VECTOR CALCULUS AND GEOMETRY Prepared By Dr. Sanjay Jain Asstt. Prof. (Maths) Govt. Holkar Science College, Indore 1 UNIT : 1 PRODUCT OF VECTORS AND VECTOR DIFFERENTIATION 2 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 formulae are given to recall the matter. In the second phase of this block we introduce the concepts of product of three and four vectors in a simple manner followed by illustrated examples. Object :- At the end of this block the reader would be able to understand the said concept. 3 UNIT : 1 Unit-I : Product of Three & Four Vectors And Vector Differentiation STRUCTURE 1.0 Introduction 1.1 Brief Review 1.2 Scalar Triple Product 1.3 Vector Triple Product 1.4 Scalar Product of Four Vectors 1.5 Vector Product of Four Vectors 1.6 Reciprocal System of Vectors 1.7 Vector Differentiation 1.8 Gradient, Divergence & Curl 4 1.0 Introduction In Physics and Geometry we generally deal with the different physical quantities e.g. length, mass, volume, density, weight, velocity, force etc. According to the basic properties of these quantities, they are classified into two types viz. scalar and vector quantities. Physical quantities which have only magnitude but have no definite direction are defined as scalars while those having both magnitude and direction defined as vectors. 1.1 Brief Review We know : (1) a . b = abcos , a scalar quantity . (2) a b = absin . nˆ , a vector quantity where nˆ is a unit vector to the plane of vectors a and b . (3) If | | = 1 ,then a is called unit vector. (4) a.b. b.a i.e. dot product is commutative (5) b .( b c) a . b a . c (6) i .i j.j k.k 1 (7) i .j j.k k.i 0 (8) Two vectors are said to be mutually perpendicular if = 0 (9) Two vectors are said to be parallel or collinear if = where is a scalar. (10) If a b = 0 then vectors and are parallel. Also × = 0 i.e. [every vector is parallel to itself] 5 (11) i i j j k k 0 (12) i j k, j k i , k i j (13) ≠ × but × = – ( ) (14) a(b c) a b a c (15) If a = a1i a2 j a3k and b = b1i +b2j +b3k then a . b = a1b1 +a2b2+a3b3 and i j k a b a1 a2 a3 b1 b2 b3 1.2 SCALAR TRIPLE PRODUCT: The product of two vectors one of which 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 called scalar triple product. Symbolically , it is denoted by [abc] a b THINGS TO REMEMBER NOTE-1 (a.b) c is meaningless since ( a.b ) is not a vector quantity. NOTE-2 a. (b c) ( b c ) . a NOTE-3 a. (b c) (a b). c NOTE-4 Volume with coterminous edges a, b, c of a parallelopiped a.( b c ) [abc] NOTE-5 Scalar triple product of vectors also represent as [abc] =[bca] = [cab] NOTE-6 [abc] = –[bac] = – [acb] i.e. sign is changed provided the cycle order be changed . a b NOTE-7 [ijk] i .(j k) i .i 1 6 NOTE-8 Three vectors are said to be coplanar if [ ] = 0 . NOTE-9 Determinant form of scalar triple product of vector a a1 i + a 2 j + a 3k , b b1 i + b2 j + b3k and c c1 i + c2 j + c3k is given by a1 a2 a3 abc b1 b2 b3 c1 c2 c3 NOTE-10 Scalar triple product [abc] = 0 when two of them are equal i.e. [aac] =[abb] = [aba] etc. =0 SOLVED EXAMPLES EX.1 If a 2i + j +3k , b i + 2j + k and c 3i + j + 2k , find a.b c Sol. Here i j k b c 1 2 1 3i 5 j 7k 3 1 2 a, b, c therefore = (2i+j+3k ) .( 3i+5j-7k ) = 2.31.5 3.(7) i. j 0 etc. = 6+5-21 i2 1 j 2 k 2 = –10. Ans. Alternative method : 7 2 1 3 a (b c) = 1 2 1 3 1 2 = 2(4 – 1) – 1 (–2–3) + 3(–)–6) = 6 + 5 – 21 = –10 Ans. Ex.2 Show that the vectors a i 3 j k , b 2i j k and c =7 j 3k are parallel to the same plane. 1 3 1 Sol. abc 2 1 1 1 3 7 29 7 4 4 0 [See note 10] 0 7 3 so the vectors are coplanar i.e. parallel to the same plane . EX 3 Find the volume of a parallelopiped whose edges are represented by a 2i 3 j 4k , b = i 2 j k and =3i j 2k . Sol. The required volume V= [ a ] 2 3 4 a.b c 1 2 1 24 1 32 3 41 6 6 15 28 7 7 3 1 2 EX 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.(BC) = ( ) .[ ( )x( a 3b 5c )] = ( ) .[ 6( ab )-10 ( ac )+3(ba )+15(bc) 4(ca) 12(cb) ] = ( ) . [ 3( ab )+6(ca )+3(bc )] {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] Soln. LHS. =[ a b,b c,c a ] = ( a b ) . [ (b c)(c a) = (a b).[bc ba ca]cc 0 = a.(bc) a.(ba) a.(ca) b.(bc) b.(ba) b.(ca) =[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.(bc) and interpret the result [Ans 0] Q. 2 Find the value of so that the vectors 2i j k , i 2j 3k and 3i j 5k are 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 are non coplanar vectors then show that [axb bxc cxa] = [abc]2. Q. 5 Show that the four points a,b,cand d are coplanar if [bcd]+[cad]+[abd]=[abc]. 1.3 VECTOR TRIPLE PRODUCT : Definition:- The vector product of two vectors one of which is itself the vector product of two vectors is a vector quantity called vector triple product. In short 9 if a,b,c are three vectors , then a b canda bc are called vector triple product . THINGS TO REMEMBER : Note 1. a b c a.cb a.bc Way to remember : a b c= (outer . remote) adjacent – (outer . adjacent) remote Where a=outer, b= adjacent , c= remote. Note 2. b c a a b c= - [(a.c)b-(a.b)c] = (b.a)c – (c.a)b - [(a.c)b - (a.b)c] = (b.a)c - (c.a)b SOLVED 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 - 5k Soln . Here i j k b c 2 1 1 3i 13i 7k therefore 3 2 5 i j k a b c 1 2 3 25i 2 j 7k ...................(1) 3 13 7 Now a.c ( i 2j 3k ) . (3i 2j - 5k) - 8 similarly a.b (i 2j 3k) . (2i- j k ) 3 (a.c) b (a.b) c - 8 (2i- j k ) - 3(3i 2j - 5k) - 25i 2j 7k ………….(2) from (1) and (2) a (b c) = (a.c) b - (a.b) c verified 10 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 = 0 [since a.b b.a etc.] Ex. 3 Prove that i (a i ) + j (a j) + k (a k) = 2a Sol. We know that i (a i) = (i.i) a - (i.a) i = a - a1 i [i.i 1 also if a a1i a 2 j a3k then a. i a1 ] .........…(1) similarly j (a j) = a - a2 j ……….(2) k (a k) = a - a3k ……….(3) on adding (1 ),(2) & (3) we get 3a a 2a . Ex. 4 Prove that [ab bc ca] [abc]2 Soln. 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]c since [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]= [abc]2 [[[bca] [abc]] Hence proved . 1.4 SCALAR PRODUCT OF FOUR VECTORS : Definition:- Let a,b, c and d are four vectors , then (a b).(c d) is a scalar product of four vectors and in brief it is defined as a.c a.d a b.c d a.cb.d a.db.c b.c b.d 1.5 VECTOR PRODUCT OF FOUR VECTORS : Definition :- Let and are four vectors , then (a b) (c d) is a scalar product of four vectors and it defined in brief as (a b) (c d) = [abd]c [abc]d = [acd]b [bcd]a way to remember : product = c [remaining] – d [remaining].