Let `veca` and `vecb` are unit vectors inclined at an angle `alpha` to each other , if `|veca+vecb| lt 1` then

Options

A. `alpha=(pi)/(2)`
B. `alpha lt (pi)/(3)`
C. `alpha gt (2pi)/(3)`
D. `(pi)/(3) lt alpha lt (2pi)/(3)`

Answer

To solve the problem, we need to analyze the given conditions about the unit vectors \(\vec{a}\) and \(\vec{b}\) and derive the possible values for the angle \(\alpha\) between them.

### Step-by-Step Solution:

1. **Understanding the Given Information**:
- \(\vec{a}\) and \(\vec{b}\) are unit vectors, which means:
\[ |\vec{a}| = 1 \quad \text{and} \quad |\vec{b}| = 1 \]
- The angle between \(\vec{a}\) and \(\vec{b}\) is \(\alpha\).
- We are given that:
\[ |\vec{a} + \vec{b}| < 1 \]

2. **Squaring Both Sides**:
- To simplify the expression, we square both sides:
\[ |\vec{a} + \vec{b}|^2 < 1^2 \]
- This can be rewritten using the dot product:
\[ (\vec{a} + \vec{b}) \cdot (\vec{a} + \vec{b}) < 1 \]

3. **Expanding the Dot Product**:
- Expanding the left-hand side:
\[ \vec{a} \cdot \vec{a} + 2 \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{b} < 1 \]
- Since \(|\vec{a}|^2 = 1\) and \(|\vec{b}|^2 = 1\), we have:
\[ 1 + 2 \vec{a} \cdot \vec{b} + 1 < 1 \]
- This simplifies to:
\[ 2 + 2 \vec{a} \cdot \vec{b} < 1 \]

4. **Isolating the Dot Product**:
- Rearranging gives:
\[ 2 \vec{a} \cdot \vec{b} < 1 - 2 \]
- Thus:
\[ 2 \vec{a} \cdot \vec{b} < -1 \]
- Dividing by 2:
\[ \vec{a} \cdot \vec{b} < -\frac{1}{2} \]

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

6. **Finding the Angle**:
- The cosine function is negative in the second and third quadrants. The angle \(\alpha\) for which \(\cos(\alpha) < -\frac{1}{2}\) corresponds to:
\[ \alpha > \frac{2\pi}{3} \quad \text{(or 120 degrees)} \]

### Conclusion:
The required values of \(\alpha\) are:
\[ \alpha > \frac{2\pi}{3} \]