Let a → = a i i ^ + a 2 j ^ + a 3 k ^ and b → = b 1 i ^ + b 2 j ^ + b 3 k ^ be two vectors such that | a → |…

Given $\left|\overrightarrow{a}\right|=1,\left|\overrightarrow{b}\right|=4,\overrightarrow{a}·\overrightarrow{b}=2$

Also, $\overrightarrow{c}=2(\overrightarrow{a}\times \overrightarrow{b})-3\overrightarrow{b} ...\left(i\right)$

Applying dot product with $\overrightarrow{a}$ on both sides of equation $\left(i\right)$

$\Rightarrow \overrightarrow{c}·\overrightarrow{a}=-6 ...\left(ii\right)$

$\left\{\text{as} (\overrightarrow{a}\times \overrightarrow{b})·\overrightarrow{a}=(\overrightarrow{a}\times \overrightarrow{b})·\overrightarrow{b}=0\right\}$

Again, applying dot product with $\overrightarrow{b}$ on both sides of equation $\left(i\right)$

$\Rightarrow \overrightarrow{b}·\overrightarrow{c}=-48 ...\left(iii\right)$

Now, using equation $\left(i\right)$

$\Rightarrow {\left|\overrightarrow{c}\right|}^2=4|\overrightarrow{a}\times \overrightarrow{b}{|}^2+9|\overrightarrow{b}{|}^2-12\left(\overrightarrow{a}\times \overrightarrow{b}\right)·\overrightarrow{b}$

We know that, ${\left(\overrightarrow{a}\times \overrightarrow{b}\right)}^2+{\left(\overrightarrow{a}.\overrightarrow{b}\right)}^2={\left|\overrightarrow{a}\right|}^2{\left|\overrightarrow{b}\right|}^2$

$\Rightarrow |\overrightarrow{c}{|}^2=4\left[\left|\overrightarrow{a}{|}^2\right| \overrightarrow{b}{|}^2-(\overrightarrow{a}·\overrightarrow{b}{)}^2\right]+9|\overrightarrow{b}{|}^2$

$\Rightarrow |\overrightarrow{c}{|}^2=4\left[\left(1\right)(4{)}^2-(4)\right]+9(16)$

$\Rightarrow |\overrightarrow{c}{|}^2=4\times 12+144$

$\Rightarrow |\overrightarrow{c}{|}^2=48+144$

$\Rightarrow |\overrightarrow{c}{|}^2=192$

Now, $\cos \theta =\frac{\overrightarrow{b}·\overrightarrow{c}}{\left|\overrightarrow{b}\right|\left|\overrightarrow{c}\right|}$

$\Rightarrow \cos \theta =\frac{-48}{\sqrt{192}\times 4}$

$\Rightarrow \cos \theta =\frac{-48}{8\sqrt{3}.4}$

$\Rightarrow \cos \theta =\frac{-3}{2\sqrt{3}}$

$\Rightarrow \cos \theta =\frac{-\sqrt{3}}{2}$

$\Rightarrow \theta ={\cos }^{-1}\left(\frac{-\sqrt{3}}{2}\right)$