Let $\vec{a},\vec{b},$ and $\vec{c}$ be three unit vectors such that $\vec{a}\times (\vec{b}\times \vec{c})=\frac{1}{2}\vec{b}$. If the angle between $\vec{b}$ and $\vec{c}$ is $\theta$, then the value of ${\sin }^2\theta$ is equal to:

Solution

1. The vector triple product is $$\begin{array}{rl}\vec{a}\times (\vec{b}\times \vec{c})& =(\vec{a}\cdot \vec{c})\vec{b}-(\vec{a}\cdot \vec{b})\vec{c}\\ & =\frac{1}{2}\vec{b}\end{array}$$.
2. Since $\vec{b}$ and $\vec{c}$ are non-collinear, we must have $\vec{a}\cdot \vec{b}=0$ and $\vec{a}\cdot \vec{c}=1/2$.
3. Given $|\vec{a}|=1$ and $|\vec{c}|=1$, the condition $$\begin{array}{rl}\vec{a}\cdot \vec{c}& =\cos \varphi \\ & =1/2\end{array}$$ implies the angle between $\vec{a}$ and $\vec{c}$ is $60^{\circ }$.
4. Since $\vec{a}\cdot \vec{b}=0$, $\vec{a}$ is perpendicular to $\vec{b}$, and since $\vec{a}\cdot \vec{c}=1/2$, $\vec{a}$ makes $60^{\circ }$ with $\vec{c}$.
5. Using the geometry of unit vectors, $$\begin{array}{rl}{\sin }^2\theta & =1-(\vec{b}\cdot \vec{c}{)}^2\\ & =3/4\end{array}$$.
6. Hence the answer is (C).