Let `veca, vecb and vecc` be three non- coplanar vectors and `vecr` be any arbitrary vector. Then `(vecaxxvecb) xx (vecr xxvecc) + (vecbxxvecc)xx(vecrxxveca)+ (vecc xxveca) (vecrxxvecb)` is always equal to

If veca,vecb and vecc are three non coplanar vectors and vecr is any vector in space, then (vecaxxvecb)xx(vecrxxvecc)+(vecb xxvecc)xx(vecrxxveca)+(veccxxveca)xx(vecrxxvecb)=

If veca,vecb and vecc are three non coplanar vectors and vecr is any vector in space, then (vecaxxvecb),(vecrxxvecc)+(vecb xxvecc)xx(vecrxxveca)+(veccxxveca)xx(vecrxxvecb)=

If veca,vecb and vecc are three non coplanar vectors and vecr is any vector in space, then (vecxxvecb),(vecrxxvecc)+(vecb xxvecc)xx(vecrxxveca)+(veccxxveca)xx(vecrxxvecb)= (A) [veca vecb vecc] (B) 2[veca vecb vecc]vecr (C) 3[veca vecb vecc]vecr (D) 4[veca vecb vecc]vecr

If veca,vecb and vecc are three non coplanar vectors and vecr is any vector in space, then (vecxxvecb),(vecrxxvecc)+(vecb xxvecc)xx(vecrxxveca)+(veccxxveca)xx(vecrxxvecb)= (A) [veca vecb vecc] (B) 2[veca vecb vecc]vecr (C) 3[veca vecb vecc]vecr (D) 4[veca vecb vecc]vecr

If veca, vecb, vecc are non-coplanar non-zero vectors, then (vecaxxvecb)xx(vecaxxvecc)+(vecbxxvecc)xx(vecbxxveca)+(veccxxveca)xx(veccxxvecb) is equal to

veca , vecb and vecc are three non-coplanar vectors and vecr . Is any arbitrary vector. Prove that [vecbvecc vecr]veca+[vecc veca vecr]vecb+[vecavecbvecr]vecc=[veca vecb vecc]vecr .

If veca, vecb and vecc are three non-coplanar vectors, then (veca + vecb + vecc). [(veca + vecb) xx (veca + vecc)] equals

vec a , vec ba n d vec c are three non-coplanar ,non-zero vectors and vec r is any vector in space, then ( veca × vecb )×( vecr × vecc )+( vecb × vecc )×( vecr × veca )+( vecc × veca )×( vecr × vecb ) is equal to

If vecb and vecc are unit vectors, then for any arbitary vector veca, (((veca xx vecb) + (veca xx vecc))xx (vecb xx vecc)) .(vecb -vecc) is always equal to

If veca, vecb and vecc are three non-coplanar non-zero vectors, then prove that (veca.veca) vecb xx vecc + (veca.vecb) vecc xx veca + (veca.vecc)veca xx vecb = [vecb vecc veca] veca