If |A|=2 and |B|=4 and angle between then is `60^(@)` then |A-B|

Options

A. `sqrt(13)`
B. `3sqrt(3)`
C. `sqrt(3)`
D. `2sqrt(3)`

Explanation

To find the magnitude of the vector \( |A - B| \), we can use the formula derived from the law of cosines. The formula for the magnitude of the difference of two vectors is given by:

\[ |A - B| = \sqrt{|A|^2 + |B|^2 - 2 |A| |B| \cos(\theta)} \]

Where:
- \( |A| \) is the magnitude of vector \( A \)
- \( |B| \) is the magnitude of vector \( B \)
- \( \theta \) is the angle between vectors \( A \) and \( B \)

### Step-by-step Solution:

1. **Identify the given values:**
- \( |A| = 2 \)
- \( |B| = 4 \)
- \( \theta = 60^\circ \)

2. **Substitute the values into the formula:**
\[ |A - B| = \sqrt{2^2 + 4^2 - 2 \cdot 2 \cdot 4 \cdot \cos(60^\circ)} \]

3. **Calculate \( |A|^2 \) and \( |B|^2 \):**
- \( |A|^2 = 2^2 = 4 \)
- \( |B|^2 = 4^2 = 16 \)

4. **Calculate \( \cos(60^\circ) \):**
- \( \cos(60^\circ) = \frac{1}{2} \)

5. **Substitute these values back into the equation:**
\[ |A - B| = \sqrt{4 + 16 - 2 \cdot 2 \cdot 4 \cdot \frac{1}{2}} \]

6. **Simplify the expression:**
\[ |A - B| = \sqrt{4 + 16 - 2 \cdot 2 \cdot 4 \cdot \frac{1}{2}} = \sqrt{4 + 16 - 8} \]
\[ |A - B| = \sqrt{12} \]

7. **Further simplify \( \sqrt{12} \):**
\[ \sqrt{12} = \sqrt{4 \cdot 3} = 2\sqrt{3} \]

8. **Final answer:**
\[ |A - B| = 2\sqrt{3} \]

### Conclusion:
The magnitude of the vector \( |A - B| \) is \( 2\sqrt{3} \).