To find the scalar (dot) products of vectors, we use the formula:

$A \cdot B = |A| |B| \cos(\theta)$

where $|A|$ and $|B|$ are the magnitudes of vectors $A$ and $B$, and $\theta$ is the angle between the vectors.

Part 1: Find $A \cdot B$

Given:

• Magnitude of $A = 5.50$ m
• Magnitude of $B = 10.0$ m
• Angle between $A$ and $B$ is $\theta = 30^\circ$

The scalar product $A \cdot B$ is:

$A \cdot B = |A| |B| \cos(\theta) = 5.50 \times 10.0 \times \cos(30^\circ)$

$\cos(30^\circ) = \frac{\sqrt{3}}{2}$

$A \cdot B = 5.50 \times 10.0 \times \frac{\sqrt{3}}{2} = 55 \times \frac{\sqrt{3}}{2}$

Let's calculate this.

Part 2: Find $A \cdot C$

Given:

• Magnitude of $A = 5.50$ m
• Magnitude of $C = 13.1$ m
• Angle between $A$ and $C$ is $\theta = 37^\circ$

The scalar product $A \cdot C$ is:

$A \cdot C = |A| |C| \cos(\theta) = 5.50 \times 13.1 \times \cos(37^\circ)$

$\cos(37^\circ) \approx 0.7986$

$A \cdot C = 5.50 \times 13.1 \times 0.7986$

Part 3: Find $2B \cdot C$

Given:

• Magnitude of $B = 10.0$ m
• Magnitude of $C = 13.1$ m
• Angle between $B$ and $C$ is $\theta = 60^\circ$

The scalar product $2B \cdot C$ is:

$2B \cdot C = 2 \times |B| |C| \cos(\theta) = 2 \times 10.0 \times 13.1 \times \cos(60^\circ)$

$\cos(60^\circ) = \frac{1}{2}$

$2B \cdot C = 2 \times 10.0 \times 13.1 \times \frac{1}{2} = 10.0 \times 13.1$

$2B \cdot C = 131$

Let's calculate these values.Here are the results for each part:

Part 1: $A \cdot B$

$A \cdot B = 47.63 \, \text{m}^2$

Part 2: $A \cdot C$

$A \cdot C = 57.54 \, \text{m}^2$

Part 3: $2B \cdot C$

$2B \cdot C = 131.00 \, \text{m}^2$

These calculations use the dot product formula and the given magnitudes and angles.

Do you need more details or have any questions?

Here are some related questions that you might find helpful:

1. What is the dot product of two perpendicular vectors?
2. How do you find the angle between two vectors using the dot product?
3. How can the scalar product be used in physics to calculate work done by a force?
4. What is the geometric interpretation of a scalar product?
5. How does the dot product differ from the cross product in vector mathematics?

Tip: Remember that the dot product can tell you whether two vectors are orthogonal (perpendicular) when the result is zero!