If the magnitudes of two vectors are `2 and 3` and the magnitude of their scalar product is `3 sqrt 2`, then find the angle between the vectors.

Options

A. `45`
B. `60`
C. `120`
D. `150`

Explanation

To solve the problem, we need to find the angle between two vectors given their magnitudes and the magnitude of their scalar (dot) product.

### Step-by-Step Solution:

1. **Identify the given values:**
- Magnitude of vector A (|A|) = 2
- Magnitude of vector B (|B|) = 3
- Magnitude of the scalar product (A · B) = 3√2

2. **Use the formula for the dot product:**
The dot product of two vectors A and B can be expressed as:
\[ A \cdot B = |A| |B| \cos \theta \]
where θ is the angle between the vectors.

3. **Substitute the known values into the dot product formula:**
\[ 3\sqrt{2} = (2)(3) \cos \theta \]

4. **Simplify the equation:**
\[ 3\sqrt{2} = 6 \cos \theta \]

5. **Solve for cos θ:**
\[ \cos \theta = \frac{3\sqrt{2}}{6} \]
\[ \cos \theta = \frac{\sqrt{2}}{2} \]

6. **Find the angle θ:**
The angle whose cosine is \(\frac{\sqrt{2}}{2}\) is:
\[ \theta = 45^\circ \]

### Final Answer:
The angle between the vectors is \(45^\circ\).

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