angle between $\vec{b}$ and $\vec{c}$

Question

If $\vec{a}$ and $\vec{b}$ are two vectors such that $|\vec{a}|=1,|\vec{b}|=4,\vec{a}.\vec{b}=2$. If $\vec{c}=(2\vec{a}\times \vec{b})-3\vec{b},$ then angle between $\vec{b}$ and $\vec{c}$ is

solution i tried $|\vec{a}\times \vec{b}|^2=|\vec{a}|^2|\vec{b}|^2-(\vec{a}\cdot\vec{b})^2$ $|\vec{a}\times \vec{b}|^2=12$ from $|3\vec{b}+\vec{c}|^2=|2\vec{a}\times\vec{b}|^2==48$ how to find $\vec{c}$ magnitude in this question

Answer 1

Just correcting some typos. $$ |\vec a\times\vec b|^2=|\vec a|^2 |\vec b|^2-(\vec a\cdot\vec b)^2=12$$ Therefore $$\vec b\cdot \vec c=-3\vec b\cdot \vec b=-48$$ since $(2\vec a\times\vec b)\perp\vec b$, and $$ |\vec c|^2=4|\vec a\times\vec b|^2+9|\vec b|^2=48+144=192 \;.$$ Hence $$ \sphericalangle(\vec b,\vec c)=\arccos\left(\frac{\vec b\cdot\vec c}{|\vec b|\,|\vec c|}\right)=\arccos\left(\frac{-48}{4\sqrt{196}}\right)=\frac{\pi}{6}\;.$$

Answer 2

Outline: Note that $2 \vec a \times \vec b$ is perpendicular to $3 \vec b$ (why?), so $$\|c\|^2 = 4 \|\vec a \times \vec b\|^2 + 9 \|\vec b\|^2.$$ You now can find the magnitude of $c$, and then taking the dot product of $\vec b$ and $\vec c$ (which simplifies substantially due to the same fact about perpendicular vectors!) will get you almost to the answer.