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 \( \vec{a} + \vec{b} \) is also a unit vector.

### Step-by-Step Solution:

1. **Understanding the Given Information**:
We know that \( \vec{a} \) and \( \vec{b} \) are unit vectors. Therefore, we have:
\[ |\vec{a}| = 1 \quad \text{and} \quad |\vec{b}| = 1 \]
The angle between them is \( \alpha \).

2. **Setting Up the Equation**:
We need to find when \( \vec{a} + \vec{b} \) is a unit vector. This means:
\[ |\vec{a} + \vec{b}| = 1 \]

3. **Using the Magnitude Formula**:
The magnitude of the sum of two vectors can be expressed as:
\[ |\vec{a} + \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + 2 \vec{a} \cdot \vec{b} \]
Substituting the magnitudes of the unit vectors:
\[ |\vec{a} + \vec{b}|^2 = 1^2 + 1^2 + 2 \vec{a} \cdot \vec{b} \]
This simplifies to:
\[ |\vec{a} + \vec{b}|^2 = 1 + 1 + 2 \vec{a} \cdot \vec{b} = 2 + 2 \vec{a} \cdot \vec{b} \]

4. **Setting the Equation Equal to 1**:
Since we want \( |\vec{a} + \vec{b}| = 1 \), we square both sides:
\[ 1 = 2 + 2 \vec{a} \cdot \vec{b} \]
Rearranging gives:
\[ 2 \vec{a} \cdot \vec{b} = 1 - 2 \]
\[ 2 \vec{a} \cdot \vec{b} = -1 \]

5. **Finding the Dot Product**:
The dot product \( \vec{a} \cdot \vec{b} \) can also be expressed in terms of the angle \( \alpha \):
\[ \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \alpha = 1 \cdot 1 \cdot \cos \alpha = \cos \alpha \]
Substituting this into our equation gives:
\[ 2 \cos \alpha = -1 \]
Thus:
\[ \cos \alpha = -\frac{1}{2} \]

6. **Finding the Angle \( \alpha \)**:
The angle \( \alpha \) for which \( \cos \alpha = -\frac{1}{2} \) is:
\[ \alpha = 120^\circ \quad \text{or} \quad \alpha = 240^\circ \]
However, since we are typically interested in angles between \( 0^\circ \) and \( 180^\circ \) for the angle between two vectors, we take:
\[ \alpha = 120^\circ \]

### Final Answer:
The angle \( \alpha \) between the two unit vectors \( \vec{a} \) and \( \vec{b} \) such that \( \vec{a} + \vec{b} \) is a unit vector is:
\[ \alpha = 120^\circ \]