Browse Questions

The projection of $\overrightarrow{OP}$ on a unit vector $\overrightarrow{OQ}$ equals thrice the area of parallelogram $OPRQ$. Then $\angle{POQ}$ is

$\begin {array}{1 1}(1)\tan^{-1}\frac{1}{3}&(2)\cos^{-1}(\frac{3}{10})\\(3)\sin^{-1}(\frac{3}{\sqrt{10}})&(4)\sin^{-1}\frac{1}{3}\end{array}$