Question

$ABCD$ is a quadrilateral with $\overrightarrow{AB}=\overrightarrow{a},\overrightarrow{AD}=\overrightarrow{b},\overrightarrow{AC}=2\overrightarrow{a}+3\overrightarrow{b}$. If the area of the quadrilateral is $p$ times the area of parallelogram with $AB$ and $AD$ as adjacent side, then value of $p$ is

• A. $3$
• B. $\frac{1}{2}$
• C. $\frac{5}{2}$
• D. $\frac{7}{2}$

Answer 1

The correct option is C $\frac{5}{2}$

$\overrightarrow{AB}=\overrightarrow{a},\overrightarrow{AD}=\overrightarrow{b},\overrightarrow{AC}=2\overrightarrow{a}+3\overrightarrow{b}$
$\overrightarrow{BD}=\overrightarrow{AD}-\overrightarrow{AB}=\overrightarrow{b}-\overrightarrow{a}$
Area of quadrilateral $=\frac{1}{2}|\overrightarrow{BD}\times \overrightarrow{AC}|=\frac{1}{2}|(\overrightarrow{b}-\overrightarrow{a})\times (2\overrightarrow{a}+3\overrightarrow{b})|$
$\Rightarrow \frac{1}{2}|2(\overrightarrow{b}\times \overrightarrow{a})-3(\overrightarrow{a}\times \overrightarrow{b})|=\frac{5}{2}|\overrightarrow{a}\times \overrightarrow{b}|\cdots \left(i\right)$
Area of parallelogram $=|\overrightarrow{AB}\times \overrightarrow{AD}|=|\overrightarrow{a}\times \overrightarrow{b}|\cdots \left(ii\right)$
From $\left(i\right)$ and $\left(ii\right)$
Area of quadrilateral = $\frac{5}{2}$ (Area of parallelogram)
Hence $p=\frac{5}{2}$