If the vectors `veca,vecb,vecc` form the sides BC,CA and AB respectively of a triangle ABC then (A) `veca.(vecbxxvecc)=vec0` (B) `vecaxx(vecbxvecc)=vec0` (C) `veca.vecb=vecc=vecc=veca.a!=0` (D) `vecaxxvecb+vecbxxvecc+veccxxvecavec0`

Options

A. `veca.vecb+ vecb.vecc+vecc.veca=0`
B. `vecaxxvecb = vecbxxvecc =veccxxveca`
C. `veca.vecb = vecb.vecc = vecc.veca`
D. `vecaxxvecb + vecbxxvecc +veccxxveca=vec0`

Answer

b

Explanation

Given `veca +vecb +vecc= vec0` (by triangle law). Therefore,
`veca xx (veca +vecb+vecc)=veca xxvec0`
`veca xx veca +veca xx vecb+ veca xx vecc=vec0`
`veca xx vecb = vecc xx veca`
Similarly by taking cross product with `vecb`. We get `veca xx vecb = vecb xx vecc`
`veca xx vecb = vecb xx vecc= vecc xx veca`