Q60 / 150 · MHT CET 2024 (11 May Shift 2) · Mathematics
Let $\bar{a}, \bar{b}$ and $\bar{c}$ be three unit vectors such that $\bar{a} \times(\bar{b} \times \bar{c})=\frac{\sqrt{3}}{2}(\bar{b}+\bar{c})$.
If $\bar{b}$ is not parallel to $\bar{c}$, then the angle between $\bar{a}$ and $\bar{b}$ is
- A$\frac{3 \pi}{4}$
- B$\frac{\pi}{2}$
- C$\frac{2 \pi}{3}$
- Correct answerD$\frac{5 \pi}{6}$
Solution
$\begin{aligned} & \overline{\mathrm{a}} \times(\overline{\mathrm{b}} \times \overline{\mathrm{c}})=\frac{\sqrt{3}}{2}(\overline{\mathrm{~b}}+\overline{\mathrm{c}}) \\ & \Rightarrow(\overline{\mathrm{a}} \cdot \overline{\mathrm{c}}) \overline{\mathrm{b}}-(\overline{\mathrm{a}} \cdot \overline{\mathrm{b}}) \overline{\mathrm{c}}=\left(\frac{\sqrt{3}}{2}\right) \overline{\mathrm{b}}+\left(\frac{\sqrt{3}}{2}\right) \overline{\mathrm{c}} \\ & \Rightarrow \overline{\mathrm{a}} \cdot \overline{\mathrm{c}}=\frac{\sqrt{3}}{2} \text { and } \overline{\mathrm{a}} \cdot \overline{\mathrm{b}}=\frac{-\sqrt{3}}{2} \\ & \Rightarrow|\overline{\mathrm{a}}||\overline{\mathrm{b}}| \cos \theta=\frac{-\sqrt{3}}{2} \\ & \Rightarrow \cos \theta=\frac{-\sqrt{3}}{2}=\cos \frac{5 \pi}{6} \\ & \Rightarrow \theta=\frac{5 \pi}{6}\end{aligned}$
Hence, the correct answer is option: 'D'.