If the non zero vectors `veca and vecb` are perpendicular to each other then the solution the equation `vecrxxveca=vecb` is (A) `vecralphavecb- 1/(|vecb|^2)(vecaxxvecb)` (B) `vecralphavecb+ 1/(|veca|^2)(vecaxxvecb)` (C) `vecralphavecb+ 1/(|vecb|^2)(vecaxxvecb)` (D) none of these

If the non zero vectors veca and vecb are perpendicular to each other, then the solution of the equation vecr xx veca=vecb is given by

If veca is perpendicular to vecb then the vector vecaxx[vecaxx{vecaxx(vecaxxvecb)}] is equla (A) |veca|^2vecb (B) |veca|vecb (C) |veca|^3vecb (D) |veca|^4vecb

The vector (veca-vecb)xx(veca+vecb) is equal to (A) 1/2 (vecaxxvecb) (B) vecaxxvecb (C) 2(veca+vecb) (D) 2(vecaxxvecb)

If for two vector vecA and vecB , vecAxxvecB=0 then

If veca vecb be any two mutually perpendiculr vectors and vecalpha be any vector then |vecaxxvecb|^2 ((veca.vecalpha)veca)/(veca|^2)+|vecaxvecb|^2 ((vecb.vecalpha)vecb)/(|vecb|^2)-|vecaxxvecb|^2vecalpha= (A) |(veca.vecb)vecalpha|(vecaxxvecb) (B) [veca vecb vecalpha](vecbxxveca) (C) [veca vecb vecalpha](vecaxxvecb) (D) none of these

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

Prove that: (2veca-vecb)xx (veca+2vecb)=5vecaxxvecb .

Show that, (veca-vecb)xx(veca+vecb) = 2(vecaxxvecb) .

veca*vecb=abs(vecaxxvecb) then abs(veca-vecb) is

If vertices of /_\ABC are A(veca), B(vecb) and C(vecc) then length of perpendicular from C to AB is (A) (|vecbxxvecc+veccxxveca+vecaxxvecb|)/(|veca-vecb|) (B) (|vecbxxvecc+veccxxveca+vecaxxvecb|)/(|veca+vecb|) (C) (|vecbxxvecc|+|veccxxveca|+|vecaxxvecb|)/(|veca-vecb|) (D) none of these