A man A moves due to east with velocity $6m{s}^{-1}$ and another man B moves in $N-{30}^{0}E$ with $6m{s}^{-1}$. Find the velocity of B w.r.t. A.

Given, 

$v_B=v_B\cos {60}^0\widehat{i}+v_B\sin {60}^0\widehat{j}$

$=6\left(\frac{1}{2}\right)\widehat{i}+6\left(\frac{\sqrt{3}}{2}\right)\widehat{j}=3\widehat{i}+3\sqrt{3}\widehat{j}$

To find the velocity, 

$v_{BA}=v_B-v_A=(3\widehat{i}+3\sqrt{3}\widehat{j})-6\widehat{i}$

$=-3\widehat{i}+3\sqrt{3}\widehat{j}$

$\left|v_{BA}\right|=\sqrt{{(-3)}^2+{\left(3\sqrt{3}\right)}^2}=\sqrt{9+27}$

$=\sqrt{36}=6m{s}^{-1}$

Here, $\widehat{i}$  is –ve and $\widehat{j}$ is +ve. So, second quadrant is possible. 

Direction, $\tan \alpha =\frac{coefficientof\widehat{j}}{coefficientof\widehat{i}}$

$=\frac{3\sqrt{3}}{-3}=-\sqrt{3}$

$\Rightarrow \alpha ={60}^0$