Let `|a|=7, |b|=11, |a+b|=10 sqrt3`
What is `|a-b|` equal to ?

Options

A. `2 sqrt2`
B. `2 sqrt10`
C. 5
D. 10

Answer

To solve the problem, we need to find the value of \(|a-b|\) given the magnitudes \(|a|=7\), \(|b|=11\), and \(|a+b|=10\sqrt{3}\).

### Step-by-Step Solution:

1. **Use the formula for the magnitude of the sum of two vectors**:
\[ |a+b|^2 = |a|^2 + |b|^2 + 2|a||b|\cos\theta \]
where \(\theta\) is the angle between vectors \(a\) and \(b\).

2. **Substitute the known values**:
\[ |a+b|^2 = (10\sqrt{3})^2 = 300 \]
\[ |a|^2 = 7^2 = 49 \]
\[ |b|^2 = 11^2 = 121 \]
Therefore,
\[ 300 = 49 + 121 + 2 \cdot 7 \cdot 11 \cos\theta \]

3. **Calculate \(49 + 121\)**:
\[ 49 + 121 = 170 \]
Now, substitute this back into the equation:
\[ 300 = 170 + 154 \cos\theta \]

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

5. **Use the formula for the magnitude of the difference of two vectors**:
\[ |a-b|^2 = |a|^2 + |b|^2 - 2|a||b|\cos\theta \]

6. **Substituting the known values**:
\[ |a-b|^2 = 49 + 121 - 2 \cdot 7 \cdot 11 \cdot \frac{65}{77} \]
\[ |a-b|^2 = 170 - 154 \cdot \frac{65}{77} \]

7. **Calculating \(154 \cdot \frac{65}{77}\)**:
\[ 154 \cdot \frac{65}{77} = \frac{154 \cdot 65}{77} = 130 \]
Thus,
\[ |a-b|^2 = 170 - 130 = 40 \]

8. **Taking the square root**:
\[ |a-b| = \sqrt{40} = 2\sqrt{10} \]

### Final Answer:
\[ |a-b| = 2\sqrt{10} \]