Vectors \(\bar{i}\) and \(\bar{b}\) are such that \(|\bar{a}|=1 ;|\bar{b}|=4\) and \(\bar{a} \cdot \bar{b}-2\). If \(\bar{c}-2 \bar{a} \times \bar{b}-3 \bar{b}\), then the angle between \(\bar{b}\) and \(\bar{c}\) is

\(\begin{aligned}
& \text { Given: }|\overline{\mathrm{a}}|=1,|\overline{\mathrm{b}}|=4 \text { and } \overline{\mathrm{a}} \cdot \overline{\mathrm{b}}=2 \text {, } \\
& \overline{\mathrm{c}}=2 \overline{\mathrm{a}} \times \overline{\mathrm{b}}-3 \overline{\mathrm{b}}
\end{aligned}\)
Now that, \(|\overline{\mathrm{a}} \times \overline{\mathrm{b}}|^2=|\overline{\mathrm{a}}|^2|\overline{\mathrm{b}}|^2-|\overline{\mathrm{a}} \cdot \overline{\mathrm{b}}|^2\)
\(\therefore \quad|\bar{a} \times \bar{b}|^2=16-4=12\)
Given that \(\overline{\mathrm{c}}=2 \overline{\mathrm{a}} \times \overline{\mathrm{b}}-3 \overline{\mathrm{b}}\)
\(\begin{aligned}
& |\bar{c}|^2=(2 \bar{a} \times \bar{b}-3 \bar{b})^2 \\
& |\bar{c}|^2=4|\bar{a} \times \bar{b}|^2+9|\bar{b}|^2
\end{aligned}\)
\(\begin{aligned}
& |\vec{c}|^2=4(12)+9(16) \\
& |\vec{c}|^2=192 \\
& |\vec{c}|=8 \sqrt{3}
\end{aligned}\)
Now, \(\overline{\mathrm{b}} \cdot \overline{\mathrm{c}}=\mathrm{b} \cdot(2 \overline{\mathrm{a}} \times \overline{\mathrm{b}}-3 \overline{\mathrm{b}})\)
\(\therefore \quad \overline{\mathrm{b}} \cdot \overline{\mathrm{c}}=-3|\overline{\mathrm{b}}|^2=-48\)
Angle between \(b\) and \(c\) is given by
\(\cos \theta=\frac{(\bar{b} \cdot \bar{c})}{(|\mathrm{b} \| \mathrm{c}|)}=\frac{-48}{(4 \times 8 \sqrt{3})}\)
\(\cos \theta=\frac{-\sqrt{3}}{2}\)
\(\theta=\cos ^4\left(\frac{-\sqrt{3}}{2}\right)\)
\(\theta=\frac{5 \pi}{6}\)