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

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)

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

If veca, vecb and vecc be any three non coplanar vectors. Then the system of vectors veca\',vecb\' and vecc\' which satisfies veca.veca\'=vecb.vecb\'=vecc.vecc\'=1 veca.vecb\'=veca.vecb\'=vecb.veca\'=vecb.vecc\'=vecc.veca\'=vecc.vecb\'=0 is called the reciprocal system to the vectors veca,vecb, and vecc . [veca,vecb,vecc]((veca\'xxvecb\')+(vecb\'xxvecc \')+(vecc\'xxveca\'))= (A) veca+vecb+vecc (B) veca+vecb-vecc (C) 2(veca+vecb+vecc) (D) 3(veca\'+vecb\'+vecc\')

if veca xx vecb = vecc ,vecb xx vecc = veca , " where " vecc ne vec0 then (a) |veca|= |vecc| (b) |veca|= |vecb| (c) |vecb|=1 (d) |veca|=|vecb|= |vecc|=1

If veca,vecb,vecc be non coplanar vectors and vecp=(vecbxxvecc)/([veca vecb vecc]) , vecq=(veccxxveca)/([veca vecb vecc]) , vecr=(vecaxxvecb)/([veca vecb vecc]) then (A) vecp.veca=1 (B) vecp.veca+vecq.vecb+vecr.vecc=3 (C) vecp.veca+vecq.vecb+vecr.vecc=0 (D) none of these

If veca, vecb, vecc are non coplanar vectors such that vecbxxvecc=veca, vecaxxvecb=vecc and veccxxveca=vecb then (A) |veca|+|vecb|+|vecc|=3 (B) |vecb|=1 (C) |veca|=1 (D) none of these

The vectors veca and vecb are not perpendicular and vecc and vecd are two vectors satisfying : vecbxxvecc=vecbxxvecd and veca.vecd=0. Then the vecd is equal to (A) vecc+(veca.vecc)/(veca.vecb)vecb (B) vecb+(vecb.vecc)/(veca.vecb)vecc (C) vecc-(veca.vecc)/(veca.vecb)vecb (D) vecb-(vecb.vecc)/(veca.vecb)vecc

The vectors veca and vecb are not perpendicular and vecac and vecd are two vectors satisfying : vecbxxvecc=vecbxxvecd and veca.vecd=0. Then the vecd is equal to (A) vecc+(veca.vecc)/(veca.vecb))vecb (B) vecb+(vecb.vecc)/(veca.vecb))vecc (C) vecc-(veca.vecc)/(veca.vecb))vecb (D) vecb-(vecb.vecc)/(veca.vecb))vecc

If veca+2vecb+3vecc=0 , then vecaxxvecb+vecbxxvecc+veccxxveca=

If |veca|=1,|vecb|=2,|vecc|=3and veca+vecb+vecc=0 the show that veca.vecb+vecb.vecc+vecc.veca=- 7