Q93 / 155 · WBJEE 2019 · Mathematics
Let $\hat{\alpha}, \hat{\beta}, \hat{\gamma}$ be three unit vectors such that $\hat{\alpha} \times (\hat{\beta} \times \hat{\gamma})=\frac{1}{2}(\hat{\beta}+\hat{\gamma})$ where $\hat{\alpha} \times(\hat{\beta} \times \hat{\gamma})=(\hat{\alpha} \cdot \hat{\gamma}) \hat{\beta}-(\hat{\alpha} \cdot \hat{\beta}) \gamma \cdot$ If $\hat{\beta}$ is not parallel to $\hat{\gamma},$ then the angle between $\hat{\alpha}$ and $\hat{\beta}$ is
- A$\frac{5 \pi}{6}$
- B$\frac{\pi}{6}$
- C$\frac{\pi}{3}$
- Correct answerD$\frac{2 \pi}{3}$
Solution
Given,
$|\hat{\alpha}|=|\hat{\beta}|=|\hat{\gamma}|=1$
and $\quad \hat{\alpha} \times(\hat{\beta} \times \hat{\gamma})=\frac{1}{2}(\hat{\beta}+\hat{\gamma})$
$\left.\operatorname{Now}, \hat{\alpha} \times(\hat{\beta} \times \hat{\gamma})=\frac{1}{2} \hat{(\hat{\beta}}+\hat{\gamma}\right)$
$\Rightarrow \quad(\hat{\alpha} \cdot \hat{\gamma}) \hat{\beta}-(\hat{\alpha} \cdot \hat{\beta}) \hat{\gamma}=\frac{1}{2} \hat{\beta}+\frac{1}{2} \hat{\gamma}$
On comparing we get, $-(\hat{\alpha} \cdot \hat{\beta})=\frac{1}{2}$
$\Rightarrow|\hat{\alpha}||\hat{\beta}| \cos \theta=-\frac{1}{2}$
$\Rightarrow \quad \cos \theta=-\frac{1}{2} \quad[\because|\hat{\alpha}|=|\hat{\beta}|=1]$
$\Rightarrow \quad \theta=120^{\circ}$ or $\frac{2 \pi}{3}$
Hence, the correct answer is option: 'D'.