Explanation

(b)
Let the required vector be $\overrightarrow{\mathbf{c}} = x\widehat{\dot{\mathbf{i}}} + z\widehat{\mathbf{k}}$
Since, $\left| \overrightarrow{\mathbf{c}} \right| = 1 \Longrightarrow x^{2} + z^{2} = 1\ \ \ \ldots.(i)$
$\overrightarrow{\mathbf{a}}\ and\ \overrightarrow{\mathbf{c}}\$ are inclined at the angle $45{^\circ}$
$\therefore\cos{45{^\circ} = \frac{2x - z}{\sqrt{4 + 4 + 1}} \Longrightarrow 2x - z = \frac{3}{\sqrt{2}}}\ \ \ldots..(ii)$
$\overrightarrow{\mathbf{b}}\$and $\overrightarrow{\mathbf{c}}$ are inclined at an angle $60{^\circ}$
$\therefore\ - \frac{z}{\sqrt{2}} = \cos{60{^\circ}\ } \Longrightarrow z = - \frac{1}{\sqrt{2}}\ \ \ \ \ldots(iii)$
$From\ Eqs.(ii)\ and\ (iii),\ we\ get\ x = \frac{1}{\sqrt{2}}$
$Hence,\ the\ required\ \ ector\ is\frac{1}{\sqrt{2}}\widehat{\dot{\mathbf{i}}} - \frac{1}{\sqrt{2}}\widehat{\mathbf{k}}$