Question

if$\overrightarrow{a},\overrightarrow{b}$ and$\overrightarrow{c}$ are three unit vector $\overrightarrow{a}$ is perpendicular to $\overrightarrow{b}$ and $\overrightarrow{c}$ and angle between $\overrightarrow{b}$ and $\overrightarrow{c}$b is${30}^{\circ },$ prove that $\overrightarrow{a}=\pm 2\left(\overrightarrow{b}\times \overrightarrow{c}\right).$

Answer 1

Given

$$\begin{array}{c}\left|\overline{a}\right|=1,\left|\overline{b}\right|=1,\left|\overline{c}\right|=1\\ \overline{a}is{1}^{2}to\overline{b},\overline{c},\\ \left(\overline{b}\overline{c}\right)={30}^0\\ Toprove\overline{a}=\pm 2(\overline{b}\times \overline{c})\\ Weknow\overline{b}\times \overline{c}is{1}^{2}to\overline{b},\overline{c}\\ \overline{a}and\overline{b}\times \overline{c}areparallol\\ \overline{a}=\pm (\overline{b}\times \overline{c})\to 1ttScales\\ \overline{a}=\left|t\right|(\overline{b}\times \overline{c})\\ 1=\left|t\right|\left|\overline{b}\right|\left|\overline{c}\right|in\left(\overline{a}\overline{c}\right)\\ 1={\left|t\right|}_{x}1\times 1\sin {30}^0\\ \left|t\right|=2to\pm 2in\to 1\\ \overline{a}=\pm 2(\overline{b}\times \overline{c})\end{array}$$