Let \(\bar{a}, \bar{b}\) and \(\bar{c}\) be three unit vectors such that \(\bar{a} \times(\bar{b} \times \bar{c})=\frac{\sqrt{3}}{2}(\bar{b}+\bar{c})\).
If \(\bar{b}\) is not parallel to \(\bar{c}\), then the angle between \(\bar{a}\) and \(\bar{b}\) is

\(\begin{aligned} & \overline{\mathrm{a}} \times(\overline{\mathrm{b}} \times \overline{\mathrm{c}})=\frac{\sqrt{3}}{2}(\overline{\mathrm{~b}}+\overline{\mathrm{c}}) \\ & \Rightarrow(\overline{\mathrm{a}} \cdot \overline{\mathrm{c}}) \overline{\mathrm{b}}-(\overline{\mathrm{a}} \cdot \overline{\mathrm{b}}) \overline{\mathrm{c}}=\left(\frac{\sqrt{3}}{2}\right) \overline{\mathrm{b}}+\left(\frac{\sqrt{3}}{2}\right) \overline{\mathrm{c}} \\ & \Rightarrow \overline{\mathrm{a}} \cdot \overline{\mathrm{c}}=\frac{\sqrt{3}}{2} \text { and } \overline{\mathrm{a}} \cdot \overline{\mathrm{b}}=\frac{-\sqrt{3}}{2} \\ & \Rightarrow|\overline{\mathrm{a}}||\overline{\mathrm{b}}| \cos \theta=\frac{-\sqrt{3}}{2} \\ & \Rightarrow \cos \theta=\frac{-\sqrt{3}}{2}=\cos \frac{5 \pi}{6} \\ & \Rightarrow \theta=\frac{5 \pi}{6}\end{aligned}\)