A particle A is moving towards North with an acceleration of 5 ms^-2 and particle B is moving North-East direction with an acceleration of $5\sqrt{2}{\mathrm{ms}}^{-2}$. The relative  acceleration of particle A with respect to particle B is

The relative acceleration of A with respect to B is ${\mathrm{a}}_{\mathrm{AB}}={\mathrm{a}}_{\mathrm{A}}-{\mathrm{a}}_{\mathrm{B}}$

$\begin{array}{c}{\mathrm{a}}_{\mathrm{AB}}=\sqrt{{\mathrm{a}}_{\mathrm{A}}^2+{\mathrm{a}}_{\mathrm{B}}^2-2{\mathrm{a}}_A{\mathrm{a}}_{\mathrm{B}}\cos {45}^{\circ }}\\ =\sqrt{5^2+(5\sqrt{2}{)}^2-2\times 5\times 5\sqrt{2}\times \frac{1}{\sqrt{2}}}\\ =\sqrt{25+50-50}\\ =\sqrt{25}=5{\mathrm{ms}}^{-2}\end{array}$

Let a_AB makes $\theta$ angle with North direction then 

$\begin{array}{c}\tan \mathrm{\theta }=\frac{{\mathrm{a}}_{\mathrm{B}}\sin {45}^{\circ }}{{\mathrm{a}}_{\mathrm{A}}-{\mathrm{a}}_{\mathrm{B}}\cos {45}^{\circ }}\\ =\frac{5\sqrt{2}\times 1/\sqrt{2}}{5-5\sqrt{2}\times 1/\sqrt{2}}=0\end{array}$

$\Rightarrow \theta ={90}^{\circ }$ It means a_AB is directed towards West.