Let $\overline{\mathrm{a}}, \overline{\mathrm{b}}$ and $\overline{\mathrm{c}}$ be three non-zero vectors…

Let $\overline{\mathrm{a}}, \overline{\mathrm{b}}$ and $\overline{\mathrm{c}}$ be three non-zero vectors such that no two of them are collinear and $\left.(\overline{\mathrm{a}} \times \overline{\mathrm{b}}) \times \overline{\mathrm{c}}=\frac{1}{3}|\overline{\mathrm{~b}}| \overline{\mathrm{c}} \right\rvert\, \overline{\mathrm{a}}$. If $\theta$ is the angle between vectors $\bar{b}$ and $\bar{c}$, then the value of $\sin \theta$ is
  1. $\frac{2}{3}$
  2. $\frac{-2 \sqrt{2}}{3}$
  3. $\frac{2 \sqrt{2}}{3}$
  4. $\frac{-\sqrt{2}}{3}$

Solution

Given, $(\overline{\mathrm{a}} \times \overline{\mathrm{b}}) \times \overline{\mathrm{c}}=\frac{1}{3}|\overline{\mathrm{~b}}||\overline{\mathrm{c}}| \overline{\mathrm{a}}$ We know that, $(\overline{\mathrm{a}} \times \overline{\mathrm{b}}) \times \overline{\mathrm{c}}=(\overline{\mathrm{a}} \cdot \overline{\mathrm{c}}) \overline{\mathrm{b}}-(\overline{\mathrm{b}} \cdot \overline{\mathrm{c}}) \overline{\mathrm{a}}$
On comparing, we get $\begin{aligned} & \frac{1}{3}|\overline{\mathrm{~b}}||\overline{\mathrm{c}}|=-\overline{\mathrm{b}} \cdot \overline{\mathrm{c}} \\ & \Rightarrow \frac{1}{3}|\overline{\mathrm{~b}}||\overline{\mathrm{c}}|=-|\overline{\mathrm{b}}||\overrightarrow{\mathrm{c}}| \cos \theta \\ & \Rightarrow \cos \theta=\frac{-1}{3} \\ & \Rightarrow \cos ^2 \theta=\frac{1}{9} \end{aligned}$ $\begin{aligned} \sin ^2 \theta & =1-\cos ^2 \theta \\ & =1-\frac{1}{9} \\ \therefore \quad \sin ^2 \theta & =\frac{8}{9} \\ \therefore \quad \sin \theta & =\sqrt{\frac{8}{9}}=\frac{2 \sqrt{2}}{3}\end{aligned}$

Asked in: MHT CET 2024 (16 May Shift 1)

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