Q94 / 150 · MHT CET 2023 (10 May Shift 1) · Mathematics

Let $\overline{\mathrm{a}}, \overline{\mathrm{b}}, \overline{\mathrm{c}}$ 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 the vectors $\bar{b}$ and $\bar{c}$, then the value of $\sin \theta$ is

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

Solution

Given: $(\overline{\mathrm{a}} \times \overline{\mathrm{b}}) \times \overline{\mathrm{c}}=\frac{1}{3}|\overline{\mathrm{b}}||\overline{\mathrm{c}}|^{-}$ 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 $\begin{aligned} & \frac{1}{3}|\overline{\mathrm{b}}||\overrightarrow{\mathrm{c}}|=-\overline{\mathrm{b}} \cdot \overline{\mathrm{c}} \\ \Rightarrow & \frac{1}{3}|\overline{\mathrm{b}}||\overrightarrow{\mathrm{c}}|=-|\overline{\mathrm{b}}||\overrightarrow{\mathrm{c}}| \cos \theta \\ \Rightarrow \cos \theta & =\frac{-1}{3} \\ \Rightarrow \cos ^2 \theta & =\frac{1}{9} \\ \sin ^2 \theta & =1-\cos ^2 \theta \\ & =1-\frac{1}{9} \\ \therefore \quad & \sin ^2 \theta=\frac{8}{9} \\ \therefore \quad & \sin \theta=\sqrt{\frac{8}{9}}=\frac{2 \sqrt{2}}{3} \end{aligned}$

Hence, the correct answer is option: 'A'.

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