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|) .

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

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

If vecb and vecc are two non-collinear such that veca ||(vecbxxvecc) . Then prove that (vecaxxvecb).(vecaxxvecc) is equal to |veca|^(2)(vecb.vecc)

If veca" and "vecb are any two vectors, then (veca xx vecb)^(2)= |veca|^(2)|vecb|^(2)-(veca* vecb)^(2) .

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