Question

Let $\overrightarrow{OA}=\overrightarrow{a},\overrightarrow{OB}=10\overrightarrow{a}+2\overrightarrow{b}$ and $\overrightarrow{OC}=\overrightarrow{b},$ where $O,A$ and $C$ are non-collinear points. Let $p$ denote the area of the quadrilateral $OABC$, and $q$ denote the area of the parallelogram with $OA$ and $OC$ as adjacent sides. If $p=Kq$, then find the value of $K$.

• A. 2
• B. 8
• C. 6
• D. 4

Answer 1

The correct option is C 6

q=area of parallelogram wih $OA$ and $OC$ as adjacent sided $=|OA\times OC|$
$=|a\times b|$
and
$p=$ area of quadrilateral OABC
$=\left(\frac{1}{2}\right)|OA\times OB|+\left(\frac{1}{2}\right)|OB\times OC|$
$=\left(\frac{1}{2}\right)|a\times (10a+2b)|+\left(\frac{1}{2}\right)|(10a+2b)\times b|=|a\times b|+5|a\times b|=6|a\times b|=6q$
Thus $K=6$