If A,B,C,D are any four points in space prove that `vec(AB)xxvec(CD)+vec(BC)xvec(AD)+vec(CA)xxvec(BD)=2vec(AB)xxvec(CA)`

If A,B,C,D are four points in space, then |vec(AB)xvec(CD)+vec(BC)xxvec(AD)+vec(CA)xxvec(BD)|=k (are of /_\ABC) where k= (A) 5 (B) 4 (C) 2 (D) none of these

If A ,B ,C ,D be any four points in space, prove that | vec A Bxx vec C D+ vec B Cxx vec A D+ vec C Axx vec B D|=4 (Area of triangle ABC)

A , B , Ca n dD are any four points in the space, then prove that | vec A Bxx vec C D+ vec B Cxx vec A D+ vec C Axx vec B D|=4 (area of A B C .)

A , B , Ca n dD are any four points in the space, then prove that | vec A Bxx vec C D+ vec B Cxx vec A D+ vec C Axx vec B D|=4 (area of A B C .)

A , B , Ca n dD are any four points in the space, then prove that | vec A Bxx vec C D+ vec B Cxx vec A D+ vec C Axx vec B D|=4 (area of A B C .)

A , B , C , D are any four points, prove that vec A Bdot vec C D+ vec B Cdot vec A D+ vec C Adot vec B D=0.

In a regular hexagon ABCDEF, prove that vec(AB)+vec(AC)+vec(AD)+vec(AE)+vec(AF)=3vec(AD)

A , B , C , D are any four points, prove that vec A Bdot vec C D+ vec B Cdot vec A D+ vec C Adot vec B D=4(Area \ of triangle ABC).

If ABCDEF is a regular hexagon, prove that vec(AC)+vec(AD)+vec(EA)+vec(FA)=3vec(AB)

Given vec(A)xxvec(B)=vec(0) and vec(B)xxvec(C )=vec(0) Prove that vec(A)xxvec(C )=vec(0)