Question

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

• A. $\frac{-1}{3}$
• B. $\frac{-2\sqrt{2}}{3}$
• C. $\frac{2}{3}$
• D. $\frac{2\sqrt{2}}{3}$

Answer 1

The correct option is D $\frac{2\sqrt{2}}{3}$

We have $\frac{1}{3}|\overrightarrow{b}||\overrightarrow{c}|\overrightarrow{a}=(\overrightarrow{a}\times \overrightarrow{b})\times \overrightarrow{c}$
We know that $(\overrightarrow{a}\times \overrightarrow{b})\times \overrightarrow{c}=(\overrightarrow{a}\cdot \overrightarrow{c})\overrightarrow{b}-(\overrightarrow{b}\cdot \overrightarrow{c})\overrightarrow{a}$ $\therefore \frac{1}{3}|\overrightarrow{b}||\overrightarrow{c}|\overrightarrow{a}=(\overrightarrow{a}\cdot \overrightarrow{c})\overrightarrow{b}-(\overrightarrow{b}\cdot \overrightarrow{c})\overrightarrow{a}$
on comparing both sides, we get
$\Rightarrow \overrightarrow{b}\cdot \overrightarrow{c}=-\frac{1}{3}|\overrightarrow{b}||\overrightarrow{c}|$
$\therefore \cos \theta =-\frac{1}{3}\Rightarrow \sin \theta =\frac{2\sqrt{2}}{3}$