Explanation

(d) ${\overrightarrow{v}}_{man}$ = $\frac{v}{\sqrt{2}}\widehat{i} + \frac{v}{\sqrt{2}}\widehat{j}$
Let ${\overrightarrow{v}}_{wind} = a\widehat{i} + b\widehat{j}$
$\Rightarrow \ \ {\overrightarrow{v}}_{wind/man}$= $\left( a - \frac{v}{\sqrt{2}} \right)\widehat{i} + \left( b - \frac{v}{\sqrt{2}} \right)\widehat{j}$
$\Rightarrow \ \tan\theta = \frac{b - \frac{v}{\sqrt{2}}}{a - \frac{v}{\sqrt{2}}} = \tan{270{^\circ}}$
$\Rightarrow \ a - \frac{v}{\sqrt{2}} = 0$
$\Rightarrow a = \frac{v}{\sqrt{2}}\ \Rightarrow {\overrightarrow{v}}_{wind} = \frac{v}{\sqrt{2}}\widehat{i} + b\widehat{j}$
when the man doubles his speed
${\overrightarrow{v}}_{man}^{'} = 2\left( \frac{v}{\sqrt{2}}\widehat{i} + \frac{v}{\sqrt{2}}\widehat{j} \right) = \sqrt{2\ }(v\widehat{i} + v\widehat{j})$
$\Rightarrow \ {\overrightarrow{v}'}_{wind/man}$=$\left( \frac{v}{\sqrt{2}} - \sqrt{2\ }v \right)\widehat{i} + (b - \sqrt{2\ }v)\widehat{j}$
$\Rightarrow \ \tan\theta^{'} = \frac{b - \sqrt{2\ }}{\frac{v}{\sqrt{2} - \sqrt{2\ }v}} = \frac{2v - \sqrt{2\ b}}{v}$
But $\theta^{'} = 270{^\circ} - \cot^{- 1}(2)$
$\Rightarrow$ tan [$270{^\circ} - \cot^{- 1}(2)$]$= \ \frac{2v - \sqrt{2b}}{v}$
$\Rightarrow$ cot[$\cot^{- 1}{(2)\rbrack} = \frac{2v - \sqrt{2b}}{v}$
$\Rightarrow 2v = 2v - \sqrt{2\ b} = b = 0$
$\therefore\ {\overrightarrow{v}}_{wind}$ =$\frac{v}{\sqrt{2}}\widehat{i}$