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If veca, vecb, vecc are three non colana...

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[vec a vecb vec c]vecr`

C

`[veca vecb vec c]vecr`

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
A
`underset(p)ubrace((veca xx vecb)) xx (vecr xx vecc) + underset(w)ubrace((vecb xx vecc)) xx (vec r xx veca) + (vecc xx veca) xx underset(V)ubrace((vecr xx vecb))`
`= vecP xx (vecr xx vecc) + vecW xx ( vecr xx veca) + ( vecc xx veca) xx vecV`
`=(vecP. vecc) vecr - (vecP. vecr) vecc + (vecW . veca) vecr - (vecW . vecr)veca + (vecc . vecV)veca - (veca.vecV)vecc`
`=[veca vecb vecc]vecr - [veca vecb vecr]vecc + [vecb vecc veca]vecr - [vecb vecc vecr]veca = [vecc vecr vecb ] veca - [veca vecr vecb]vecc=2[veca vecb vecc]vecr`
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