Q139 / 225 · JEE Main 2004 · Mathematics
Let $\overline{\mathrm{a}}, \overline{\mathrm{b}}$ and $\overline{\mathrm{c}}$ be non-zero vectors such that $(\overline{\mathrm{a}} \times \overline{\mathrm{b}}) \times \overline{\mathrm{c}}=\frac{1}{3}|\overline{\mathrm{b}}||\overline{\mathrm{c}}| \overline{\mathrm{a}}$. If $\theta$ is the acute angle between the vectors $\overline{\mathrm{b}}$ and $\overline{\mathrm{c}}$, then $\sin \theta$ equals
- A$\frac{1}{3}$
- B$\frac{\sqrt{2}}{3}$
- C$\frac{2}{3}$
- Correct answerD$\frac{2 \sqrt{2}}{3}$
Solution
$(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}) \times \overrightarrow{\mathrm{c}}=\frac{1}{3}|\overrightarrow{\mathrm{b}}||\overline{\mathrm{c}}| \overline{\mathrm{a}} \Rightarrow(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}) \overrightarrow{\mathrm{b}}-(\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}) \overrightarrow{\mathrm{a}}=\frac{1}{3}|\overline{\mathrm{b}}||\overline{\mathrm{c}}| \overline{\mathrm{a}}$
$\Rightarrow(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}) \overrightarrow{\mathrm{b}}=\left(\frac{1}{3}|\overline{\mathrm{b}}||\overline{\mathrm{c}}|+(\overline{\mathrm{b}} \cdot \overline{\mathrm{c}})\right) \overrightarrow{\mathrm{a}} \Rightarrow \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}=0$ and $\frac{1}{3}|\overline{\mathrm{b}}||\overline{\mathrm{c}}|+(\overline{\mathrm{b}} \cdot \overline{\mathrm{c}})=0$
$\Rightarrow|\overline{\mathrm{b}}||\overline{\mathrm{c}}|\left(\frac{1}{3}+\cos \theta\right)=0 \Rightarrow \cos \theta=-1 / 3 \Rightarrow \sin \theta=\frac{2 \sqrt{2}}{3}$
Hence, the correct answer is option: 'D'.