If the unit vectors ` vec(e_(1))` and `vec(e_(2))` are inclined at an angle `2theta` and `|vec(e_(1)) - vec(e_(2))| lt 2` , then for `theta in [0,pi], theta` may lie in the interval

Options

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

Answer

To solve the problem step by step, we will analyze the given conditions and derive the range for θ.

### Step 1: Understand the Given Information
We have two unit vectors \( \vec{e_1} \) and \( \vec{e_2} \) such that:
- The angle between them is \( 2\theta \).
- The magnitude of their difference is less than 2, i.e., \( |\vec{e_1} - \vec{e_2}| < 2 \).

### Step 2: Express the Magnitude Condition
Since both vectors are unit vectors, we know:
\[ |\vec{e_1}| = 1 \quad \text{and} \quad |\vec{e_2}| = 1 \]
Now, we can express the magnitude of the difference:
\[ |\vec{e_1} - \vec{e_2}|^2 < 4 \]
Expanding this, we have:
\[ (\vec{e_1} - \vec{e_2}) \cdot (\vec{e_1} - \vec{e_2}) < 4 \]
This expands to:
\[ |\vec{e_1}|^2 + |\vec{e_2}|^2 - 2 \vec{e_1} \cdot \vec{e_2} < 4 \]

### Step 3: Substitute the Magnitudes
Since \( |\vec{e_1}|^2 = 1 \) and \( |\vec{e_2}|^2 = 1 \), we substitute these values:
\[ 1 + 1 - 2 \vec{e_1} \cdot \vec{e_2} < 4 \]
This simplifies to:
\[ 2 - 2 \vec{e_1} \cdot \vec{e_2} < 4 \]

### Step 4: Rearranging the Inequality
Rearranging gives:
\[ -2 \vec{e_1} \cdot \vec{e_2} < 2 \]
Dividing by -2 (and reversing the inequality):
\[ \vec{e_1} \cdot \vec{e_2} > -1 \]

### Step 5: Relate the Dot Product to the Angle
The dot product of two vectors can be expressed in terms of the angle between them:
\[ \vec{e_1} \cdot \vec{e_2} = |\vec{e_1}| |\vec{e_2}| \cos(2\theta) = \cos(2\theta) \]
Thus, we have:
\[ \cos(2\theta) > -1 \]

### Step 6: Analyzing the Cosine Condition
The cosine function has a range of \([-1, 1]\). The condition \( \cos(2\theta) > -1 \) is always satisfied for any angle \( 2\theta \) except when \( 2\theta = \pi \) (where \( \cos(2\theta) = -1 \)). Therefore, we need to exclude this case:
\[ 2\theta \neq \pi \implies \theta \neq \frac{\pi}{2} \]

### Step 7: Determine the Range of θ
Since \( \theta \) is defined in the interval \([0, \pi]\), we conclude:
\[ \theta \in [0, \pi] \text{ but } \theta \neq \frac{\pi}{2} \]
Thus, the valid interval for \( \theta \) is:
\[ \theta \in [0, \frac{\pi}{2}) \cup (\frac{\pi}{2}, \pi] \]

### Final Answer
The interval for \( \theta \) is:
\[ \theta \in [0, \frac{\pi}{2}) \cup (\frac{\pi}{2}, \pi] \]