The vector `a + 3b` is perpendicular to `7a-5b` and `a-5b` is perpendicular to `7a +3b`. The angle between a and b is

Options

A. `pi//4`
B. `pi//6`
C. `pi//2`
D. none

Answer

To solve the problem, we need to use the property that two vectors are perpendicular if their dot product is zero. We will derive two equations based on the given conditions and then solve them to find the angle between the vectors \( \mathbf{a} \) and \( \mathbf{b} \).

### Step-by-Step Solution:

1. **Set up the first condition**:
The vector \( \mathbf{a} + 3\mathbf{b} \) is perpendicular to \( 7\mathbf{a} - 5\mathbf{b} \). Therefore, we can write:
\[ (\mathbf{a} + 3\mathbf{b}) \cdot (7\mathbf{a} - 5\mathbf{b}) = 0 \]

2. **Expand the dot product**:
\[ \mathbf{a} \cdot (7\mathbf{a}) + \mathbf{a} \cdot (-5\mathbf{b}) + 3\mathbf{b} \cdot (7\mathbf{a}) + 3\mathbf{b} \cdot (-5\mathbf{b}) = 0 \]
This simplifies to:
\[ 7|\mathbf{a}|^2 - 5(\mathbf{a} \cdot \mathbf{b}) + 21(\mathbf{b} \cdot \mathbf{a}) - 15|\mathbf{b}|^2 = 0 \]
Since \( \mathbf{a} \cdot \mathbf{b} = \mathbf{b} \cdot \mathbf{a} \), we can combine terms:
\[ 7|\mathbf{a}|^2 + 16(\mathbf{a} \cdot \mathbf{b}) - 15|\mathbf{b}|^2 = 0 \tag{1} \]

3. **Set up the second condition**:
The vector \( \mathbf{a} - 5\mathbf{b} \) is perpendicular to \( 7\mathbf{a} + 3\mathbf{b} \). Therefore:
\[ (\mathbf{a} - 5\mathbf{b}) \cdot (7\mathbf{a} + 3\mathbf{b}) = 0 \]

4. **Expand the dot product**:
\[ \mathbf{a} \cdot (7\mathbf{a}) + \mathbf{a} \cdot (3\mathbf{b}) - 5\mathbf{b} \cdot (7\mathbf{a}) - 5\mathbf{b} \cdot (3\mathbf{b}) = 0 \]
This simplifies to:
\[ 7|\mathbf{a}|^2 + 3(\mathbf{a} \cdot \mathbf{b}) - 35(\mathbf{b} \cdot \mathbf{a}) - 15|\mathbf{b}|^2 = 0 \]
Again, combining terms gives:
\[ 7|\mathbf{a}|^2 - 32(\mathbf{a} \cdot \mathbf{b}) - 15|\mathbf{b}|^2 = 0 \tag{2} \]

5. **Solve the system of equations**:
We have two equations:
- Equation (1): \( 7|\mathbf{a}|^2 + 16(\mathbf{a} \cdot \mathbf{b}) - 15|\mathbf{b}|^2 = 0 \)
- Equation (2): \( 7|\mathbf{a}|^2 - 32(\mathbf{a} \cdot \mathbf{b}) - 15|\mathbf{b}|^2 = 0 \)

Subtract Equation (2) from Equation (1):
\[ (7|\mathbf{a}|^2 + 16(\mathbf{a} \cdot \mathbf{b}) - 15|\mathbf{b}|^2) - (7|\mathbf{a}|^2 - 32(\mathbf{a} \cdot \mathbf{b}) - 15|\mathbf{b}|^2) = 0 \]
This simplifies to:
\[ 48(\mathbf{a} \cdot \mathbf{b}) = 0 \]

6. **Conclusion**:
Since \( 48(\mathbf{a} \cdot \mathbf{b}) = 0 \), it follows that \( \mathbf{a} \cdot \mathbf{b} = 0 \). This means that the angle \( \theta \) between \( \mathbf{a} \) and \( \mathbf{b} \) is:
\[ \cos \theta = 0 \implies \theta = \frac{\pi}{2} \]

### Final Answer:
The angle between vectors \( \mathbf{a} \) and \( \mathbf{b} \) is \( \frac{\pi}{2} \) radians.