Let $\overrightarrow{a},\overrightarrow{b}$ and $\stackrel{\rightharpoonup }{c}$ unit vectors such that $\overrightarrow{a}\cdot \overrightarrow{b}=\overrightarrow{a}\cdot \overrightarrow{c}=0$ and angle between $\stackrel{\rightharpoonup }{b}$ and $\stackrel{\rightharpoonup }{c}$ is $\frac{\pi }{6}$ then $\overrightarrow{a}=$

Given $\overline{a}\cdot \overline{b}=\overline{a}\cdot \overline{c}=0$, angle between $\overline{b}$ and $\overline{c}$ is $\frac{\pi }{6}$

$\begin{array}{l}\overline{b}\times \overline{c}=\left|\overline{b}\right|\left|\overline{c}\right|\sin \frac{\pi }{6}\widehat{n}\\ =1.1\cdot \frac{1}{2}\widehat{n}\end{array}$

$\Rightarrow \widehat{n}=\pm 2(\overline{b}\times \overline{c})$

Since $\overrightarrow{a}$ is $\perp$to $\overline{b}$and $\overline{c}$ and direction along to $\overline{b,}$ so

$\overline{a}=\pm 2(\overline{b}\times \overline{c})$