what is the dot product of two vectors of magnitudes 3 and 5,if angle between them is `60^@`?

Options

A. 5.2
B. 7.5
C. 8.4
D. 8.6

Explanation

To find the dot product of two vectors given their magnitudes and the angle between them, we can use the formula:

\[ \mathbf{A} \cdot \mathbf{B} = |\mathbf{A}| |\mathbf{B}| \cos(\theta) \]

Where:
- \(\mathbf{A} \cdot \mathbf{B}\) is the dot product of vectors \(\mathbf{A}\) and \(\mathbf{B}\).
- \(|\mathbf{A}|\) is the magnitude of vector \(\mathbf{A}\).
- \(|\mathbf{B}|\) is the magnitude of vector \(\mathbf{B}\).
- \(\theta\) is the angle between the two vectors.

### Step-by-step Solution:

1. **Identify the magnitudes and angle:**
- Magnitude of vector \(\mathbf{A}\) is \(|\mathbf{A}| = 3\)
- Magnitude of vector \(\mathbf{B}\) is \(|\mathbf{B}| = 5\)
- Angle between the vectors is \(\theta = 60^\circ\)

2. **Substitute the values into the dot product formula:**
\[ \mathbf{A} \cdot \mathbf{B} = 3 \times 5 \times \cos(60^\circ) \]

3. **Calculate \(\cos(60^\circ)\):**
- We know that \(\cos(60^\circ) = \frac{1}{2}\)

4. **Substitute \(\cos(60^\circ)\) into the equation:**
\[ \mathbf{A} \cdot \mathbf{B} = 3 \times 5 \times \frac{1}{2} \]

5. **Perform the multiplication:**
\[ \mathbf{A} \cdot \mathbf{B} = 15 \times \frac{1}{2} = 7.5 \]

6. **Conclusion:**
The dot product of the two vectors is \(7.5\).

### Final Answer:
\[ \mathbf{A} \cdot \mathbf{B} = 7.5 \]