Question

Prove that vectore[a×b b×c c×a]=0

Answer 1

in the following explanation a, b, c are vectors, . is dot product and × is cross product.

I believe that they are not in a matrix. [ ] is a vector operation called box product or scalar triple product. It must have 3 vector as operands and the result is a scalar.

The basic rules of box product are:

1. [ a b c ] is the volume of the parallelopide with a, b, c as adjacent sides.

2. [ a b c ] = (a×b)c =(a.b)×c.

3. [ a b b ] = 0 i.e. if same vectors are in the box product it's value is zero.

4. [ a b c ]=[ b c a ]=[ c a b ] i.e. if the order is changed in a cyclic manner.

Coming back to the original question,

[ a+b b+c c+a ] = ((a+b). (b+c))×(c+a) expanding

= ((a.b)+(a.c)+(b.b)+(b.c))×(c+a) expanding

= ((a.b)×c)+((a.c)×c+((b.b)×c)+((b.c)×c) + ((a.b)×a)+((a.c)×a)+((b.b)×a)+((b.c)×a)

= [ a b c ]+[ a c c ]+[ b b c ]+[ a c c ] + [ a b a ]+[ a c a ]+[ b b a ]+[ b c a ]

= [ a b c ] + [ b c a ] by rule 3

= [ a b c ]+[ a b c ] by rule 4

= 2[ a b c ]

Hence proved.

Please let me know if there is any mistake.

Hope this helps you....

Answer 2

Step-by-step explanation:

The basic rules of box product are:

1. [ a b c ] is the volume of the parallelopide with a, b, c as adjacent sides.

2. [ a b c ] = (a×b).c =(a.b)×c.

3. [ a b b ] = 0 i.e. if same vectors are in the box product it's value is zero.

4. [ a b c ]=[ b c a ]=[ c a b ] i.e. if the order is changed in a cyclic manner.

Coming back to the original question,

[ a+b b+c c+a ] = ((a+b). (b+c))×(c+a) expanding

= ((a.b)+(a.c)+(b.b)+(b.c))×(c+a) expanding

= ((a.b)×c)+((a.c)×c+((b.b)×c)+((b.c)×c) + ((a.b)×a)+((a.c)×a)+((b.b)×a)+((b.c)×a)

= [ a b c ]+[ a c c ]+[ b b c ]+[ a c c ] + [ a b a ]+[ a c a ]+[ b b a ]+[ b c a ]

= [ a b c ] + [ b c a ] by rule 3

= [ a b c ]+[ a b c ] by rule 4

= 2[ a b c ]

Hence proved.