Let ` veca and vecb` be two unit vectors and ` alpha` be the angle between them, then ` veca + vecb` is a unit vector , if ` alpha ` =

Options

A. `pi//4`
B. ` pi//3`
C. `2pi//3`
D. `pi//2`

Explanation

To solve the problem, we need to determine the angle \( \alpha \) between two unit vectors \( \vec{a} \) and \( \vec{b} \) such that the vector sum \( \vec{a} + \vec{b} \) is also a unit vector.

### Step-by-Step Solution:

1. **Understanding the Given Information:**
- Let \( \vec{a} \) and \( \vec{b} \) be two unit vectors.
- The angle between them is \( \alpha \).
- We need to find \( \alpha \) such that \( \vec{a} + \vec{b} \) is a unit vector.

2. **Using the Magnitude of the Sum of Vectors:**
- The magnitude of the sum of two vectors \( \vec{a} + \vec{b} \) can be expressed as:
\[ |\vec{a} + \vec{b}| = \sqrt{|\vec{a}|^2 + |\vec{b}|^2 + 2(\vec{a} \cdot \vec{b})} \]
- Since \( \vec{a} \) and \( \vec{b} \) are unit vectors, we have \( |\vec{a}| = 1 \) and \( |\vec{b}| = 1 \). Thus:
\[ |\vec{a} + \vec{b}| = \sqrt{1^2 + 1^2 + 2(\vec{a} \cdot \vec{b})} = \sqrt{2 + 2(\vec{a} \cdot \vec{b})} \]

3. **Finding the Dot Product:**
- The dot product \( \vec{a} \cdot \vec{b} \) can be expressed in terms of the angle \( \alpha \):
\[ \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos(\alpha) = 1 \cdot 1 \cdot \cos(\alpha) = \cos(\alpha) \]
- Therefore, we can rewrite the magnitude of the sum as:
\[ |\vec{a} + \vec{b}| = \sqrt{2 + 2\cos(\alpha)} \]

4. **Setting the Magnitude Equal to 1:**
- Since \( \vec{a} + \vec{b} \) is a unit vector, we set its magnitude equal to 1:
\[ \sqrt{2 + 2\cos(\alpha)} = 1 \]

5. **Squaring Both Sides:**
- Squaring both sides gives:
\[ 2 + 2\cos(\alpha) = 1 \]

6. **Solving for \( \cos(\alpha) \):**
- Rearranging the equation:
\[ 2\cos(\alpha) = 1 - 2 \]
\[ 2\cos(\alpha) = -1 \]
\[ \cos(\alpha) = -\frac{1}{2} \]

7. **Finding the Angle \( \alpha \):**
- The angle \( \alpha \) for which \( \cos(\alpha) = -\frac{1}{2} \) is:
\[ \alpha = \frac{2\pi}{3} \text{ or } \alpha = \frac{4\pi}{3} \]
- However, since we are typically interested in the angle between two vectors, we take:
\[ \alpha = \frac{2\pi}{3} \]

### Final Answer:
The angle \( \alpha \) between the two unit vectors \( \vec{a} \) and \( \vec{b} \) such that \( \vec{a} + \vec{b} \) is also a unit vector is:
\[ \alpha = \frac{2\pi}{3} \]