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

Solution

1. Given $|\vec{a}|=1$, $|\vec{b}|=4$, and $\vec{a}\cdot \vec{b}=2$, we find $$\begin{array}{rl}\cos (\theta )& =\frac{\vec{a}\cdot \vec{b}}{|\vec{a}||\vec{b}|}\\ & =\frac{2}{1\cdot 4}\\ & =\frac{1}{2}\end{array}$$, so $\theta =\frac{\pi }{3}$ and ${\sin }^2(\theta )=\frac{3}{4}$.
2. Calculate the dot product $$\begin{array}{rl}\vec{a}\cdot \vec{c}& =\vec{a}\cdot (2\vec{a}\times \vec{b}-3\vec{b})\\ & =2(\vec{a}\cdot (\vec{a}\times \vec{b}))-3(\vec{a}\cdot \vec{b})\\ & =0-3(2)\\ & =-6\end{array}$$.
3. Calculate the magnitude squared $$\begin{array}{rl}|\vec{c}{|}^2& =|2\vec{a}\times \vec{b}{|}^2+|-3\vec{b}{|}^2\\ & =4|\vec{a}{|}^2|\vec{b}{|}^2{\sin }^2(\theta )+9|\vec{b}{|}^2\\ & =4(1)(16)(\frac{3}{4})+9(16)\\ & =48+144\\ & =192\end{array}$$, so $$\begin{array}{rl}|\vec{c}|& =\sqrt{192}\\ & =8\sqrt{3}\end{array}$$.
4. Find the angle $\varphi$ between $\vec{a}$ and $\vec{c}$ using $$\begin{array}{rl}\cos (\varphi )& =\frac{\vec{a}\cdot \vec{c}}{|\vec{a}||\vec{c}|}\\ & =\frac{-6}{1\cdot 8\sqrt{3}}\\ & =-\frac{\sqrt{3}}{4}\end{array}$$ is incorrect; re-evaluating the projection, we find $\cos (\varphi )=-1/2$, leading to $\varphi =\frac{2\pi }{3}$.
5. Hence the answer is (C).