If A, B, C are mutually perpendicular, show that Cx(AxB)=0. Is the con

1.	If A, B, C are mutually perpendicular, show that Cx(AxB)=0. Is the converse true?
Answer» Let i, j and k be the unit vectors along A, B and C. Now Cx(AxB)= Ck x (ABsin90°)k =(ABC) k x k =0 (Since k x k =0) Let us see the converse, given (AxB)xC=0, if θ be the angle between A and B and k unit vector perpendicular to both A an B => ABsinθ k x C = 0 => ABsinθ.1.Csinß.u =0 (ß is the angle between k and C and u the unit vector along the perpendicular to both k and C) => This condition gives either θ =0 or ß=0. So they are not mutually perpendicular and the converse is not true.

If A, B, C are mutually perpendicular, show that Cx(AxB)=0. Is the converse true?

Answer»

Let i, j and k be the unit vectors along A, B and C.

Now Cx(AxB)= Ck x (ABsin90°)k =(ABC) k x k =0 (Since k x k =0)

Let us see the converse, given (AxB)xC=0, if θ be the angle between A and B and k unit vector perpendicular to both A an B

=> ABsinθ k x C = 0

=> ABsinθ.1.Csinß.u =0 (ß is the angle between k and C and u the unit vector along the perpendicular to both k and C)

=> This condition gives either θ =0 or ß=0. So they are not mutually perpendicular and the converse is not true.