In a parallelogram ABCD

MATHEMATICS · Single correct · CBSE · Mathematics · Vector Algebra · Vector Algebra

Problem

In a parallelogram ABCD,

\mathbf{AB} = \hat{i} + \hat{j} + \hat{k}

and diagonal

\mathbf{AC} = \hat{i} - \hat{j} + \hat{k}

and area of parallelogram is

\sqrt{8}

sq units, then

\angle BAC

is equal to (1)

\frac{\pi}{6}

(2)

\frac{\pi}{3}

(3)

\sin^{-1}\left(\frac{\sqrt{8}}{3}\right)

(4)

\cos^{-1}\left(\frac{\sqrt{8}}{3}\right)