Question

The area of the parallelogram constructed on the Vectors $\overrightarrow{a}=\overrightarrow{p}+2\overrightarrow{q}$ and $\overrightarrow{b}=2\overrightarrow{p}+\overrightarrow{q}$ as sides, where $\overrightarrow{p},\overrightarrow{q}$ are unit Vectors forming an angle of ${60}^0$ in square units is

• A. $\frac{3}{2}$
• B. $\frac{3\sqrt{3}}{2}$
• C. $\frac{3\sqrt{3}}{4}$
• D. $\frac{1}{2}$

Answer 1

The correct option is B $\frac{3\sqrt{3}}{2}$

$Area=|\left(\overrightarrow{a}\right)\times \left(\overrightarrow{b}\right)|$

$=|(\overrightarrow{p}+2\overrightarrow{q})\times (2\overrightarrow{p}+\overrightarrow{q})|$

$=|(\overrightarrow{p}\times \overrightarrow{q})+(2\overrightarrow{q}\times 2\overrightarrow{p})|$

$=|-(\overrightarrow{q}\times \overrightarrow{p})+4(\overrightarrow{q}\times \overrightarrow{p})|$

$=|3(\overrightarrow{q}\times \overrightarrow{p})|$

$=3|\overrightarrow{p}||\overrightarrow{q}|.\sin \theta$

$=3.\sin \left({60}^0\right)$

$=\frac{3\sqrt{3}}{2}$