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 quadrilateral $OABC$, and $q$ denote the area of the parallelogram with $OA$ and $OC$ as adjecent sides, then $\frac{p}{q}$ is equal to

• A. $4$
• B. $6$
• C. $\frac{1}{2}\frac{|\overrightarrow{a}-\overrightarrow{b}|}{\overrightarrow{a}}$
• D. None of these

Answer 1

The correct option is B $6$

In quadrilateral $OABC$, adjacent sides are $OA$ and $AB$, i.e. $\overrightarrow{a}$ and $9\overrightarrow{a}+2\overrightarrow{b}$
Area of a quadrilateral is half the product of its diagonals.

So, Area $=\frac{1}{2}|(\overrightarrow{b}-\overrightarrow{a})\times (10\overrightarrow{a}+2\overrightarrow{b})|$

$=\frac{1}{2}|10\overrightarrow{b}\times \overrightarrow{a}+2\overrightarrow{b}\times \overrightarrow{a}|$

$=6|\overrightarrow{a}\times \overrightarrow{b}|=p$

Now the area of parallelogram with adjacent sides $OA,OC$ will be $|\overrightarrow{a}\times \overrightarrow{b}|=q$
Thus the ratio between the two areas $=\frac{p}{q}=6$