Question

Let $ \vec{a}, \vec{b}, \vec{c} $ be unit vectors such that: - $ \vec{a} \perp \text{plane of } \vec{b}, \vec{c} $ - Angle between $ \vec{b} $ and $ \vec{c} $ is $ \frac{\pi}{3} $ Find: $$ |\vec{a} + \vec{b} + \vec{c}| $$

• A. 3
• B. 1
• C. 2
• D. 4

Hint:

Use dot product identities with geometric constraints (like perpendicularity or known angles) to expand squared magnitude expressions.

Answer 1

The Correct Option is C

Let’s square the magnitude: \[ |\vec{a} + \vec{b} + \vec{c}|^2 = (\vec{a} + \vec{b} + \vec{c}) \cdot (\vec{a} + \vec{b} + \vec{c}) \] Expand: \[ = \vec{a} \cdot \vec{a} + \vec{b} \cdot \vec{b} + \vec{c} \cdot \vec{c} + 2(\vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c} + \vec{b} \cdot \vec{c}) \] Since all are unit vectors: \[ \vec{a} \cdot \vec{a} = \vec{b} \cdot \vec{b} = \vec{c} \cdot \vec{c} = 1 \Rightarrow \text{Sum of squares} = 3 \] Given \( \vec{a} \perp \vec{b}, \vec{c} \Rightarrow \vec{a} \cdot \vec{b} = \vec{a} \cdot \vec{c} = 0 \) Also given angle between \( \vec{b} \) and \( \vec{c} \) is \( \frac{\pi}{3} \), so: \[ \vec{b} \cdot \vec{c} = \cos\left( \frac{\pi}{3} \right) = \frac{1}{2} \] So: \[ |\vec{a} + \vec{b} + \vec{c}|^2 = 3 + 2(0 + 0 + \frac{1}{2}) = 3 + 1 = 4 \Rightarrow |\vec{a} + \vec{b} + \vec{c}| = \boxed{2} \]