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

Let $B(x,\;y)$ be the final position of the girl and $O$ be the initial point of departure.

Then,
$AL = AB\cos {60^o} = \cfrac{3}{2}$ km and $BL = AB\sin {60^o} = \cfrac{{3\sqrt 3 }}{2}$ km

$\therefore$ $OL = OA - AL = \left( {4 - \cfrac{3}{2}} \right) = \cfrac{5}{2}$ and $BL = \cfrac{{3\sqrt 3 }}{2}$

Clearly, $B(x,\;y)$ lies in second quadrant.

So, coordinates of $B$ are $\left( { - \cfrac{5}{2},\cfrac{{3\sqrt 3 }}{2}} \right)$

Hence, position vector of $B$ is $- \cfrac{5}{2}\hat i + \cfrac{{3\sqrt 3 }}{2}\hat j$