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 $(\overline{\mathrm{a}} \times \overline{\mathrm{b}}) \times \overline{\mathrm{c}}=\frac{1}{3}|\overline{\mathrm{~b}}||\overline{\mathrm{c}}| \overline{\mathrm{a}}$. If $\theta$ is the angle between vectors $\overline{\mathrm{b}}$ and $\overline{\mathrm{c}}$, then the value of $\operatorname{cosec} \theta$ is
  1. $\frac{3 \sqrt{3}}{2}$
  2. $\frac{2 \sqrt{2}}{3}$
  3. $\frac{2}{\sqrt{3}}$
  4. $\frac{3}{2 \sqrt{2}}$

Solution

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

Asked in: MHT CET 2024 (10 May Shift 1)

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