Two vector A and B have equal magnitudes. Then the vector `A+B` is perpendicular to

Options

A. `AxxB`
B. `A-B`
C. `3A-3B`
D. all of these

Explanation

To solve the problem, we need to determine which of the given vectors is perpendicular to the vector \( \mathbf{A} + \mathbf{B} \) when the vectors \( \mathbf{A} \) and \( \mathbf{B} \) have equal magnitudes.

### Step-by-Step Solution:

1. **Understanding Perpendicular Vectors:**
Two vectors are perpendicular if their dot product is zero. Therefore, we need to check the dot product of \( \mathbf{A} + \mathbf{B} \) with each of the given vectors.

2. **Checking the first option: \( \mathbf{A} + \mathbf{B} \cdot (\mathbf{A} \times \mathbf{B}) \):**
- We calculate the dot product:
\[ (\mathbf{A} + \mathbf{B}) \cdot (\mathbf{A} \times \mathbf{B}) = \mathbf{A} \cdot (\mathbf{A} \times \mathbf{B}) + \mathbf{B} \cdot (\mathbf{A} \times \mathbf{B}) \]
- The dot product of any vector with a vector that is perpendicular to it (like \( \mathbf{A} \times \mathbf{B} \)) is zero:
\[ \mathbf{A} \cdot (\mathbf{A} \times \mathbf{B}) = 0 \quad \text{and} \quad \mathbf{B} \cdot (\mathbf{A} \times \mathbf{B}) = 0 \]
- Therefore, the total is:
\[ 0 + 0 = 0 \]
- Thus, \( \mathbf{A} + \mathbf{B} \) is perpendicular to \( \mathbf{A} \times \mathbf{B} \).

3. **Checking the second option: \( \mathbf{A} + \mathbf{B} \cdot (\mathbf{A} - \mathbf{B}) \):**
- We calculate the dot product:
\[ (\mathbf{A} + \mathbf{B}) \cdot (\mathbf{A} - \mathbf{B}) = \mathbf{A} \cdot \mathbf{A} - \mathbf{A} \cdot \mathbf{B} + \mathbf{B} \cdot \mathbf{A} - \mathbf{B} \cdot \mathbf{B} \]
- Since \( \mathbf{A} \cdot \mathbf{B} = \mathbf{B} \cdot \mathbf{A} \), we can simplify:
\[ = \|\mathbf{A}\|^2 - \|\mathbf{B}\|^2 \]
- Given that \( \|\mathbf{A}\| = \|\mathbf{B}\| \), we have:
\[ = \|\mathbf{A}\|^2 - \|\mathbf{A}\|^2 = 0 \]
- Thus, \( \mathbf{A} - \mathbf{B} \) is also perpendicular to \( \mathbf{A} + \mathbf{B} \).

4. **Checking the third option: \( \mathbf{A} + \mathbf{B} \cdot (3\mathbf{A} - 3\mathbf{B}) \):**
- Factor out the 3:
\[ (\mathbf{A} + \mathbf{B}) \cdot (3(\mathbf{A} - \mathbf{B})) = 3(\mathbf{A} + \mathbf{B}) \cdot (\mathbf{A} - \mathbf{B}) \]
- Since we already established that \( (\mathbf{A} + \mathbf{B}) \cdot (\mathbf{A} - \mathbf{B}) = 0 \):
\[ 3 \cdot 0 = 0 \]
- Therefore, \( 3(\mathbf{A} - \mathbf{B}) \) is also perpendicular to \( \mathbf{A} + \mathbf{B} \).

5. **Conclusion:**
All three options \( \mathbf{A} \times \mathbf{B} \), \( \mathbf{A} - \mathbf{B} \), and \( 3(\mathbf{A} - \mathbf{B}) \) are perpendicular to \( \mathbf{A} + \mathbf{B} \).

### Final Answer:
All of the given vectors are perpendicular to \( \mathbf{A} + \mathbf{B} \).