Question

Let $\vec{a}$ and $\vec{b}$ be two unit vectors and $\theta$ is the angle between them. Then $\vec{a} + \vec{b}$ is a unit vector if:

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

Answer 1

The Correct Option is C

1. Understand the condition:

Given two unit vectors \( \bar{a} \) and \( \bar{b} \) with angle \( \theta \) between them, we need to find when \( \bar{a} + \bar{b} \) is also a unit vector.

2. Compute the magnitude of \( \bar{a} + \bar{b} \):

\[ |\bar{a} + \bar{b}|^2 = |\bar{a}|^2 + |\bar{b}|^2 + 2 \bar{a} \cdot \bar{b} = 1 + 1 + 2 \cos \theta = 2 + 2 \cos \theta \]

For \( \bar{a} + \bar{b} \) to be a unit vector:

\[ |\bar{a} + \bar{b}| = 1 \implies 2 + 2 \cos \theta = 1 \implies \cos \theta = -\frac{1}{2} \]

3. Solve for \( \theta \):

\[ \cos \theta = -\frac{1}{2} \implies \theta = \frac{2\pi}{3} \text{ (or } 120^\circ) \]

Correct Answer: (C) \( \theta = \frac{2\pi}{3} \)

Answer 2

For $ \mathbf{a} + \mathbf{b} $ to be a unit vector, its magnitude must be 1: $ \|\mathbf{a} + \mathbf{b}\| = 1 $.

We can use the formula for the magnitude of the sum of two vectors:

$$ \|\mathbf{a} + \mathbf{b}\|^2 = \|\mathbf{a}\|^2 + \|\mathbf{b}\|^2 + 2 (\mathbf{a} \cdot \mathbf{b}) $$

Since $ \mathbf{a} $ and $ \mathbf{b} $ are unit vectors, $ \|\mathbf{a}\| = \|\mathbf{b}\| = 1 $. Also, the dot product can be expressed in terms of the angle $ \theta $ between them:

$$ \mathbf{a} \cdot \mathbf{b} = \|\mathbf{a}\| \|\mathbf{b}\| \cos \theta = \cos \theta $$

Substituting into the magnitude equation:

$$ \|\mathbf{a} + \mathbf{b}\|^2 = 1^2 + 1^2 + 2 \cos \theta = 2 + 2 \cos \theta $$

If $ \|\mathbf{a} + \mathbf{b}\| = 1 $, then:

$$ \|\mathbf{a} + \mathbf{b}\|^2 = 1 $$ $$ 2 + 2 \cos \theta = 1 $$ $$ 2 \cos \theta = -1 $$ $$ \cos \theta = -\frac{1}{2} $$

The angle $ \theta $ whose cosine is $ -\frac{1}{2} $ is $ \theta = \frac{2\pi}{3} $.

Therefore, $ \mathbf{a} + \mathbf{b} $ is a unit vector if $ \theta = \frac{2\pi}{3} $.