If `veca` is perpendiculasr to both `vecb and vecc` then (A) `veca.(vecbxxvecc)=vec0` (B) `vecaxx(vecbxvecc)=vec0` (C) `vecaxx(vecb+vecc)=vec0` (D) `veca+(vecb+vecc)=vec0`

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

If vecc=vecaxxvecb and vecb=veccxxveca then (A) veca.vecb=vecc^2 (B) vecc.veca.=vecb^2 (C) veca_|_vecb (D) veca||vecbxxvecc

If vecax(vecaxxvecb)=vecbxx(vecbxxvecc) and veca.vecb!=0 , and [(veca,vecb,vecc)]=

Prove that vecaxx(vecb+vecc)+vecbxx(vecc+veca)+veccxx(veca+vecb)=0

Let veca,vecb,vecc be unit such that veca+vecb+vecc=vec0 . Which one of the following is correct? (A) vecaxxvecb=vecbxxvecc=veccxxveca=vec0 (B) vecaxxvecb=vecbxxvecc=veccxxveca!=vec0 (C) vecaxxvecb=vecbxxvecc=vecxxvecc!=vec0 (D) vecaxxvecb, vecbxxvecc, veccxxveca are mutually perpendicular

If vecaxx(vecbxxvecc)=vecbxx(veccxxveca) and [(vec, vecb, vecc)]!=0 then vecaxx(vecbxxvecc) is equal to

The distance of the point A(veca) from the line vecr=vecb+tvecc is (A) |(veca-vecb)xxvecc| (B) (|(veca-vecb)xxvecc|)/(|(veca-vecb)|) (C) (|(veca-vecb)xxvecc|)/(|vecc|) (D) (|vecaxx(vecb-vecc)|)/(|veca|)

If veca, vecb, vecc are vectors such that |vecb|=|vecc| then {(veca+vecb)xx(veca+vecc)}xx(vecbxxvecc).(vecb+vecc)=

If veca.vecb=veca.vecc, vecaxxvecb=vecaxxvecc and veca!=vec0, then prove that vecb=vecc.

If vecd=vecaxxvecb+vecbxxvecc+veccxxveca is a on zero vector and |(vecd.vecc)(vecaxxvecb)+(vecd.veca)(vecbxxvecc)+(vecd.vecb)(veccxxveca)|=0 then (A) |veca|+|vecb|+|vecc|=|vecd| (B) |veca|=|vecb|=|vecc| (C) veca,vecb,vecc are coplanar (D) veca+vecc=vec(2b)