Q.Let $\vec a,\vec b,\vec c$ be unit vectors such that $\vec a\cdot\vec b=0=\vec a\cdot\vec c$ and the angle between $\vec b$ and $\vec c$ is $\dfrac\pi3$. Prove that $\vec a=\pm\dfrac{2}{\sqrt3}(\vec b\times\vec c)$.

Step 1. $\vec a\cdot\vec b=0$ and $\vec a\cdot\vec c=0$ mean $\vec a$ is perpendicular to both $\vec b$ and $\vec c$.

Step 2. The vector $\vec b\times\vec c$ is also perpendicular to both $\vec b$ and $\vec c$. Since (generically) there is only one line perpendicular to both $\vec b$ and $\vec c$, $\vec a$ must be a scalar multiple of $\vec b\times\vec c$: $\vec a=\lambda(\vec b\times\vec c)$.

Step 3. $|\vec b\times\vec c|=|\vec b||\vec c|\sin\dfrac\pi3=(1)(1)\dfrac{\sqrt3}2=\dfrac{\sqrt3}2$ (since $\vec b,\vec c$ are unit vectors). …

Argue $\vec a\parallel(\vec b\times\vec c)$ from the two perpendicularity conditions, then fix the scalar by …

• Forgetting the $\pm$ (both directions along $\vec b\times\vec c$ are perpendicular to both $\vec b,\vec c$). …