courses students have already been studied the concepts of dot and cross product of two vectors. In the present block initially a brief review of basic concepts and some useful .

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