Question

If $ \vec{a}, \vec{b}, \vec{c} $ are three unit vectors such that $ \vec{a} \times (\vec{b} \times \vec{c}) = \frac{\sqrt{3}}{2} \vec{b} + \frac{1}{2} \vec{c} $, and $ \alpha, \beta $ are the angles between $ \vec{a}, \vec{c} $ and $ \vec{a}, \vec{b} $ respectively, then $ \alpha + \beta = ? $

• A. \( \frac{\pi}{2} \)
• B. \( \frac{7\pi}{6} \)
• C. \( \frac{\pi}{6} \)
• D. \( \frac{5\pi}{6} \)

Hint:

Use vector triple product identity to equate coefficients and extract dot products.

Answer 1

The Correct Option is D

Using vector triple product identity: \[ \vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c} \] Comparing with: \[ \frac{\sqrt{3}}{2} \vec{b} + \frac{1}{2} \vec{c} \Rightarrow \vec{a} \cdot \vec{c} = \frac{\sqrt{3}}{2},\quad -(\vec{a} \cdot \vec{b}) = \frac{1}{2} \Rightarrow \vec{a} \cdot \vec{b} = -\frac{1}{2} \] Now: \[ \cos \alpha = \frac{\sqrt{3}}{2} \Rightarrow \alpha = \frac{\pi}{6},\quad \cos \beta = -\frac{1}{2} \Rightarrow \beta = \pi - \frac{\pi}{3} = \frac{2\pi}{3} \Rightarrow \alpha + \beta = \frac{\pi}{6} + \frac{2\pi}{3} = \frac{5\pi}{6} \]