If \(\bar{a}, \bar{b}, \bar{c}\) are non-coplanar unit vectors such that \(\overline{\mathrm{a}} \times(\overline{\mathrm{b}} \times \overline{\mathrm{c}})=\frac{(\overline{\mathrm{b}}+\overline{\mathrm{c}})}{\sqrt{2}}\) then the angle between \(\overline{\mathrm{a}}\) and \(\bar{b}\) is

\(\begin{aligned} & \bar{a} \times(\bar{b} \times \bar{c})=\frac{\bar{b}+\bar{c}}{\sqrt{2}} \\ & \Rightarrow(\bar{a} \cdot \bar{c}) \bar{b}-(\bar{a} \cdot \bar{b}) \bar{c}=\frac{\bar{b}+\bar{c}}{\sqrt{2}} \\ & \Rightarrow\left(\overline{\mathrm{a}} \cdot \overline{\mathrm{c}}-\frac{1}{\sqrt{2}}\right) \bar{b}-\left(\bar{a} \cdot \bar{b}+\frac{1}{\sqrt{2}}\right) \bar{c}=0\end{aligned}\)
Since \(\overline{\mathrm{a}}, \overline{\mathrm{b}}, \overline{\mathrm{c}}\) are non-coplanar unit vectors,
\(\begin{aligned}
& \overline{\mathrm{a}} \cdot \overline{\mathrm{~b}}+\frac{1}{\sqrt{2}}=0 \\
& \Rightarrow \overline{\mathrm{a}} \cdot \overline{\mathrm{~b}}=-\frac{1}{\sqrt{2}} \\
& \Rightarrow|\overline{\mathrm{a}}||\overrightarrow{\mathrm{b}}| \cos \theta=-\frac{1}{\sqrt{2}} \\
& \Rightarrow \cos \theta=-\frac{1}{\sqrt{2}} \\
& \Rightarrow \theta=\frac{3 \pi}{4}
\end{aligned}\)