Question

A girl walks $4\text{km}$ towards west, then she walks $3\text{km}$ in a direction ${30}^{\circ }$ east of north and stops. Girl's displacement from her initial point of departure is

• A. $-\frac{5}{2}\hat{i}+\frac{3\sqrt{3}}{2}\hat{j}$
• B. $\frac{5}{2}\hat{i}+\frac{3\sqrt{3}}{2}\hat{j}$
• C. $-\frac{5}{2}\hat{i}+\frac{2\sqrt{3}}{2}\hat{j}$
• D. $\frac{5}{2}\hat{i}+\frac{2\sqrt{3}}{2}\hat{j}$

Answer 1

The correct option is A $-\frac{5}{2}\hat{i}+\frac{3\sqrt{3}}{2}\hat{j}$


Let $B(x,y)$ be the final position of the girl and $O$ be the initial point of departure.
Then, $\overrightarrow{AL}=\overrightarrow{AB}\cos {60}^{\circ }=\frac{3}{2}$ and,
$\overrightarrow{BL}=\overrightarrow{AB}\sin {60}^{\circ }=\frac{3\sqrt{3}}{2}$
$\therefore \overrightarrow{OL}=\overrightarrow{OA}-\overrightarrow{AL}=(4-\frac{3}{2})=\frac{5}{2}$ and $\overrightarrow{BL}=\frac{3\sqrt{3}}{2}$
Clearly, $B(x,y)$ lies in second quadrant.
So, coordinates of $B$ are $(-\frac{5}{2},\frac{3\sqrt{3}}{2})$.
Hence, position vector of $B$ is $-\frac{5}{2}\hat{i}+\frac{3\sqrt{3}}{2}\hat{j}$