# A girl walks 4 km towards west, then she walks 3 km in a direction 30° east of north and stops. Determine the girl’s displacement from her initial point of departure.

Toolbox:
• By triangle law of vectors : $\overrightarrow {AB}+\overrightarrow{BC}+\overrightarrow{CA}=0$
• The cosine of the angle made by the vector with the area is the direction cosines.
Step 1:
A girl walks 4km toward west (i.e) along $OX$ axis.
Therefore $\overrightarrow{OP}=-4\hat i$----(1)
Next she goes in the direction $30^{\large\circ}$ east of north (i.e) she moves along $OQ$ and stops at $Q$
This implies $PQ$ makes an angle $60^{\large\circ}$ with $OP$.
Step 2:
Hence the scalar component of $PQ$ along $OX=OQ\cos 60^{\large \circ}$
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad=3\cos60^{\large\circ}$
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad=3\times \large\frac{1}{2}=\frac{3}{2}$
Step 3:
Scalar vertical component of $PQ$ along $OY=OQ\sin 60^{\large\circ}$
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad=3\sin 60^{\large\circ}$
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad=\large\frac{3\sqrt 3}{2}$
Therefore $\overrightarrow{PQ}=\large\frac{3}{2}$$\hat i+\large\frac{3\sqrt 3}{2}$$\hat j$------(2)
Step 4:
The girl walks along $OP$ and then along $PQ$
$\overrightarrow {OP}+\overrightarrow{PQ}=\overrightarrow{OQ}$ (By triangle law of vectors)
By adding eq(1) and eq(2) we get
$\overrightarrow{OQ}=-4\hat i+\large\frac{3}{2}$$\hat i+\large\frac{3\sqrt 3}{2}$$\hat j$
$\overrightarrow{OQ}=-4\hat i+\large\frac{3}{2}$$\hat i+\big(\large\frac{3\sqrt 3}{2}$$\hat j\big)$
$\quad\quad=\large\frac{5}{2}$$\hat i+\large\frac{3\sqrt 3}{2}$$\hat j$
$|\overrightarrow {OQ}|=\sqrt{\big(\large\frac{5}{2}\big)^2+\big(\frac{3\sqrt 3}{2}\big)^2}$
$\qquad=\sqrt{\big(\large\frac{25}{4}\big)+\big(\large\frac{27}{4}\big)}$
$\qquad=\sqrt{\big(\large\frac{52}{4}\big)}$
$\qquad=\sqrt{13}$
Step 5:
Hence the distance travelled by the girl from her initial position is $\sqrt{13}$ units.
edited May 23, 2013