Show that the equation `vecrxxvecb=vec0(vecb!=vec0)` has a solution iff `veca.vecb!=0`.

Find vector vecr if vecr.veca=m and vecrxxvecb=vecc, where veca.vecb!=0

Three vectors veca , vecb and vecc satisfy the condition veca+ vecb+ vecc= vec0 . Evaluate the quantity mu= veca* vecb+vecb* vecc+vecc* veca , if | veca|=3,| vecb|=4 and |vecc|=2 .

If is given that vecrxxvecb=veccxxvecb, vecr.veca=0 and veca.vecb!=0 . What is the geometrical meaning of these equation separately? If the abvoe three statements hold good simultaneously, determine the vector vecr in terms of veca,vecb and vecc .

a,b are the magnitudes of vectors veca & vecb . If vecaxxvecb=0 the value of veca.vecb is

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

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

If veca,vecb and vecc are unit vectors such that veca+vecb+vecc=vec0 then the value of veca.vecb+vecb.vecc+vecc.veca is a) 1 b) 0 c) 3 d) -3/2

If a,b are the magnitudes of vectors veca & vecb and veca.vecb=0 the value of |vecaxxvecb| is

Prove that (veca+ vecb)*( veca+ vecb)=|veca|^2+| vecb|^2 , if and only if veca , vecb are perpendicular, given veca!= vec0, vecb!= vec0

Let veca=hati-hatj,vecb=hati+hatj+hatk and vecc be a vector such that vecaxxvecc+vecb=vec0 and veca.vecc=4 , then |vecc|^(2) is equal to: