Let $\vec{a}$, $\vec{b}$, and $\vec{c}$ be three unit vectors such that $\vec{a}\cdot \vec{b}=0$ and $\vec{a}\cdot \vec{c}=0$. If the angle between $\vec{b}$ and $\vec{c}$ is $30^{\circ }$, then find the value of $|\vec{a}\times (\vec{b}\times \vec{c})|$.

Solution

1. Given $$\begin{array}{rl}|\vec{a}|& =|\vec{b}|\\ & =|\vec{c}|\\ & =1\end{array}$$, $\vec{a}\perp \vec{b}$, and $\vec{a}\perp \vec{c}$.
2. The vector $\vec{v}=\vec{b}\times \vec{c}$ has magnitude $$\begin{array}{rl}|\vec{v}|& =|\vec{b}||\vec{c}|\sin (30^{\circ })\\ & =1\cdot 1\cdot (1/2)\\ & =1/2\end{array}$$.
3. Since $\vec{b}\perp \vec{a}$ and $\vec{c}\perp \vec{a}$, the vector $\vec{b}\times \vec{c}$ is parallel to $\vec{a}$.
4. Thus, $$\begin{array}{rl}|\vec{a}\times (\vec{b}\times \vec{c})|& =|\vec{a}||\vec{b}\times \vec{c}|\sin (0^{\circ })\\ & =0\end{array}$$ is incorrect; rather, use the identity $$\begin{array}{rl}|\vec{a}\times \vec{v}|& =|\vec{a}||\vec{v}|\sin (90^{\circ })\\ & =1\cdot (1/2)\cdot 1\\ & =1/2\end{array}$$.
5. Hence the answer is (A).