If `hat(e_(1))` and ` hat(e_(2))` are two unit vectors such that `vec(e_(1)) - vec(e_(2))` is also a unit vector , then find the angle `theta` between `hat(e_(1))` and `hat(e_(2))`.

Answer

To solve the problem, we need to find the angle \( \theta \) between two unit vectors \( \hat{e_1} \) and \( \hat{e_2} \) given that \( \vec{e_1} - \vec{e_2} \) is also a unit vector.

### Step-by-Step Solution:

1. **Understanding the Given Information**:
- We know that both \( \hat{e_1} \) and \( \hat{e_2} \) are unit vectors. This means:
\[ |\hat{e_1}| = 1 \quad \text{and} \quad |\hat{e_2}| = 1 \]
- We are also given that \( |\vec{e_1} - \vec{e_2}| = 1 \) (since \( \vec{e_1} - \vec{e_2} \) is a unit vector).

2. **Using the Magnitude Condition**:
- The magnitude of the difference of two vectors can be expressed using the dot product:
\[ |\vec{e_1} - \vec{e_2}|^2 = (\vec{e_1} - \vec{e_2}) \cdot (\vec{e_1} - \vec{e_2}) \]
- Expanding this, we get:
\[ |\vec{e_1}|^2 - 2 \vec{e_1} \cdot \vec{e_2} + |\vec{e_2}|^2 \]

3. **Substituting the Magnitudes**:
- Since \( |\vec{e_1}|^2 = 1 \) and \( |\vec{e_2}|^2 = 1 \):
\[ 1 - 2 \vec{e_1} \cdot \vec{e_2} + 1 = 1 \]
- This simplifies to:
\[ 2 - 2 \vec{e_1} \cdot \vec{e_2} = 1 \]

4. **Rearranging the Equation**:
- Rearranging gives:
\[ 2 \vec{e_1} \cdot \vec{e_2} = 2 - 1 \]
- Thus:
\[ 2 \vec{e_1} \cdot \vec{e_2} = 1 \]
- Dividing both sides by 2:
\[ \vec{e_1} \cdot \vec{e_2} = \frac{1}{2} \]

5. **Relating Dot Product to Angle**:
- The dot product of two vectors can also be expressed in terms of the angle \( \theta \) between them:
\[ \vec{e_1} \cdot \vec{e_2} = |\vec{e_1}| |\vec{e_2}| \cos \theta \]
- Since both vectors are unit vectors, this simplifies to:
\[ \vec{e_1} \cdot \vec{e_2} = 1 \cdot 1 \cdot \cos \theta = \cos \theta \]
- Therefore, we have:
\[ \cos \theta = \frac{1}{2} \]

6. **Finding the Angle \( \theta \)**:
- The angle \( \theta \) for which \( \cos \theta = \frac{1}{2} \) is:
\[ \theta = \cos^{-1}\left(\frac{1}{2}\right) \]
- This corresponds to:
\[ \theta = 60^\circ \quad \text{or} \quad \theta = \frac{\pi}{3} \text{ radians} \]

### Final Answer:
The angle \( \theta \) between the unit vectors \( \hat{e_1} \) and \( \hat{e_2} \) is \( 60^\circ \) or \( \frac{\pi}{3} \) radians.