Let ` vec a\ a n d\ vec b` be two unit vectors and `alpha` be the angle between them, then ` vec a+ vec b` is a unit vectors, if

Options

A. `theta = (pi)/(4)`
B. `theta = (pi)/(3)`
C. `theta = (pi)/(2)`
D. `theta = (2pi)/(3)`

Explanation

To determine the conditions under which the vector \( \vec{a} + \vec{b} \) is a unit vector, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Unit Vectors**:
Given that \( \vec{a} \) and \( \vec{b} \) are unit vectors, we know that:
\[ |\vec{a}| = 1 \quad \text{and} \quad |\vec{b}| = 1 \]

2. **Magnitude of the Sum**:
We want to find the condition under which \( |\vec{a} + \vec{b}| = 1 \).

3. **Squaring Both Sides**:
Squaring the magnitude gives us:
\[ |\vec{a} + \vec{b}|^2 = 1^2 = 1 \]

4. **Expanding the Left Side**:
Using the property of magnitudes, we expand the left side:
\[ |\vec{a} + \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + 2 \vec{a} \cdot \vec{b} \]
Substituting the magnitudes:
\[ 1^2 + 1^2 + 2 \vec{a} \cdot \vec{b} = 1 \]
This simplifies to:
\[ 1 + 1 + 2 \vec{a} \cdot \vec{b} = 1 \]

5. **Simplifying the Equation**:
Rearranging gives:
\[ 2 + 2 \vec{a} \cdot \vec{b} = 1 \]
\[ 2 \vec{a} \cdot \vec{b} = 1 - 2 \]
\[ 2 \vec{a} \cdot \vec{b} = -1 \]
\[ \vec{a} \cdot \vec{b} = -\frac{1}{2} \]

6. **Using the Dot Product**:
The dot product can also be expressed in terms of the angle \( \alpha \) between the vectors:
\[ \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos(\alpha) \]
Since both are unit vectors:
\[ \vec{a} \cdot \vec{b} = 1 \cdot 1 \cdot \cos(\alpha) = \cos(\alpha) \]
Therefore, we have:
\[ \cos(\alpha) = -\frac{1}{2} \]

7. **Finding the Angle**:
The angle \( \alpha \) that satisfies \( \cos(\alpha) = -\frac{1}{2} \) is:
\[ \alpha = 120^\circ \quad \text{or} \quad \alpha = \frac{2\pi}{3} \text{ radians} \]

### Conclusion:
The vector \( \vec{a} + \vec{b} \) is a unit vector if the angle \( \alpha \) between the two unit vectors \( \vec{a} \) and \( \vec{b} \) is \( 120^\circ \) or \( \frac{2\pi}{3} \) radians.