Let \(\overline{\mathrm{a}}, \overline{\mathrm{b}}, \overline{\mathrm{c}}\) be three non-zero vectors, such that no two of them are collinear and \((\overline{\mathrm{a}} \times \overline{\mathrm{b}}) \times \overline{\mathrm{c}}=\frac{1}{3}|\overline{\mathrm{b}}||\overline{\mathrm{c}}| \overline{\mathrm{a}}\). If \(\theta\) is the angle between the vectors \(\bar{b}\) and \(\bar{c}\), then the value of \(\sin \theta\) is

Given: \((\overline{\mathrm{a}} \times \overline{\mathrm{b}}) \times \overline{\mathrm{c}}=\frac{1}{3}|\overline{\mathrm{b}}||\overline{\mathrm{c}}|^{-}\)
We know that,
\((\overline{\mathrm{a}} \times \overline{\mathrm{b}}) \times \overline{\mathrm{c}}=(\overline{\mathrm{a}} \cdot \overline{\mathrm{c}}) \overline{\mathrm{b}}-(\overline{\mathrm{b}} \cdot \overline{\mathrm{c}}) \overline{\mathrm{a}}^{-}\)
On comparing, we get
\(\begin{aligned}
& \frac{1}{3}|\overline{\mathrm{b}}||\overrightarrow{\mathrm{c}}|=-\overline{\mathrm{b}} \cdot \overline{\mathrm{c}} \\
\Rightarrow & \frac{1}{3}|\overline{\mathrm{b}}||\overrightarrow{\mathrm{c}}|=-|\overline{\mathrm{b}}||\overrightarrow{\mathrm{c}}| \cos \theta \\
\Rightarrow \cos \theta & =\frac{-1}{3} \\
\Rightarrow \cos ^2 \theta & =\frac{1}{9} \\
\sin ^2 \theta & =1-\cos ^2 \theta \\
& =1-\frac{1}{9} \\
\therefore \quad & \sin ^2 \theta=\frac{8}{9} \\
\therefore \quad & \sin \theta=\sqrt{\frac{8}{9}}=\frac{2 \sqrt{2}}{3}
\end{aligned}\)