Let $\overline{\mathrm{a}}, \overline{\mathrm{b}}, \overline{\mathrm{c}}$ be three vectors having mag­nitudes 1,1,2 respectively. If $\overline{\mathrm{a}}\times (\overline{\mathrm{a}}\times \overline{\mathrm{c}})+\overline{\mathrm{b}}=\overline{0}$ then the angle between $\overline{\mathrm{a}}$ and $\overline{\mathrm{c}}$ is

$$\begin{gathered} \left|\overline{\mathrm{a}}\right|=1=\left|\overline{\mathrm{b}}\right| \left|\overline{\mathrm{c}}\right|=2 \\ \begin{array}{c}\begin{array}{c}\overline{\mathrm{a}}\times (\overline{\mathrm{a}}\times \overline{\mathrm{c}})=-\overline{\mathrm{b}}\end{array}\\ (\overline{\mathrm{a}}·\overline{\mathrm{c}}) \overline{\mathrm{a}}-\overline{\mathrm{c}}=-\overline{\mathrm{b}}\end{array} \overline{) \begin{array}{c}\overline{\mathrm{a}}·\overline{\mathrm{c}}=1\times 2\times \cos \mathrm{\theta }\\ =2\cos \mathrm{\theta }\end{array}} \\ {\left|(2\cos \mathrm{\theta }) \overline{\mathrm{a}}-\overline{\mathrm{c}}\right|}^2=|\overline{\mathrm{b}}{|}^2 \\ 4{\cos }^2\mathrm{\theta }+4-4\mathrm{cos\theta }(\overline{\mathrm{a}}·\overline{\mathrm{c}})=1 \\ 4{\cos }^2\mathrm{\theta }+4-8{\cos }^2\mathrm{\theta }=1 \Rightarrow 4{\cos }^2\mathrm{\theta }=3 \\ \Rightarrow {\cos }^2\mathrm{\theta }=\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$4$}\right. \Rightarrow \mathrm{\theta }=\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$6$}\right. \end{gathered}$$