Question

$ABCD$ is a quadrilateral with $\overrightarrow{ AB }=\vec{ a }, \overrightarrow{ AD }=\vec{ b }$ and $\overrightarrow{ AC }=2 \vec{ a }+3 \vec{ b }$. If its area is $\alpha$ times the area of the parallelogram $gm$ with $AB , AD$ as adjacent sides, then find the value of $\alpha$.

Answer 1

Area of quadrilateral $ABCD$
= area of $\triangle ABC +$ area of $\triangle ACD$
$=\frac{1}{2}[\vec{a} \times(2 \vec{a}+3 \vec{b})]+\frac{1}{2}[(2 \vec{a}+3 \vec{b}) \times \vec{b}]$
$=\frac{1}{2}(3 \vec{a} \times \vec{b}+2 \vec{a} \times \vec{b})=\frac{5}{2}(\vec{a} \times \vec{b})$
$=\frac{5}{2}$ area of parallelogram $ABCD$
$\Rightarrow \alpha=\frac{5}{2}$