Question

If $\overrightarrow{OA}=\overrightarrow{a},\overrightarrow{OB}=10\overrightarrow{a}+2\overrightarrow{b}$ and $\overrightarrow{OC}=\overrightarrow{b}$, where $A$ and $C$ are non collinear points. Let $p$ denote the area of the Quadrillateral $OABC$ and $q$ denote the area of a parallelogram with $\overrightarrow{OA}$ and $\overrightarrow{OC}$ as adjacent sides. The $pq=$

• A. $4$
• B. $6$
• C. $8$
• D. $10$

Answer 1

The correct option is B $6$

$p=$ area of parallelogram with $OA$ and $OC$ as adjacent sides

$=|OA\times OC|=|\overrightarrow{a}\times \overrightarrow{b}|.$

and $q=$ area of quadrilateral $OABC$

$=\frac{1}{2}|OA\times OB|+\frac{1}{2}|OB\times OC|$

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

$=|a\times \overrightarrow{b}|+5|\overrightarrow{a}\times \overrightarrow{b}|=6|\overrightarrow{a}\times \overrightarrow{b}|.$

$\therefore q=6p\Rightarrow k=6.$