$\overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{b}}$are two vectors such that $\left|\overrightarrow{\mathrm{a}}\right|=1, \left|\overrightarrow{\mathrm{b}}\right|=4$ and $\overrightarrow{\mathrm{a}}\overrightarrow{\mathrm{b}}=2.$ If $\overrightarrow{\mathrm{c}}=(2\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}})-3\overrightarrow{\mathrm{b}}$ then find the angle between $\overrightarrow{\mathrm{b}}$ and  $\overrightarrow{\mathrm{a}}$

$$\begin{gathered} \left|\overline{\mathrm{a}}\right|=1 \left|\overline{\mathrm{b}}\right|=4 \overline{\mathrm{a}}·\overline{\mathrm{b}}=2 \\ 1\times 4\times \mathrm{cos\theta }=2 \\ \mathrm{cos\theta }=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right. \Rightarrow \mathrm{\theta }=\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$3$}\right. \\ \overline{\mathrm{c}}=2\overline{\mathrm{a}}\times \overline{\mathrm{b}} -3\overline{\mathrm{b}} \\ |\overline{\mathrm{c}}{|}^2=|2\overline{\mathrm{a}}\times \overline{\mathrm{b}}-3\overline{\mathrm{b}}{|}^2=4|\overline{\mathrm{a}}\times \overline{\mathrm{b}}{|}^2+9|\overline{ }{|}^2-0 \\ =4\times 1\times 16\times \frac{3}{4}+9\times 16=16(3+9)=16\times 12 \\ 3\overline{\mathrm{b}}+\overline{\mathrm{c}}=2(\overline{\mathrm{a}}\times \overline{\mathrm{b}}) \\ |3\overline{\mathrm{b}}+\overline{\mathrm{c}}{|}^2=4|\overline{\mathrm{a}}\times \overline{\mathrm{b}}{|}^2 \\ \Rightarrow 9\times 16+16\times 12+6\times 4\times 4\times \times 2\sqrt{3} \mathrm{cos\theta }=4\times 4\times 3 \\ 9+12+12\sqrt{3} \mathrm{cos\theta }=3 \\ \mathrm{cos\theta }=\frac{-18}{12\sqrt{3}}=\raisebox{1ex}{$-\sqrt{3}$}\!\left/ \!\raisebox{-1ex}{$2$}\right. \Rightarrow \mathrm{\theta }=\raisebox{1ex}{$5\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$6$}\right. \end{gathered}$$