A girl walks 4 km towards west, then she walks 3 km in a direction 30° east of north and stops.

Let O and B be the initial and final positions of the girl respectively.

Then, the girl’s position can be shown as:

Now, we have:

$\overrightarrow{\mathrm{OA}}=-4 \hat{i}$

$\overrightarrow{\mathrm{AB}}=\hat{i}|\overrightarrow{\mathrm{AB}}| \cos 60^{\circ}+\hat{j}|\overrightarrow{\mathrm{AB}}| \sin 60^{\circ}$

$=\hat{i} 3 \times \frac{1}{2}+\hat{j} 3 \times \frac{\sqrt{3}}{2}$

$=\frac{3}{2} \hat{i}+\frac{3 \sqrt{3}}{2} \hat{j}$

By the triangle law of vector addition, we have:

$\overrightarrow{\mathrm{OB}}=\overrightarrow{\mathrm{OA}}+\overrightarrow{\mathrm{AB}}$

$=(-4 \hat{i})+\left(\frac{3}{2} \hat{i}+\frac{3 \sqrt{3}}{2} \hat{j}\right)$

$=\left(-4+\frac{3}{2}\right) \hat{i}+\frac{3 \sqrt{3}}{2} \hat{j}$

$=\left(\frac{-8+3}{2}\right) \hat{i}+\frac{3 \sqrt{3}}{2} \hat{j}$

$=\frac{-5}{2} \hat{i}+\frac{3 \sqrt{3}}{2} \hat{j}$

Hence, the girl’s displacement from her initial point of departure is

$\frac{-5}{2} \hat{i}+\frac{3 \sqrt{3}}{2} \hat{j}$