Question

Let $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ be unit vectors such that $\overrightarrow{a}.\overrightarrow{b}=\overrightarrow{a}.\overrightarrow{c}=0$ & the angle between $\overrightarrow{b}$ & $\overrightarrow{c}$ is $\frac{\pi }{6}$. Prove that $\overrightarrow{a}=\pm 2(\overrightarrow{b}\times \overrightarrow{c})$.

Answer 1

Given that
$|\overrightarrow{a}|=1,|\overrightarrow{b}|=1$ and $|\overrightarrow{c}|=1$
$\because \overrightarrow{a}.\overrightarrow{b}=0$
$\Rightarrow$ vector $\overrightarrow{a}$ is perpendicular to vector $\overrightarrow{b}$ and $\overrightarrow{a}.\overrightarrow{c}=0$
$\Rightarrow$ vector $\overrightarrow{a}$ is perpendicular to vector $\overrightarrow{b}$
$\because$ vector $\overrightarrow{a}$ is perpendicular to both vector $\overrightarrow{a}$ and $\overrightarrow{b}$.
$\therefore \overrightarrow{a}$ will be parallel to perpendicular vector $b\times c$.
$\therefore$ Let $\overrightarrow{a}=\lambda (\overrightarrow{b}\times \overrightarrow{c})$ ....(1)
where $\lambda$ is any scalar
But $\overrightarrow{b}\times \overrightarrow{c}=|\overrightarrow{b}||\overrightarrow{c}|\sin \frac{\pi }{6}\hat{n}$
$\Rightarrow \overrightarrow{b}\times \overrightarrow{c}=\frac{1}{2}\hat{n}$ $[\because |\overrightarrow{b}|=|\overrightarrow{c}|=1]$
$\therefore$ From equation $\left(1\right)$,
$\overrightarrow{a}=\frac{\lambda }{2}\hat{n}$
$\Rightarrow (\overrightarrow{a}{)}^2={\left(\frac{\lambda }{2}\hat{n}\right)}^2$
$\Rightarrow |\overrightarrow{a}{|}^2=\frac{{\lambda }^2}{4}|\hat{n}{|}^2$
$\Rightarrow 1=\frac{{\lambda }^2}{4}.1$
$\Rightarrow {\lambda }^2=4$
$\therefore \lambda =\pm 2$
$\therefore$ From equation $\left(1\right)$
$\therefore \overrightarrow{a}=\pm 2(\overrightarrow{b}\times \overrightarrow{c})$