Question

If `veca, vecb, vecc` are three non-zero non-null vectors are `vecr` is any vector in space then
`[(vecb, vecc, vecr)]veca+[(vecc, veca, vecr)]vecb+[(veca, vecb, vecr)]vecc` is equal to
A. `2[(veca, vecb, vecc)]vecr`
B. `3[(veca, vecb, vecc)]vecr`
C. `[(veca, vecb, vecc)]`
D. None of these

Answer 1

Correct Answer - C
We have `vecr=xveca+yvecb+zvecc`………………i
Taking product successively with `vecbxxvecc, veccxxveca` and `vecaxxvecb` we obtain
`x=([(vecb, vecc, vecr)])/([(veca, vecb, vecc)]),y=([(vecc, veca, vecr)])/([(veca, vecb, vecc)]),z=([(veca, vecb, vecr)])/([(veca, vecb, vecc)])`
Substituting the values of `x,y,z` in (i) we get
`[(vecb, vecc, vecr)]veca+[(vecc, veca, vecr)]vecb+[(veca, vecb, vecr)]=vecc=[(veca, vecb, vecc)]vecr`