• # question_answer If $\vec{p}$ and q are two diagonals of a quadrilateral such that $\left| \vec{p}-\vec{q} \right|=\vec{p}.\,\vec{q}$, $\left| {\vec{p}} \right|=1$, $\left| {\vec{q}} \right|=\sqrt{2}$, then the area of quadrilateral is equal to A)  1                     B)  $\frac{1}{2}$ C)  $\frac{3}{2}$                                      D)  $\sqrt{3}$

$|\vec{p}-\vec{q}|\,=\vec{p}.\vec{q}$ $\Rightarrow \,\,\,|\vec{p}{{|}^{2}}\,+|\vec{q}{{|}^{2}}\,-2\,\vec{p}.\vec{q}\,=\sqrt{2}\,\cos \theta$ $\Rightarrow \,\,\cos \theta =\frac{1}{\sqrt{2}}$ Area $=\frac{1}{2}|\vec{p}||\vec{q}|\sin \theta =\frac{1}{2}$