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 the vectors $\overline{\mathrm{b}}$ and $\overline{\mathrm{c}}$, then value of $\sin \theta$ is
  1. $\frac{2}{3}$
  2. $\frac{-\sqrt{2}}{3}$
  3. $-\frac{1}{3}$
  4. $\frac{2 \sqrt{2}}{3}$

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 $\begin{aligned} & \frac{1}{3}||\overline{\mathrm{~b}}| \overline{\mathrm{c}}|=-\overline{\mathrm{b}} \cdot \overline{\mathrm{c}} \\ & \Rightarrow \frac{1}{3}|\overline{\mathrm{~b}}||\overline{\mathrm{c}}|=-|\overline{\mathrm{b}}||\overline{\mathrm{c}}| \cos \theta \\ & \Rightarrow \cos \theta=\frac{-1}{3} \\ & \Rightarrow \cos ^2 \theta=\frac{1}{9} \\ & \sin ^2 \theta=1-\cos ^2 \theta \\ & \quad=1-\frac{1}{9} \end{aligned}$ $\begin{array}{ll}\therefore & \sin ^2 \theta=\frac{8}{9} \\ \therefore & \sin \theta=\sqrt{\frac{8}{9}}=\frac{2 \sqrt{2}}{3}\end{array}$

Asked in: MHT CET 2024 (11 May Shift 1)

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