Question

Let \( \vec{p} \) and \( \vec{q} \) be two unit vectors and \( \alpha \) be the angle between them. Then \( (\vec{p} + \vec{q}) \) will be a unit vector for what value of \( \alpha \)?

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

Answer 1

The Correct Option is D

Given that \( \vec{p} \) and \( \vec{q} \) are unit vectors and the angle between them is \( \alpha \), we are told that \( \vec{p} + \vec{q} \) is also a unit vector.

We use the identity: \[ |\vec{p} + \vec{q}| = \sqrt{|\vec{p}|^2 + |\vec{q}|^2 + 2|\vec{p}||\vec{q}|\cos\alpha} \] Since \( \vec{p} \) and \( \vec{q} \) are unit vectors: \[ |\vec{p}| = |\vec{q}| = 1 \] So the expression becomes: \[ |\vec{p} + \vec{q}| = \sqrt{1^2 + 1^2 + 2 \cdot 1 \cdot 1 \cdot \cos\alpha} = \sqrt{2 + 2\cos\alpha} \] Since \( \vec{p} + \vec{q} \) is a unit vector, we set: \[ \sqrt{2 + 2\cos\alpha} = 1 \] Squaring both sides: \[ 2 + 2\cos\alpha = 1 \Rightarrow \cos\alpha = \frac{-1}{2} \Rightarrow \alpha = \frac{2\pi}{3} \]

Answer:

Option (D) \( \frac{2\pi}{3} \)