Explanation

(d)
Let the unit vector in $xy$-plane be $\overrightarrow{\mathbf{a}} = x\widehat{\mathbf{i}} + y\widehat{\mathbf{j}}$.
$\therefore\cos{45{^\circ} = \frac{\left( x\widehat{\mathbf{i}} + y\widehat{\mathbf{j}} \right)(\widehat{\mathbf{i}}\mathbf{+}\widehat{\mathbf{j}})\ }{\sqrt{x^{2} + y^{2}}\sqrt{1^{2} + 1^{2}}}}$
$\Rightarrow \ \frac{1}{\sqrt{2}} = \frac{x + y}{\sqrt{2}\sqrt{x^{2} + y^{2}}}$
$\Rightarrow 1 = \frac{x + y}{\sqrt{x^{2} + y^{2}}}$
$\Rightarrow x + y = \sqrt{x^{2} + y^{2}}$
Since, $\overrightarrow{a}$ is a unit vector.
$\therefore\left| \overrightarrow{\mathbf{a}} \right| = \sqrt{x^{2} + y^{2}}$=1
$\Rightarrow x + y = 1$ …(i)
Again $\cos{60{^\circ} = \frac{\left( x\widehat{\mathbf{i}} + y\widehat{\mathbf{j}} \right) \bullet (3\widehat{\mathbf{i}} - \widehat{4\mathbf{j}})}{\sqrt{x^{2} + y^{2}}\sqrt{3^{2} + 4^{2}}}}$
$\Rightarrow \frac{1}{2} = \frac{3x - 4y}{1 \bullet 5}\ \Rightarrow \frac{5}{2} = 3x - 4y$
$5 = 6x - 8y$ …..(ii)
On solving Eqs. (i) and (ii), we get
$x = \frac{13}{14},\ y = \frac{1}{14}$
$\therefore\overrightarrow{\mathbf{a}} = \frac{1}{14}\left( 13\widehat{\mathbf{i}} + \widehat{\mathbf{j}} \right)$
No value in the given options satisfies the above relations.
Thus, option (d) is correct.