If $\vec a,\vec b,\vec c$ are unit vectors such that $\vec a ·\vec b =\vec a ·\vec c = 0$, and the angle between $\vec b$ and $\vec c$ is $\frac{\pi}{6}$, then

$\vec a = ±2(\vec b× \vec c)$

The correct answer is Option (3) → $\vec a = ±2(\vec b× \vec c)$

Given: $\vec{a},\vec{b},\vec{c}$ are unit vectors such that $\vec{a}\cdot\vec{b}=0$ and $\vec{a}\cdot\vec{c}=0$.

Hence $\vec{a}$ is perpendicular to both $\vec{b}$ and $\vec{c}$, so $\vec{a}\parallel (\vec{b}\times\vec{c})$.

Let $\theta$ be the angle between $\vec{b}$ and $\vec{c}$.

$|\vec{b}\times\vec{c}|=|\vec{b}||\vec{c}|\sin\theta=\sin\theta$ (since both are unit vectors).

Given $\theta=\frac{\pi}{6}$, so $|\vec{b}\times\vec{c}|=\frac{1}{2}$.

Since $\vec{a}$ is a unit vector and parallel to $(\vec{b}\times\vec{c})$:

$\vec{a}=\pm\frac{\vec{b}\times\vec{c}}{|\vec{b}\times\vec{c}|}=\pm\frac{\vec{b}\times\vec{c}}{1/2}=\pm2(\vec{b}\times\vec{c})$

$\vec{a}=\pm2(\vec{b}\times\vec{c})$