Question

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

• A. $2$
• B. $3$
• C. $6$
• D. $4$

Answer 1

The correct option is C $6$

$p=$ area of $OABC$
$=$area of $(△OAB+△OBC)$
$=|\frac{1}{2}\overrightarrow{OA}\times \overrightarrow{OB}+\frac{1}{2}\overrightarrow{OB}\times \overrightarrow{OC}|$
$=\frac{1}{2}|(\overrightarrow{OA}-\overrightarrow{OC})\times \overrightarrow{OB}|$
$=\frac{1}{2}|(\overrightarrow{a}-\overrightarrow{b})\times (10\overrightarrow{a}+2\overrightarrow{b})|$
$=\frac{1}{2}|12\overrightarrow{a}\times \overrightarrow{b}|$
$=6|\overrightarrow{a}\times \overrightarrow{b}|=6q$
$\frac{p}{q}=6$