If `vecasxxvecb=0 and veca.vecb=0` then (A) `veca_|_vecb` (B) `veca||vecb` (C) `veca=0 and vecb=0` (D) `veca=0 or vecb=0`

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

For two vectors veca and vecb,veca,vecb=|veca||vecb| then (A) veca||vecb (B) veca_|_vecb (C) veca=vecb (D) none of these

If veca.vecb=0 and vecaxxvecb=0 prove that veca=0 or vecb=vec0 .

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 |veca+vecb|=|veca-vecb| show that veca_|_vecb .

For non zero vectors veca,vecb,vecc|(vecaxxvecb).vecc|=|veca||vecb||vec| holds if and only if (A) veca.vecb=0,vecb.vecc=0 (B) vecb.vecc=0,vecc.veca=0 (C) vecc.veca=0,veca.vecb=0 (D) veca.vecb=vecb.vecc=vecc.veca=0

The plane contaning the two straight lines vecr=veca+lamda vecb and vecr=vecb+muveca is (A) [vecr vecas vecb]=0 (B) [vecr veca veca xxvecb]=0 (C) [vecr vecb vecaxxvecb]=0 (D) [vecr veca+vecb vecaxxvecb]=0

If veca + 2 vecb + 3 vecc = vec0 " then " veca xx vecb + vecb xx vecc + vecc xx veca=

If |veca|=|vecb| , then (veca+vecb).(veca-vecb) is equal to

l veca . m vecb = lm(veca . vecb) and veca . vecb = 0 then veca and vecb are perpendicular if veca and vecb are not null vector