Question

The area of the parallelogram constructed on the vectors $\overrightarrow{a}=\overrightarrow{p}+3\overrightarrow{q}$ and $\overrightarrow{b}=3\overrightarrow{p}+\overrightarrow{q}$ where $\overrightarrow{p},\overrightarrow{q}$ are unit vectors enclosing an angle of ${30}^{\circ }$ is

• A. $4$
• B. $\frac{7}{2}$
• C. $\frac{3}{2}$
• D. None of these

Answer 1

The correct option is A $4$

Area of the above parallelogram $=|\overrightarrow{a}\times \overrightarrow{b}|$
$\Rightarrow Area=\left|\right(\overrightarrow{p}+3\overrightarrow{q})\times (3\overrightarrow{p}+\overrightarrow{q})|$
$\Rightarrow Area=|\overrightarrow{p}\times \overrightarrow{q}+9\overrightarrow{q}\times \overrightarrow{p}|$
$\Rightarrow Area=|8\overrightarrow{p}\times \overrightarrow{q}|$
$\Rightarrow Area=8\times \sin {30}^o=4$ sq. units