Question

If $ \mathbf{a} $ and $ \mathbf{b} $ are non-coplanar unit vectors such that $ \mathbf{a} \times (\mathbf{b} \times \mathbf{c}) = \frac{\mathbf{b}}{2} $ then the angle between $ \mathbf{a} $ and $ \mathbf{b} $ is:

• A. \( \frac{\pi}{4} \)
• B. \( \frac{\pi}{3} \)
• C. \( \frac{\pi}{2} \)
• D. \( \frac{\pi}{6} \)

Hint:

When solving vector equations involving triple products, use the vector triple product identity to simplify the expression and find relationships between the angles.

Answer 1

The Correct Option is A

We are given the vector equation: \[ \mathbf{a} \times (\mathbf{b} \times \mathbf{c}) = \frac{\mathbf{b}}{2} \] We will use the vector triple product identity to simplify this equation: \[ \mathbf{a} \times (\mathbf{b} \times \mathbf{c}) = (\mathbf{a} \cdot \mathbf{c}) \mathbf{b} - (\mathbf{a} \cdot \mathbf{b}) \mathbf{c} \] Thus, the equation becomes: \[ (\mathbf{a} \cdot \mathbf{c}) \mathbf{b} - (\mathbf{a} \cdot \mathbf{b}) \mathbf{c} = \frac{\mathbf{b}}{2} \] Now, since \( \mathbf{a} \) and \( \mathbf{b} \) are unit vectors, we focus on the coefficients of \( \mathbf{b} \) on both sides of the equation: \[ \mathbf{a} \cdot \mathbf{c} = \frac{1}{2} \quad \text{(coefficient of } \mathbf{b} \text{)} \] This implies that the angle between \( \mathbf{a} \) and \( \mathbf{c} \) is \( \cos^{-1} \left( \frac{1}{2} \right) = \frac{\pi}{3} \). Next, we need to find the angle between \( \mathbf{a} \) and \( \mathbf{b} \). From the equation, we can deduce that the angle between \( \mathbf{a} \) and \( \mathbf{b} \) is \( \frac{\pi}{4} \). 
Thus, the correct answer is \( \frac{\pi}{4} \).