Let `|a|=7, |b|=11, |a+b|=10 sqrt3`
What is the angle between `(a+b) and (a-b)` ?

Options

A. `(pi)/(2)`
B. `(pi)/(4)`
C. `(pi)/(6)`
D. None of these

Answer

To find the angle between the vectors \( (a+b) \) and \( (a-b) \), we can use the properties of vectors and the information provided in the question.

### Step-by-Step Solution:

1. **Given Information**:
- \( |a| = 7 \)
- \( |b| = 11 \)
- \( |a + b| = 10\sqrt{3} \)

2. **Use the Law of Cosines**:
The magnitude of the sum of two vectors can be expressed using the Law of Cosines:
\[ |a + b|^2 = |a|^2 + |b|^2 + 2|a||b|\cos\theta \]
where \( \theta \) is the angle between vectors \( a \) and \( b \).

3. **Substituting the Known Values**:
\[ |a + b|^2 = (10\sqrt{3})^2 = 300 \]
\[ |a|^2 = 7^2 = 49 \]
\[ |b|^2 = 11^2 = 121 \]
Now substitute these values into the equation:
\[ 300 = 49 + 121 + 2 \cdot 7 \cdot 11 \cdot \cos\theta \]
Simplifying this gives:
\[ 300 = 170 + 154 \cos\theta \]

4. **Solving for \( \cos\theta \)**:
Rearranging the equation:
\[ 300 - 170 = 154 \cos\theta \]
\[ 130 = 154 \cos\theta \]
\[ \cos\theta = \frac{130}{154} = \frac{65}{77} \]

5. **Finding the Angle Between \( (a+b) \) and \( (a-b) \)**:
To find the angle between \( (a+b) \) and \( (a-b) \), we can use the fact that:
\[ |a-b|^2 = |a|^2 + |b|^2 - 2|a||b|\cos\theta \]
Substituting the known values:
\[ |a-b|^2 = 49 + 121 - 2 \cdot 7 \cdot 11 \cdot \frac{65}{77} \]
Simplifying gives:
\[ |a-b|^2 = 170 - 154 \cdot \frac{65}{77} \]
Now, we can find the angle \( \phi \) between \( (a+b) \) and \( (a-b) \) using the dot product:
\[ \cos\phi = \frac{(a+b) \cdot (a-b)}{|a+b||a-b|} \]
However, we can also conclude that since \( (a+b) \) and \( (a-b) \) are derived from the same vectors \( a \) and \( b \), and they are perpendicular due to their geometric arrangement, the angle \( \phi \) is \( 90^\circ \) or \( \frac{\pi}{2} \) radians.

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

---