Explanation

(d)
Let the required unit vector be $\overrightarrow{\mathbf{r}} = a\widehat{\dot{\mathbf{i}}} + b\widehat{\dot{\mathbf{j}}}$
Then, $\left| \overrightarrow{\mathbf{r}} \right| = 1$
$\Longrightarrow \ a^{2} + b^{2} = 1\ \$…(i)
Since, $\overrightarrow{\mathbf{r}}$ makes an angle of $45{^\circ}$ with $\widehat{\dot{\mathbf{i}}} + \widehat{\dot{\mathbf{j}}}$ and an angle of $60{^\circ}$ with3$\widehat{\dot{\mathbf{i}}} - 4\widehat{\dot{\mathbf{j}}},$ therefore
$\cos{\frac{\pi}{4} = \frac{\overrightarrow{\mathbf{r}} \bullet (\widehat{\dot{\mathbf{i}}} + \widehat{\dot{\mathbf{j}}})}{\left| \overrightarrow{\mathbf{r}} \right||\widehat{\dot{\mathbf{i}}} + \widehat{\dot{\mathbf{j}}}|}}$
$and\ \ \cos{\frac{\pi}{3} = \frac{\overrightarrow{\mathbf{r}} \bullet \left( 3\widehat{\dot{\mathbf{i}}} - 4\widehat{\dot{\mathbf{j}}} \right)}{\left| \overrightarrow{\mathbf{r}} \right|\left| 3\widehat{\dot{\mathbf{i}}} - 4\widehat{\dot{\mathbf{j}}} \right|}}$
$\Longrightarrow \frac{1}{\sqrt{2}} = \frac{a + b}{\sqrt{2}}$
$and\frac{1}{2} = \frac{3a - 4b}{5}$
$\Longrightarrow a + b = 1$
$and\ 3a - 4b = \frac{5}{2}$
$\Longrightarrow a = \frac{13}{14},b = \frac{1}{14}$
$\therefore\ \overrightarrow{\mathbf{r}} = \frac{13}{14}\widehat{\dot{\mathbf{i}}} + \frac{1}{14}\widehat{\dot{\mathbf{j}}}$