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 vectors and the conditions provided. Let’s break it down step by step:

### Step 1: Understand the vectors and their properties
Given that \(\vec{a}\) and \(\vec{b}\) are unit vectors, we know:
\[ |\vec{a}| = 1 \quad \text{and} \quad |\vec{b}| = 1 \]
The angle between them is \(\alpha\), where \(0 < \alpha < \pi\).

### Step 2: Write the expression for the magnitude of \(\vec{a} + 2\vec{b}\)
We need to find the magnitude of the vector sum \(\vec{a} + 2\vec{b}\):
\[ |\vec{a} + 2\vec{b}| = \sqrt{|\vec{a}|^2 + |2\vec{b}|^2 + 2|\vec{a}||2\vec{b}|\cos(\alpha)} \]
Substituting the magnitudes:
\[ |\vec{a} + 2\vec{b}| = \sqrt{1^2 + (2 \cdot 1)^2 + 2 \cdot 1 \cdot 2 \cos(\alpha)} = \sqrt{1 + 4 + 4\cos(\alpha)} = \sqrt{5 + 4\cos(\alpha)} \]

### Step 3: Apply the condition on the magnitude
We are given that:
\[ |\vec{a} + 2\vec{b}| < 5 \]
Substituting the expression we derived:
\[ \sqrt{5 + 4\cos(\alpha)} < 5 \]

### Step 4: Square both sides to eliminate the square root
Squaring both sides gives:
\[ 5 + 4\cos(\alpha) < 25 \]

### Step 5: Rearranging the inequality
Rearranging the inequality:
\[ 4\cos(\alpha) < 20 \]
Dividing by 4:
\[ \cos(\alpha) < 5 \]

### Step 6: Analyze the range of \(\cos(\alpha)\)
Since \(\cos(\alpha)\) for \(\alpha \in (0, \pi)\) can only take values from -1 to 1, the condition \(\cos(\alpha) < 5\) is always satisfied for all \(\alpha\) in the interval \(0 < \alpha < \pi\).

### Conclusion
Thus, the angle \(\alpha\) lies in the interval:
\[ (0, \pi) \]