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 are any four points in space prove that vec(AB)xxvec(CD)+vec(BC)xvec(AD)+vec(CA)xxvec(BD)=2vec(AB)xxvec(CA)

A,B,C and D are any four points in the space,then prove that |vec AB xxvec CD+vec BC xxvec AD+vec CA xxvec BD|=4 (area of o+ABC.)

A,B,C,D are any four points,prove that vec AB*vec CD+BC*vec AD+vec CA*vec BD=0

For any four points A,B,C,D |vec (AB) xx vec (CD)-vec (BC) xx vec (AD)+vec (CA) xx vec (BD)| is equal to 4 times the area of triangle- (1) ABC (2) BCD (3) ACD (4) ABD

A, B, C, D are four points in the space and satisfy |vec(AB)|=3, |vec(BC)|=7, |vec(CD)|=11 and |vec(DA)|=9 . Then find the value of vec(AC)*vec(BD) .

(vec and xxvec b) * (vec and xxvec c) =

If vec(A)xxvec(B)= vec(C )+vec(D) , then select the correct alternative.

[(vec a xxvec b) (vec a xxvec c)] * vec d equals