If `veca and vecb` are unit vectors making an angle `alpha` with each other, such that `alpha in (0, pi)` and `|veca+2vecb|lt 5`, then `alpha` lies in the interval

Options

A. `(0, pi)`
B. `(0, (pi)/(2))`
C. `((pi)/(2), (2pi)/(3))`
D. `((pi)/(6), (5pi)/(6))`

Answer

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

### Step-by-Step Solution:

1. **Understanding the Problem**:
We have two unit vectors \( \vec{a} \) and \( \vec{b} \) that make an angle \( \alpha \) with each other. We know that \( |\vec{a} + 2\vec{b}| < 5 \).

2. **Using the Magnitude Formula**:
The magnitude of the vector sum can be expressed using the formula:
\[ |\vec{a} + 2\vec{b}|^2 = |\vec{a}|^2 + |2\vec{b}|^2 + 2(\vec{a} \cdot 2\vec{b}) \]
Since \( |\vec{a}| = 1 \) and \( |\vec{b}| = 1 \), we have:
\[ |\vec{a} + 2\vec{b}|^2 = 1^2 + (2 \cdot 1)^2 + 2 \cdot 2(\vec{a} \cdot \vec{b}) \]
This simplifies to:
\[ |\vec{a} + 2\vec{b}|^2 = 1 + 4 + 4(\vec{a} \cdot \vec{b}) = 5 + 4(\vec{a} \cdot \vec{b}) \]

3. **Setting Up the Inequality**:
Given that \( |\vec{a} + 2\vec{b}| < 5 \), we can write:
\[ 5 + 4(\vec{a} \cdot \vec{b}) < 25 \]
Simplifying this gives:
\[ 4(\vec{a} \cdot \vec{b}) < 20 \]
Dividing both sides by 4, we find:
\[ \vec{a} \cdot \vec{b} < 5 \]

4. **Using the Dot Product**:
The dot product \( \vec{a} \cdot \vec{b} \) can be expressed in terms of the angle \( \alpha \):
\[ \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos(\alpha) = \cos(\alpha) \]
Therefore, we have:
\[ \cos(\alpha) < 5 \]

5. **Analyzing the Cosine Function**:
The cosine function \( \cos(\alpha) \) for \( \alpha \in (0, \pi) \) always lies between -1 and 1. Thus, the condition \( \cos(\alpha) < 5 \) is always satisfied for any \( \alpha \) in the interval \( (0, \pi) \).

6. **Conclusion**:
Since the condition holds for all \( \alpha \) in the interval \( (0, \pi) \), we conclude that:
\[ \alpha \in (0, \pi) \]

### Final Answer:
The angle \( \alpha \) lies in the interval \( (0, \pi) \).