Given data

$|\vec a|=1,\qquad \vec a\cdot \vec b=2,\qquad |\vec b|=4$

and

$\vec c=2(\vec a\times \vec b)-3\vec b$

We need the angle between $\vec b$ and $\vec c$.

Find $|\vec a\times \vec b|$

Using $|\vec a\times \vec b|^2=|\vec a|^2|\vec b|^2-(\vec a\cdot \vec b)^2$

we get $|\vec a\times \vec b|^2=(1)^2(4)^2-(2)^2=16-4=12$

Hence, $|\vec a\times \vec b|=\sqrt{12}=2\sqrt{3}$

Compute $\vec b\cdot \vec c$

$\vec b\cdot \vec c=\vec b\cdot\left(2(\vec a\times \vec b)-3\vec b\right)$

$=2\,\vec b\cdot(\vec a\times \vec b)-3\,\vec b\cdot\vec b$

Now $\vec a\times \vec b$ is perpendicular to $\vec b$, so $\vec b\cdot(\vec a\times \vec b)=0$

and $\vec b\cdot\vec b=|\vec b|^2=16$.

Therefore, $\vec b\cdot \vec c=-3\cdot 16=-48$

Compute $|\vec c|$

Since $(\vec a\times \vec b)$ is perpendicular to $\vec b$, the two terms in $\vec c$ are perpendicular:

$\vec c=2(\vec a\times \vec b)-3\vec b$

So, $|\vec c|^2=|2(\vec a\times \vec b)|^2+|-3\vec b|^2$

$=4|\vec a\times \vec b|^2+9|\vec b|^2$

$=4(12)+9(16)=48+144=192$

Hence, $|\vec c|=\sqrt{192}=8\sqrt{3}$

Use the formula for angle between vectors

If $\theta$ is the angle between $\vec b$ and $\vec c$, then

$\cos\theta=\frac{\vec b\cdot \vec c}{|\vec b|\,|\vec c|}$

Substituting,

$\cos\theta=\frac{-48}{4\cdot 8\sqrt{3}}=\frac{-48}{32\sqrt{3}}=-\frac{3}{2\sqrt{3}}=-\frac{\sqrt{3}}{2}$

Therefore,

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

Match with options

This is Option D.

Comparison with stored answer

Stored correct answer: D

Our derived answer: D

So they agree.