Question

Let $\overrightarrow{a},\overrightarrow{b}$ and $\overrightarrow{c}$ be three non-zero vectors such that no two of them are collinear and $(\overrightarrow{a}\times \overrightarrow{b})\times \overrightarrow{c}=\frac{1}{3}|\overrightarrow{b}||\overrightarrow{c}|\overrightarrow{a}.$ If $\theta$ is the angle between vectors $\overrightarrow{b}$ and $\overrightarrow{c}$ , then a value of sin $\theta$ is

• A. $\frac{2\sqrt{2}}{3}$
• B. $\frac{-\sqrt{2}}{3}$
• C. $\frac{2}{3}$
• D. $\frac{-2\sqrt{3}}{3}$

Answer 1

Answer: (A)

Given, $(\overrightarrow{a}\times \overrightarrow{b})\times \overrightarrow{c}=\frac{1}{3}|\overrightarrow{b}||\overrightarrow{c}|\overrightarrow{a}.$
$\Rightarrow -\overrightarrow{c}\times (\overrightarrow{a}\times \overrightarrow{b})=\frac{1}{3}|\overrightarrow{b}||\overrightarrow{c}|\overrightarrow{a}$
$\Rightarrow -(\overrightarrow{c}.\overrightarrow{b}).\overrightarrow{a}+(\overrightarrow{c}.\overrightarrow{a})\overrightarrow{b}=\frac{1}{3}|\overrightarrow{b}||\overrightarrow{c}|\overrightarrow{a}$
$\Rightarrow [\frac{1}{3}|\overrightarrow{b}||\overrightarrow{c}|+(\overrightarrow{c}.\overrightarrow{b})]a=(\overrightarrow{c}.\overrightarrow{a})\overrightarrow{b}$
Since, $a$ and $b$ are not collinear.
$\therefore \overrightarrow{c}.\overrightarrow{b}+\frac{1}{3}|\overrightarrow{b}||\overrightarrow{c}|=0$ and $\overrightarrow{c}.\overrightarrow{a}=0$
$\Rightarrow |\overrightarrow{b}||\overrightarrow{c}|\cos \theta +\frac{1}{3}|\overrightarrow{b}||\overrightarrow{c}|=0$
$\Rightarrow |\overrightarrow{b}||\overrightarrow{c}|(\cos \theta +\frac{1}{3})=0$
$\Rightarrow \cos \theta +\frac{1}{3}=0[\because |b|\ne 0,|c|\ne 0]$
$\Rightarrow \cos \theta =-\frac{1}{3}\Rightarrow \sin \theta =\frac{\sqrt{8}}{3}=\frac{2\sqrt{2}}{3}$