for any four vectors `veca,vecb, vecc and vecd` prove that `vecd. (vecaxx(vecbxx(veccxxvecd)))=(vecb.vecd)[veca vecc vecd]`

If the vectors veca, vecb, vecc, vecd are coplanar show that (vecaxxvecb)xx(veccxxvecd)=vec0

Prove that: [(vecaxxvecb)xx(vecaxxvecc)].vecd=[veca vecb vecc](veca.vecd)

For any four vectors veca, vecb, vecc, vecd the expressions (vecb xx vecc).(veca xx vecd) +(vecc xx veca).(vecb xx vecd)+(veca xx vecb).(vecc xx vecd) is always equal to:

If the vectors veca, vecb, vecc and vecd are coplanar vectors, then (vecaxxvecb)xx(veccxxvecd) is equal to

Prove that vecaxx{vecbxx(veccxxvecd)}=(vecb.vecd)(vecaxxvecc)-(vecb.vecc)(vecaxxvecd)

for any three vectors, veca, vecb and vecc , (veca-vecb) . (vecb -vecc) xx (vecc -veca) = 2 veca.vecb xx vecc .

for any three vectors, veca, vecb and vecc , (veca-vecb) . (vecb -vecc) xx (vecc -veca) =

If vectors, vecb, vecc and vecd are not coplanar, the prove that vector (veca xx vecb) xx (vecc xx vecd) + ( veca xx vecc) xx (vecd xx vecb) + (veca xx vecd) xx (vecb xx vecc) is parallel to veca .

If veca,vecb,vecc and vecd are unit vectors such that (vecaxxvecb)*(veccxxvecd)=1 and veca.vecc=1/2, then

Prove that vecaxx(vecb+vecc)+vecbxx(vecc+veca)+veccxx(veca+vecb)=0