If $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ are unit vectors equally inclined to each other at an angle $\alpha \left(\ne 0\right),$ then the angle between $\overrightarrow{a}\times \overrightarrow{b}$ and the plane containing $\overrightarrow{b}$ and $\overrightarrow{c}$ is

Let $\theta$  be the required angle, then

$\sin \theta =\cos \left({90}^{\theta }-\theta \right)=\left|\frac{\left(\overrightarrow{a}\times \overrightarrow{b}\right).\overrightarrow{b}\times \overrightarrow{c}}{\left|\overrightarrow{a}\times \overrightarrow{b}\right|.\left|\overrightarrow{b}\times \overrightarrow{c}\right|}\right|=\left|\frac{\left[\left(\overrightarrow{a}\times \overrightarrow{b}\right)\times \overrightarrow{b}\right]\overrightarrow{c}}{{\sin }^2\alpha }\right|=\left|\frac{\left[\left(\overrightarrow{a}.\overrightarrow{b}\right)\overrightarrow{b}-\left(\overrightarrow{b}.\overrightarrow{b}\right)\overrightarrow{a}\right]\overrightarrow{c}}{{\sin }^2\theta }\right|$

$=\left|\frac{\cos \alpha \overrightarrow{b.}\overrightarrow{c}-\overrightarrow{a}.\overrightarrow{c}}{{\sin }^2\alpha }\right|=\left|\frac{{\cos }^2\alpha -\cos \alpha }{{\sin }^2\alpha }\right|=\left|-\frac{\cos \alpha \left(1-\cos \alpha \right)}{\sin \alpha \sin \alpha }\right|=\left|-\frac{\cos \alpha 2{\sin }^2\frac{\alpha }{2}}{\sin \alpha 2\sin \frac{\alpha }{2}\cos \frac{\alpha }{2}}\right|$

$\begin{array}{l}=\left|-\cot \alpha \tan \frac{\alpha }{2}\right|=\left|\cot \alpha \tan \frac{\alpha }{2}\right|\\ \therefore \theta ={\sin }^{-1}\left(\tan \frac{\alpha }{2}\left|\cot \alpha \right|\right)\end{array}$