Simplify : `{(vecb-vecc)xx(vecc-veca)}odot(veca-vecb)`

Simplify : (vecb+vecc)*{(vecc+veca)xx(veca+vecb)} .

Prove that {(vecb+vecc)xx(vecc+veca)}.(veca+vecb)=2[veca,vecb,vecc]

Simplify : [veca-vecb,vecb-vecc,vecc-veca] .

If veca,vecb,vecc are any three vectors then prove that. vecaxx(vecb+vecc)+vecbxx(vecc+veca)+veccxx(veca+vecb)=0

Show that the perpendicular distance of the point vecc from the line joining veca and vecb is : (|vecbxxvecc+veccxxveca+vecaxxvecb|)/(|vecb-veca|) .

If veca,vecb,vecc are three vectors such that vecaxxvecb=vecc,vecbxxvecc=veca , prove that veca,vecb,vecc are mutually at right angles and |vecb|=1,|vecc|=|veca| .

If veca, vecb and vecc are non-coplanar vectors, prove that the four points 2veca+3vecb-vecc, veca-2vecb+3vecc, 3veca+4vecb-2vecc and veca-6vecb+ 6 vecc are coplanar.

Let veca and vecc be unit vectors inclined at pi//3 with each other. If (veca xx (vecb xx vecc)). (veca xx vecc)=5 , then [vecavecbvecc] is equal to

If [vecaxxvecb vecbxxvecc veccxxveca]=lambda[veca vecb vecc]^2 , then lambda is equal to :