If `|a+b|gt|a-b|`, then the angle between a and b is

Options

A. acute
B. obtuse
C. `(pi)/(2)`
D. `pi`

Answer

To solve the problem, we need to determine the angle between two vectors \( \mathbf{a} \) and \( \mathbf{b} \) given the inequality \( |\mathbf{a} + \mathbf{b}| > |\mathbf{a} - \mathbf{b}| \).

### Step-by-Step Solution:

1. **Start with the given inequality**:
\[ |\mathbf{a} + \mathbf{b}| > |\mathbf{a} - \mathbf{b}| \]

2. **Square both sides**:
Squaring both sides helps eliminate the absolute values:
\[ |\mathbf{a} + \mathbf{b}|^2 > |\mathbf{a} - \mathbf{b}|^2 \]

3. **Expand both sides using the formula for the magnitude of a vector**:
\[ (\mathbf{a} + \mathbf{b}) \cdot (\mathbf{a} + \mathbf{b}) > (\mathbf{a} - \mathbf{b}) \cdot (\mathbf{a} - \mathbf{b}) \]
This expands to:
\[ |\mathbf{a}|^2 + 2\mathbf{a} \cdot \mathbf{b} + |\mathbf{b}|^2 > |\mathbf{a}|^2 - 2\mathbf{a} \cdot \mathbf{b} + |\mathbf{b}|^2 \]

4. **Simplify the inequality**:
The \( |\mathbf{a}|^2 \) and \( |\mathbf{b}|^2 \) terms cancel out:
\[ 2\mathbf{a} \cdot \mathbf{b} > -2\mathbf{a} \cdot \mathbf{b} \]

5. **Combine like terms**:
Adding \( 2\mathbf{a} \cdot \mathbf{b} \) to both sides gives:
\[ 4\mathbf{a} \cdot \mathbf{b} > 0 \]

6. **Divide by 4**:
\[ \mathbf{a} \cdot \mathbf{b} > 0 \]

7. **Relate the dot product to the angle**:
The dot product can be expressed in terms of the angle \( \theta \) between the vectors:
\[ \mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \cos \theta \]
Thus, the inequality becomes:
\[ |\mathbf{a}| |\mathbf{b}| \cos \theta > 0 \]

8. **Analyze the inequality**:
Since \( |\mathbf{a}| \) and \( |\mathbf{b}| \) are both positive (as they are magnitudes of vectors), we conclude:
\[ \cos \theta > 0 \]

9. **Determine the angle**:
The cosine of an angle is positive when the angle \( \theta \) is acute, which means:
\[ 0 < \theta < 90^\circ \]

### Conclusion:
The angle between the vectors \( \mathbf{a} \) and \( \mathbf{b} \) is acute.