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

\(\overline{\mathrm{a}} \times(\overline{\mathrm{b}} \times \overline{\mathrm{c}})=\frac{\overline{\mathrm{b}}+\overline{\mathrm{c}}}{\sqrt{2}}\)
\(\begin{aligned} & \Rightarrow(\overline{\mathrm{a}} \cdot \overline{\mathrm{c}}) \overline{\mathrm{b}}-(\overline{\mathrm{a}} \cdot \overline{\mathrm{b}}) \overline{\mathrm{c}}=\frac{\overline{\mathrm{b}}+\overline{\mathrm{c}}}{\sqrt{2}} \\ & \Rightarrow\left(\overline{\mathrm{a}} \cdot \overline{\mathrm{c}}-\frac{1}{\sqrt{2}}\right) \overline{\mathrm{b}}-\left(\overline{\mathrm{a}} \cdot \overline{\mathrm{b}}+\frac{1}{\sqrt{2}}\right) \overline{\mathrm{c}}=0\end{aligned}\)
Since \(\overline{\mathrm{a}}, \overline{\mathrm{b}}, \overline{\mathrm{c}}\) are non-coplanar unit vectors,
\(\begin{aligned} & \bar{a} \cdot \bar{b}+\frac{1}{\sqrt{2}}=0 \\ & \Rightarrow \bar{a} \cdot \bar{b}=-\frac{1}{\sqrt{2}} \\ & \Rightarrow|\bar{a}||\bar{b}| \cos \theta=-\frac{1}{\sqrt{2}} \\ & \Rightarrow \cos \theta=-\frac{1}{\sqrt{2}} \\ & \Rightarrow \theta=\frac{3 \pi}{4}\end{aligned}\)
[Note: In the question, \(\frac{\overline{\mathrm{b}} \times \overline{\mathrm{c}}}{\sqrt{2}}\) is changed to \(\frac{\overline{\mathrm{b}}+\overline{\mathrm{c}}}{\sqrt{2}}\) to apply appropriate textual concepts.]