Let $$\overrightarrow{\mathrm{a}}=\mathrm{a}_1 \hat{i}+\mathrm{a}_2 \hat{j}+\mathrm{a}_3 \hat{k}$$ and $$\overrightarrow{\mathrm{b}}=\mathrm{b}_1 \hat{i}+\mathrm{b}_2 \hat{j}+\mathrm{b}_3 \hat{k}$$ be two vectors such that $|\overrightarrow{\mathrm{a}}|=1, \vec{a} \cdot \vec{b}=2$ and $|\vec{b}|=4$. If $\vec{c}=2(\vec{a} \times \vec{b})-3 \vec{b}$, then the angle between $\vec{b}$ and $\vec{c}$ is equal to:

<p>Given $|\vec{a}|=1,|\vec{b}|=4, \vec{a} \cdot \vec{b}=2$</p> <p>$\vec{\mathrm{c}}=2(\vec{\mathrm{a}} \times \vec{\mathrm{b}})-3 \vec{\mathrm{b}}$</p> <p>Dot product with $\overrightarrow{\mathrm{a}}$ on both sides</p> <p>$\overrightarrow{\mathrm{c}} . \overrightarrow{\mathrm{a}}=-6$ ..... (1)</p> <p>Dot product with $\vec{b}$ on both sides</p> <p>$$\begin{aligned} & \overrightarrow{\mathrm{b}} \overrightarrow{\mathrm{c}}=-48 \quad \text{... (2)}\\ & \overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{c}}=4|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|^2+9|\overrightarrow{\mathrm{b}}|^2 \\ & |\overrightarrow{\mathrm{c}}|^2=4\left[|\mathrm{a}|^2|\mathrm{~b}|^2-(\mathrm{a} \cdot \overrightarrow{\mathrm{b}})^2\right]+9|\overrightarrow{\mathrm{b}}|^2 \\ & |\overrightarrow{\mathrm{c}}|^2=4\left[(1)(4)^2-(4)\right]+9(16) \\ & |\overrightarrow{\mathrm{c}}|^2=4[12]+144 \\ & |\overrightarrow{\mathrm{c}}|^2=48+144 \\ & |\overrightarrow{\mathrm{c}}|^2=192 \\ & \therefore \cos \theta=\frac{\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}}{|\overrightarrow{\mathrm{b}}| \overrightarrow{\mathrm{c}} \mid} \\ & \therefore \cos \theta=\frac{-48}{\sqrt{192} \cdot 4} \\ & \therefore \cos \theta=\frac{-48}{8 \sqrt{3} .4} \\ & \therefore \cos \theta=\frac{-3}{2 \sqrt{3}} \\ & \therefore \cos \theta=\frac{-\sqrt{3}}{2} \Rightarrow \theta=\cos ^{-1}\left(\frac{-\sqrt{3}}{2}\right) \end{aligned}$$</p>