If the unit vectors ` veca and vecb ` are inclined of an angle ` 2 theta` such that ` |veca -vecb| lt 1 and 0 le theta le pi` then ` theta` in the interval

Options

A. `[0, pi//6)`
B. `(5 pi//6, pi]`
C. `[pi//6, pi//2]`
D. `(pi//2, 5pi//6]`

Explanation

To solve the problem, we need to analyze the conditions given for the unit vectors \( \vec{a} \) and \( \vec{b} \) which are inclined at an angle \( 2\theta \). We are also given that \( |\vec{a} - \vec{b}| < 1 \) and \( 0 \leq \theta \leq \pi \).

### Step-by-Step Solution:

1. **Understanding the Magnitude of the Difference of Vectors**:
Since \( \vec{a} \) and \( \vec{b} \) are unit vectors, we can express their magnitudes as:
\[ |\vec{a}| = 1 \quad \text{and} \quad |\vec{b}| = 1 \]
The magnitude of the difference 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:
\[ |\vec{a} - \vec{b}|^2 = 1 + 1 - 2 \vec{a} \cdot \vec{b} = 2 - 2 \vec{a} \cdot \vec{b} \]

2. **Using the Dot Product**:
The dot product \( \vec{a} \cdot \vec{b} \) can be expressed in terms of the angle between them:
\[ \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos(2\theta) = \cos(2\theta) \]
Thus, we can rewrite the expression:
\[ |\vec{a} - \vec{b}|^2 = 2 - 2\cos(2\theta) = 2(1 - \cos(2\theta)) \]

3. **Simplifying the Magnitude**:
The expression for the magnitude becomes:
\[ |\vec{a} - \vec{b}| = \sqrt{2(1 - \cos(2\theta))} \]

4. **Applying the Given Condition**:
We know from the problem statement that:
\[ |\vec{a} - \vec{b}| < 1 \]
Therefore:
\[ \sqrt{2(1 - \cos(2\theta)) < 1 \]
Squaring both sides gives:
\[ 2(1 - \cos(2\theta)) < 1 \]
Simplifying this inequality:
\[ 1 - \cos(2\theta) < \frac{1}{2} \]
\[ -\cos(2\theta) < -\frac{1}{2} \]
\[ \cos(2\theta) > \frac{1}{2} \]

5. **Finding the Range for \( 2\theta \)**:
The condition \( \cos(2\theta) > \frac{1}{2} \) implies:
\[ 2\theta < \frac{\pi}{3} \quad \text{or} \quad 2\theta > \frac{5\pi}{3} \]
Dividing by 2 gives:
\[ \theta < \frac{\pi}{6} \quad \text{or} \quad \theta > \frac{5\pi}{6} \]

6. **Final Interval for \( \theta \)**:
Therefore, the final intervals for \( \theta \) are:
\[ \theta \in \left[0, \frac{\pi}{6}\right) \cup \left(\frac{5\pi}{6}, \pi\right] \]

### Conclusion:
Thus, the values of \( \theta \) satisfying the given conditions are in the intervals:
\[ \theta \in \left[0, \frac{\pi}{6}\right) \cup \left(\frac{5\pi}{6}, \pi\right] \]