If `|veca+ vecb| lt | veca- vecb|`, then the angle between `veca and vecb` can lie in the interval

Options

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

Explanation

To solve the problem, we need to analyze the given condition:

\[ |\vec{a} + \vec{b}| < |\vec{a} - \vec{b}| \]

### Step 1: Square both sides
We start by squaring both sides of the inequality:

\[ |\vec{a} + \vec{b}|^2 < |\vec{a} - \vec{b}|^2 \]

### Step 2: Expand both sides
Using the formula for the magnitude of vectors, we can expand both sides:

\[ |\vec{a} + \vec{b}|^2 = \vec{a} \cdot \vec{a} + 2 \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{b} = |\vec{a}|^2 + 2 \vec{a} \cdot \vec{b} + |\vec{b}|^2 \]

\[ |\vec{a} - \vec{b}|^2 = \vec{a} \cdot \vec{a} - 2 \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{b} = |\vec{a}|^2 - 2 \vec{a} \cdot \vec{b} + |\vec{b}|^2 \]

### Step 3: Set up the inequality
Now we substitute these expansions back into the inequality:

\[ |\vec{a}|^2 + 2 \vec{a} \cdot \vec{b} + |\vec{b}|^2 < |\vec{a}|^2 - 2 \vec{a} \cdot \vec{b} + |\vec{b}|^2 \]

### Step 4: Simplify the inequality
Cancel out the common terms \(|\vec{a}|^2\) and \(|\vec{b}|^2\):

\[ 2 \vec{a} \cdot \vec{b} < -2 \vec{a} \cdot \vec{b} \]

### Step 5: Combine like terms
Now, we can combine the terms:

\[ 4 \vec{a} \cdot \vec{b} < 0 \]

### Step 6: Interpret the dot product
Since the dot product \(\vec{a} \cdot \vec{b}\) can be expressed in terms of the angle \(\theta\) between the vectors:

\[ \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta \]

Thus, we have:

\[ 4 |\vec{a}| |\vec{b}| \cos \theta < 0 \]

### Step 7: Determine the angle
Since \(|\vec{a}|\) and \(|\vec{b}|\) are both positive, we can conclude that:

\[ \cos \theta < 0 \]

This implies that the angle \(\theta\) must lie in the intervals where the cosine function is negative:

\[ \theta \in \left(\frac{\pi}{2}, \frac{3\pi}{2}\right) \]

### Final Answer
Thus, the angle between \(\vec{a}\) and \(\vec{b}\) can lie in the interval:

\[ \left(\frac{\pi}{2}, \frac{3\pi}{2}\right) \]