The vectors `veca=3hati-2hatj+2hatkandvecb=-hati-2hatk` are the adjacent sides of a paralleogram. The angle between its diagonals is……. .

Options

A. `cos^(-1)((2)/(sqrt(13)))`
B. `(pi)/(3)`
C. `(4pi)/(3)`
D. `(2pi)/(3)`

Answer

B

Explanation

Here , `AC=AB+BC`
`=(3hat(i)-2hat(j)2hat(k))+(-hat(j)-2hat(k))`
`=3hat(i)-3hat(j)`
Also `BC=AD`
and `AB=AD+DB`
`DB=AB=AD`
`=(3hat(i)-2hat(j)+2hat(k))-(-hat(j)-2hat(k))`
`=3hat(i)-hat(j)+4hat(k)`
`therefore` Angle between AC and BD is
`theta = cos^(-1)(AC.BD)/(|AC|.AD|)`
`=cos^(-1).((3hat(i)-3hat(j)).(3hat(i)-hat(j)+4hat(k)))/(sqrt(3^2+(-3)^2)sqrt(3^2+1^2+4^2))`
`=cos^(-1).((9+3)/(sqrt(18)sqrt(20)))=cos^(-1).((12)/(3sqrt(2)sqrt(2)sqrt(3)))`
`=cos^(-1)((2)/(sqrt(13)))` .