Let \(\mathrm{O}\) and \(\mathrm{B}\) be the initial and final positions of the girl respectively.
Then, the girl's position can be shown as in the figure.
Now, we have \(\mathrm{OA}=-4 \hat{\mathrm{i}}\)
\(\mathrm{AB}=\hat{\mathrm{i}}|\mathrm{AB}| \cos 60^{\circ}+\hat{\mathrm{j}}|\mathrm{AB}| \sin 60^{\circ}\)


(AB \(\cos 60^{\circ}\) is component of \(\mathrm{AB}\) along \(\mathrm{X}\)-axis and \(\mathrm{AB} \sin 60^{\circ}\) is component of \(\mathrm{AB}\) along \(\mathrm{Y}\) axis)
\(=\hat{\mathrm{i}}\left(3 \times \frac{1}{2}\right)+\hat{\mathrm{j}}\left(3 \times \frac{\sqrt{3}}{2}\right)=\frac{3}{2} \hat{\mathrm{i}}+\frac{3 \sqrt{3}}{2} \hat{\mathrm{j}}\)
By the triangle law of vector addition, we have
\(\begin{aligned}
& \mathrm{OB}=\mathrm{OA}+\mathrm{AB}=(-4 \hat{\mathrm{i}})+\left(\frac{3}{2} \hat{\mathrm{i}}+\frac{3 \sqrt{3}}{2} \hat{\mathrm{j}}\right) \\
& =\left(-4+\frac{3}{2}\right) \hat{\mathrm{i}}+\frac{3 \sqrt{3}}{2} \hat{\mathrm{j}} \\
& =\left(\frac{-8+3}{2}\right) \hat{\mathrm{i}}+\frac{3 \sqrt{3}}{2} \hat{\mathrm{j}}=\frac{-5}{2} \hat{\mathrm{i}}+\frac{3 \sqrt{3}}{2} \hat{\mathrm{j}}
\end{aligned}\)
Hence the girl's displacement from her initial point of departure is \(\frac{-5}{2} \hat{i}+\frac{3 \sqrt{3}}{2} \hat{j}\)