If a and b include an angle of `120^(@)` and their magnitudes are 2 and `sqrt(3)` then a.b is equal to

Options

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

Answer

To solve the problem of finding the dot product \( \mathbf{a} \cdot \mathbf{b} \) given the magnitudes of the vectors \( \mathbf{a} \) and \( \mathbf{b} \) and the angle between them, we can follow these steps:

### Step-by-Step Solution

1. **Identify the given values**:
- Magnitude of vector \( \mathbf{a} = 2 \)
- Magnitude of vector \( \mathbf{b} = \sqrt{3} \)
- Angle \( \theta = 120^\circ \)

2. **Use the dot product formula**:
The formula for the dot product of two vectors is given by:
\[ \mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \cos(\theta) \]
where \( |\mathbf{a}| \) and \( |\mathbf{b}| \) are the magnitudes of the vectors and \( \theta \) is the angle between them.

3. **Substitute the values into the formula**:
\[ \mathbf{a} \cdot \mathbf{b} = 2 \cdot \sqrt{3} \cdot \cos(120^\circ) \]

4. **Calculate \( \cos(120^\circ) \)**:
- We know that \( \cos(120^\circ) = \cos(180^\circ - 60^\circ) = -\cos(60^\circ) \)
- Since \( \cos(60^\circ) = \frac{1}{2} \), we have:
\[ \cos(120^\circ) = -\frac{1}{2} \]

5. **Substitute \( \cos(120^\circ) \) back into the equation**:
\[ \mathbf{a} \cdot \mathbf{b} = 2 \cdot \sqrt{3} \cdot \left(-\frac{1}{2}\right) \]

6. **Simplify the expression**:
\[ \mathbf{a} \cdot \mathbf{b} = 2 \cdot \sqrt{3} \cdot -\frac{1}{2} = -\sqrt{3} \]

7. **Final answer**:
\[ \mathbf{a} \cdot \mathbf{b} = -\sqrt{3} \]