Explanation

(d)
Let $\overrightarrow{\mathbf{a}} = \widehat{\dot{\mathbf{i}}} + \widehat{\dot{\mathbf{j}}}\ and\ \overrightarrow{\mathbf{b}} = \widehat{\dot{\mathbf{j}}} + \widehat{\mathbf{k}}$
$Now,\ \overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}} = \left| \begin{matrix}
\widehat{\dot{\mathbf{i}}} & \widehat{\dot{\mathbf{j}}} & \widehat{\mathbf{k}} \\
1 & 1 & 0 \\
0 & 1 & 1
\end{matrix} \right|$
$= \widehat{\dot{\mathbf{i}}} - \widehat{\dot{\mathbf{j}}} + \widehat{\mathbf{k}}$
$and\ \left| \overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}} \right| = \sqrt{1^{2} + ( - 1)^{2} + 1^{2}} = \sqrt{3}\$
∴Required unit vector
$= \frac{\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}}{|\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}|} = \frac{\widehat{\dot{\mathbf{i}}} - \widehat{\dot{\mathbf{j}}} + \widehat{\mathbf{k}}}{\sqrt{3}}$
Alternate Let $\overrightarrow{\mathbf{a}} = x\widehat{\dot{\mathbf{i}}} + y\widehat{\dot{\mathbf{j}}} + z\widehat{\mathbf{k}}$
Since, $\overrightarrow{\mathbf{a}} \bullet \left( \widehat{\dot{\mathbf{i}}} + \widehat{\dot{\mathbf{j}}} \right) = 0\ and\ \overrightarrow{\mathbf{a}} \bullet \left( \widehat{\dot{\mathbf{j}}} + \widehat{\mathbf{k}} \right) = 0$
$\Longrightarrow x + y = 0\ and\ y + z = 0$
Also $x^{2} + y^{2} + z^{2} = 1$
$\Longrightarrow x = 1,\ y = - 1\ and\ z = 1$
$\therefore\ \overrightarrow{\mathbf{a}} = \frac{\widehat{\dot{\mathbf{i}}} - \widehat{\dot{\mathbf{j}}} + \widehat{\mathbf{k}}}{\sqrt{3}}$