# If $\overrightarrow {OA}=\overrightarrow a,\:\overrightarrow {AB}=6\overrightarrow a+2\overrightarrow b\:and\:\overrightarrow {OC}=\overrightarrow b$, $p$ is area of quadrateral $OACB$ and $q$ is area of parallelogram with $OA$ and $OC$ as adjacent sides, and if $p=x.q$, then $x=?$

$\begin{array}{1 1} 4 \\ 5 \\ 6 \\ 7 \end{array}$

In the quadrilateral $OACB$, Area of $OACB=p=$Area of $\Delta OAC+$ Area of $\Delta OBC$
$\Rightarrow\:p=\large\frac{1}{2}$$(\overrightarrow {OA}\times \overrightarrow {OC})+\large\frac{1}{2}$$(\overrightarrow {OC}\times \overrightarrow {OB})$
$=\large\frac{1}{2}$$(-\overrightarrow {OC} \times\overrightarrow {OA}+\overrightarrow OC\times\overrightarrow {OB}) =\large\frac{1}{2}$$\overrightarrow {OC}\times(\overrightarrow {OB}-\overrightarrow {OA})$
$=\large\frac{1}{2}$$(\overrightarrow {OC}\times \overrightarrow {AB}) =\large\frac{1}{2}$$(\overrightarrow b\times(6\overrightarrow a+2\overrightarrow b)$)
$=\large\frac{1}{2}$.$6(\overrightarrow b\times\overrightarrow a)$
Area of parallelogram with sides $\overrightarrow {OA}$ and $\overrightarrow {OC}$
$=q=\large\frac{1}{2}$$(\overrightarrow b\times\overrightarrow b)$
$\therefore\:p=6q$
$\therefore\:x=6$