ABCD is a quadrilateral with $\overline{\mathrm{AB}}=\bar{a}, \overline{\mathrm{AD}}=\overline{\mathrm{b}}$…
ABCD is a quadrilateral with $\overline{\mathrm{AB}}=\bar{a}, \overline{\mathrm{AD}}=\overline{\mathrm{b}}$ and $\overline{\mathrm{AC}}=2 \bar{a}+3 \overline{\mathrm{~b}}$. If its area is $\alpha$ times the area of the parallelogram with $\mathrm{AB}, \mathrm{AD}$ as adjacent sides, then the value of $\alpha$ is
$\frac{1}{2}$
$\frac{5}{2}$
$\frac{3}{2}$
2
Solution
Define $\mathrm{AB} = \mathrm{a}$ and $\mathrm{AD} = \mathrm{b}$. The area of the parallelogram formed by these vectors is $A_{parallelogram} = |\mathrm{a} \times \mathrm{b}|$.
The quadrilateral ABCD is divided by diagonal $\mathrm{AC} = 2\mathrm{a} + 3\mathrm{b}$ into two triangles: $\triangle ABC$ and $\triangle ADC$.
The area of $\triangle ABC$ is $\frac{1}{2} |\mathrm{a} \times (2\mathrm{a} + 3\mathrm{b})| = \frac{1}{2} |3(\mathrm{a} \times \mathrm{b})| = \frac{3}{2} |\mathrm{a} \times \mathrm{b}|$, as $\mathrm{a} \times \mathrm{a} = \mathrm{0}$.
The area of $\triangle ADC$ is $\frac{1}{2} |\mathrm{b} \times (2\mathrm{a} + 3\mathrm{b})| = \frac{1}{2} |2(\mathrm{b} \times \mathrm{a})| = \frac{1}{2} | -2(\mathrm{a} \times \mathrm{b}) | = |\mathrm{a} \times \mathrm{b}|$, using $\mathrm{b} \times \mathrm{b} = \mathrm{0}$ and $\mathrm{b} \times \mathrm{a} = -(\mathrm{a} \times \mathrm{b})$.
Summing these gives the total area: $\frac{3}{2} |\mathrm{a} \times \mathrm{b}| + |\mathrm{a} \times \mathrm{b}| = \frac{5}{2} |\mathrm{a} \times \mathrm{b}|$.
Given that this equals $\alpha \times |\mathrm{a} \times \mathrm{b}|$, it follows that $\alpha = \frac{5}{2}$.
The correct choice is $\boxed{\text{B}}$.