If $\vec{a}, \vec{b}$ and $\vec{c}$ are three non-parallel unit vectors such that $\vec{a} \times (\vec{b} \times \vec{c}) = \frac{1}{2}\vec{b}$, the angle between $\vec{b}$ and $\vec{c}$ is:
Step-by-Step Solution
Key Concept: The vector triple product formula combined with coefficient comparison for non-parallel vectors yields the dot products with $\vec{a}$, which constrains the angle between $\vec{b}$ and $\vec{c}$.
Using the vector triple product formula: $\vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}$. Given this equals $\frac{1}{2}\vec{b}$, we have $(\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c} = \frac{1}{2}\vec{b}$. Since $\vec{b}$ and $\vec{c}$ are non-parallel, comparing coefficients: $(\vec{a} \cdot \vec{c}) = \frac{1}{2}$ and $(\vec{a} \cdot \vec{b}) = 0$. From $\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos(\theta_{ab}) = 0$, we get $\vec{a} \perp \vec{b}$. From $\vec{a} \cdot \vec{c} = |\vec{a}||\vec{c}|\cos(\theta_{ac}) = \cos(\theta_{ac}) = \frac{1}{2}$, and using the constraint that $\vec{b}, \vec{c}$ lie in a plane perpendicular to $\vec{a}$, the angle between $\vec{b}$ and $\vec{c}$ is $\pi/3$.
Correct Answer: 2