Question

If `veca, vecb, vecc` are three non colanar, non =null vectors, and `vecr` is any vector in space, then
`(vecaxxvecb)xx(vecrxxvecc)+(vecbxxvecc)xx(vecrxxveca)+(veccxxveca)xx(vecrxxvecb)` is equal to
A. `2[(veca, vecb, vecc)]vecr`
B. `3[(veca, vecb, vecc)]vecr`
C. `[(veca, vecb, vecc)]vecr`
D. none of these

Answer 1

Correct Answer - A
We have
`(vecaxxvecb)xx(vecrxxvecc)=[(veca, vecb, vecc)]vecr-[(veca, vecb, vecr)]vecc`
`(vecbxxvecc)xx(vecrxxveca)=[(vecab, vecc, veca)]vecc-[(vecb, vecc, vecr)]veca`
`(veccxxveca)xx(vecrxxvecc)=[(vecc, veca, vecb)]vecr-[(vecc, veca, vecr)]vecb`
`:.(vecaxxvecb)xx(vecrxxvecc)+(vecbxxvecc)xx(vecrxxveca)+(veccxxveca)xx(vecrxxvecb)`
`=3[(veca, vecb, vecc)]vecr-[(veca, vecb, vecc)]vecc+[(vecb, vecc, vecr)]veca+[(vecc, veca, vecr)]vecb`
`=3[(veca, vecb, vecc)]vecr=[(veca, vecb, vecc)]vecr=2[(veca, vecb, vecc)]vecr`.