To a man walking at the rate of $3 \mathrm{km}/\mathrm{h}$ the rain appears to fall vertically downwards. When he increases his speed to $6 \mathrm{km}/\mathrm{h}$ it appears to meet him at an angle of $45°$ with vertical. Find the speed of rain.

Let $\hat{\mathbf{i}} and \hat{\mathbf{j}}$ be the until vectors in horizontal and vertical directions respectively. 

Let velocity of rain ${\mathbf{v}}_r=a\hat{\mathbf{i}}+b\hat{\mathbf{j}} -------\left(i\right)$

Then, speed of rain will be, $\left|{\mathbf{v}}_r\right|=\sqrt{a^2+b^2} -----\left(ii\right)$

In the first case,  $$\begin{gathered} {\mathbf{v}}_m=\text{ velocity of }man=3\hat{\mathbf{i}} \\ {\mathrm{v}}_{\mathrm{m}}={\mathrm{v}}_{\mathrm{r}}-{\mathrm{v}}_{\mathrm{m}}=(\mathrm{a}-3)\hat{\mathbf{i}}+\mathrm{b}\hat{\mathbf{j}} \end{gathered}$$

It seems to be in vertical direction, 

Hence, $a-3=0 or a =3$.

In the second case, 

$$\begin{gathered} {\mathbf{v}}_m=6\hat{\mathrm{i}} \\ Therefore {\mathbf{v}}_{rm}=(a-6)\hat{\mathbf{i}}+b\hat{\mathbf{j}}=-3\hat{\mathbf{i}}+b\hat{\mathbf{j}} \end{gathered}$$

This seems to be at ${45}^o$ with vertical. Hence $\left|b\right|=3$.

Therefore, from Eq. (ii) speed of rain is $$\begin{gathered} \left|v_r\right|=\sqrt{(3{)}^2+(3{)}^2} \\ =3\sqrt{2}km/h \end{gathered}$$