If `|a|=7, |b|=11 and |a+b|=10 sqrt3`, then `|a-b|` is equal to

Options

A. 40
B. 10
C. `4 sqrt10`
D. `2 sqrt10`

Answer

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

### Step-by-Step Solution

1. **Square the Magnitudes**:
We start by squaring the magnitude of \( |a + b| \):
\[ |a + b|^2 = (10\sqrt{3})^2 = 300 \]

2. **Use the Formula for the Magnitude of the Sum of Vectors**:
The formula for the magnitude of the sum of two vectors is:
\[ |a + b|^2 = |a|^2 + |b|^2 + 2(a \cdot b) \]
Substituting the known values:
\[ 300 = |a|^2 + |b|^2 + 2(a \cdot b) \]

3. **Substitute the Magnitudes**:
We know \( |a| = 7 \) and \( |b| = 11 \):
\[ |a|^2 = 7^2 = 49 \quad \text{and} \quad |b|^2 = 11^2 = 121 \]
So, substituting these values:
\[ 300 = 49 + 121 + 2(a \cdot b) \]

4. **Combine and Solve for \( a \cdot b \)**:
Combine the constants:
\[ 300 = 170 + 2(a \cdot b) \]
Rearranging gives:
\[ 2(a \cdot b) = 300 - 170 = 130 \]
Therefore:
\[ a \cdot b = \frac{130}{2} = 65 \]

5. **Use the Formula for the Magnitude of the Difference of Vectors**:
Now, we need to find \( |a - b| \):
\[ |a - b|^2 = |a|^2 + |b|^2 - 2(a \cdot b) \]
Substituting the values we have:
\[ |a - b|^2 = 49 + 121 - 2 \times 65 \]

6. **Calculate the Values**:
Calculate \( 49 + 121 \):
\[ 49 + 121 = 170 \]
Now calculate \( 2 \times 65 = 130 \):
\[ |a - b|^2 = 170 - 130 = 40 \]

7. **Take the Square Root**:
Finally, we take the square root to find \( |a - b| \):
\[ |a - b| = \sqrt{40} = \sqrt{4 \times 10} = 2\sqrt{10} \]

Thus, the final answer is:
\[ |a - b| = 2\sqrt{10} \]