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

Given, \(|\vec{a}|=1,|\bar{b}|=1\) and \(|\bar{c}|=2\)
\(\begin{aligned}
& \text { Also, } \overline{\mathrm{a}} \times(\overline{\mathrm{a}} \times \overline{\mathrm{c}})+\overline{\mathrm{b}}=\overline{0} \\
& \Rightarrow(\overline{\mathrm{a}} \cdot \overline{\mathrm{c}}) \overline{\mathrm{a}}-(\overline{\mathrm{a}} \cdot \overline{\mathrm{a}}) \overline{\mathrm{c}}+\overline{\mathrm{b}}=\overline{0} \\
& \Rightarrow(\overline{\mathrm{a}} \cdot \overline{\mathrm{c}}) \overline{\mathrm{a}}-\overline{\mathrm{c}}+\overline{\mathrm{b}}=\overline{0} \quad \ldots\left[\because \overline{\mathrm{a}} \cdot \overline{\mathrm{a}}=|\overline{\mathrm{a}}|^2=1\right] \\
& \Rightarrow(\overline{\mathrm{a}} \cdot \overline{\mathrm{c}}) \overline{\mathrm{a}}-\overline{\mathrm{c}}=-\overline{\mathrm{b}}
\end{aligned}\)
\(\begin{aligned} & \Rightarrow|(\overline{\mathrm{a}} \cdot \overline{\mathrm{c}}) \overline{\mathrm{a}}-\overline{\mathrm{c}}|=|-\overline{\mathrm{b}}| \\ & \Rightarrow|(\overline{\mathrm{a}} \cdot \overline{\mathrm{c}}) \overline{\mathrm{a}}-\overline{\mathrm{c}}|^2=|\overline{\mathrm{b}}|^2 \\ & \Rightarrow|(\overline{\mathrm{a}} \cdot \overline{\mathrm{c}}) \overline{\mathrm{a}}|^2+|\overline{\mathrm{c}}|^2-2\{(\overline{\mathrm{a}} \cdot \overline{\mathrm{c}}) \overline{\mathrm{a}} \cdot \overline{\mathrm{c}}\}=|\overline{\mathrm{b}}|^2 \\ & \Rightarrow(\overline{\mathrm{a}} \cdot \overline{\mathrm{c}})^2|\overline{\mathrm{a}}|^2+|\overline{\mathrm{c}}|^2-2(\overline{\mathrm{a}} \cdot \overline{\mathrm{c}})(\overline{\mathrm{a}} \cdot \overline{\mathrm{c}})=|\overline{\mathrm{b}}|^2 \\ & \Rightarrow(\overline{\mathrm{a}} \cdot \overline{\mathrm{c}})^2\left\{|\overline{\mathrm{a}}|^2-2\right\}+|\overline{\mathrm{c}}|^2=|\overline{\mathrm{b}}|^2\end{aligned}\)
\(\begin{aligned}
& \Rightarrow-(\overline{\mathrm{a}} \cdot \overline{\mathrm{c}})^2+4=1 \quad \ldots\left[\because|\overline{\mathrm{~b}}|^2=1,|\overline{\mathrm{c}}|^2=4\right] \\
& \Rightarrow(\overline{\mathrm{a}} \cdot \overline{\mathrm{c}})^2=3 \\
& \Rightarrow \overline{\mathrm{a}} \cdot \overline{\mathrm{c}}= \pm \sqrt{3} \\
& \Rightarrow|\overline{\mathrm{a}}||\overline{\mathrm{c}}| \cos \theta=\sqrt{3}
\end{aligned}\)
where \(\theta\) is an acute angle between \(\overline{\mathrm{a}}\) and \(\overline{\mathrm{c}}\)
\(\Rightarrow \cos \theta=\frac{\sqrt{3}}{2} \Rightarrow \theta=\frac{\pi}{6}\)