Two vectors have magnitudes 2 m and 3m. The angle between them is `60^0`. Find a the scalar product of the two vectors b. the magnitude of their vector product.

Explanation

To solve the problem, we will find the scalar product (dot product) and the magnitude of the vector product (cross product) of the two vectors step by step.

### Step 1: Find the Scalar Product (Dot Product)

The formula for the scalar product (dot product) of two vectors \( \vec{A} \) and \( \vec{B} \) is given by:

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

Where:
- \( |\vec{A}| = 2 \, \text{m} \) (magnitude of the first vector)
- \( |\vec{B}| = 3 \, \text{m} \) (magnitude of the second vector)
- \( \theta = 60^\circ \) (angle between the two vectors)

Substituting the values into the formula:

\[ \vec{A} \cdot \vec{B} = 2 \times 3 \times \cos(60^\circ) \]

We know that \( \cos(60^\circ) = \frac{1}{2} \), so:

\[ \vec{A} \cdot \vec{B} = 2 \times 3 \times \frac{1}{2} = 3 \]

Thus, the scalar product of the two vectors is:

\[ \vec{A} \cdot \vec{B} = 3 \]

### Step 2: Find the Magnitude of the Vector Product (Cross Product)

The formula for the magnitude of the vector product (cross product) of two vectors \( \vec{A} \) and \( \vec{B} \) is given by:

\[ |\vec{A} \times \vec{B}| = |\vec{A}| |\vec{B}| \sin(\theta) \]

Substituting the values into the formula:

\[ |\vec{A} \times \vec{B}| = 2 \times 3 \times \sin(60^\circ) \]

We know that \( \sin(60^\circ) = \frac{\sqrt{3}}{2} \), so:

\[ |\vec{A} \times \vec{B}| = 2 \times 3 \times \frac{\sqrt{3}}{2} \]

Simplifying this gives:

\[ |\vec{A} \times \vec{B}| = 3\sqrt{3} \]

Thus, the magnitude of the vector product of the two vectors is:

\[ |\vec{A} \times \vec{B}| = 3\sqrt{3} \]

### Final Answers:
a. The scalar product of the two vectors is \( 3 \).

b. The magnitude of their vector product is \( 3\sqrt{3} \).

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