If $\vec{a}$ and $\vec{b}$ are unit vectors inclined at an angle $30^\circ$ to each other, then find the area of the parallelogram with $(\vec{a} + 3\vec{b})$ and $(3\vec{a} + \vec{b})$ as adjacent sides.

$4$ sq. units

The correct answer is Option (2) → $4$ sq. units ##

We know, area of parallelogram with adjacent sides $\vec{p}$ and $\vec{q}$ is given by

$A = |\vec{p} \times \vec{q}|$

Here, Area $= |(\vec{a} + 3\vec{b}) \times (3\vec{a} + \vec{b})|$

$= |3(\vec{a} \times \vec{a}) + (\vec{a} \times \vec{b}) + 9(\vec{b} \times \vec{a}) + 3(\vec{b} \times \vec{b})|$

$= |3 \times 0 + (\vec{a} \times \vec{b}) - 9(\vec{a} \times \vec{b}) + 3 \times 0|$

$[∵\vec{a} \times \vec{a} = 0 = \vec{b} \times \vec{b} \text{ and } \vec{b} \times \vec{a} = -(\vec{a} \times \vec{b})]$

$= |-8(\vec{a} \times \vec{b})| = 8|\vec{a} \times \vec{b}| = 8|\vec{a}||\vec{b}| \sin \theta$

$= 8 \cdot 1 \cdot 1 \cdot \sin 30^\circ$ [Given, $|\vec a|=1=|\vec b|$ and $\theta=30^\circ$]

$= 8 \cdot \frac{1}{2} = 4 \text{ sq. units}$