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What is the magnitude of the x component of vector A?


What is the magnitude of the y component of vector A?


What is the magnitude of the x component of vector B?


What is the magnitude of the y component of vector B?
 

What is the overall magnitude of vector C? 

What angle above the horizontal is vector C directed?

To solve this, let's break down the problem step by step using vector components.

### Step 1: Analyze the given vectors:
- **Vector A**: 50 km at 60° northeast
- **Vector B**: 20 km at 20° northeast

We need to calculate the components of these vectors in the x and y directions using trigonometry.

### Step 2: Vector Components

1. **Magnitude of the x component of vector A**:
   The x-component of vector A is given by:
   \[
   A_x = |A| \cdot \cos(\theta_A)
   \]
   where \(|A| = 50 \, \text{km}\) and \(\theta_A = 60^\circ\).
   \[
   A_x = 50 \cdot \cos(60^\circ) = 50 \cdot 0.5 = 25.0 \, \text{km}
   \]

2. **Magnitude of the y component of vector A**:
   The y-component of vector A is:
   \[
   A_y = |A| \cdot \sin(\theta_A)
   \]
   \[
   A_y = 50 \cdot \sin(60^\circ) = 50 \cdot 0.866 = 43.3 \, \text{km}
   \]

3. **Magnitude of the x component of vector B**:
   The x-component of vector B is:
   \[
   B_x = |B| \cdot \cos(\theta_B)
   \]
   where \(|B| = 20 \, \text{km}\) and \(\theta_B = 20^\circ\).
   \[
   B_x = 20 \cdot \cos(20^\circ) = 20 \cdot 0.9397 = 18.8 \, \text{km}
   \]

4. **Magnitude of the y component of vector B**:
   The y-component of vector B is:
   \[
   B_y = |B| \cdot \sin(\theta_B)
   \]
   \[
   B_y = 20 \cdot \sin(20^\circ) = 20 \cdot 0.3420 = 6.8 \, \text{km}
   \]

### Step 3: Resultant Vector C

Now, to find the overall magnitude and direction of vector C:

- The x-component of vector C is:
  \[
  C_x = A_x + B_x = 25.0 + 18.8 = 43.8 \, \text{km}
  \]

- The y-component of vector C is:
  \[
  C_y = A_y + B_y = 43.3 + 6.8 = 50.1 \, \text{km}
  \]

5. **Magnitude of vector C**:
   The magnitude of the resultant vector C is:
   \[
   |C| = \sqrt{C_x^2 + C_y^2}
   \]
   \[
   |C| = \sqrt{43.8^2 + 50.1^2} = \sqrt{1916.64 + 2510.01} = \sqrt{4426.65} \approx 66.5 \, \text{km}
   \]

6. **Direction of vector C**:
   The angle above the horizontal (relative to the x-axis) is:
   \[
   \theta_C = \tan^{-1}\left(\frac{C_y}{C_x}\right)
   \]
   \[
   \theta_C = \tan^{-1}\left(\frac{50.1}{43.8}\right) \approx \tan^{-1}(1.14) = 48.6^\circ
   \]

### Final Answers:
1. **x-component of vector A**: \( 25.0 \, \text{km} \)
2. **y-component of vector A**: \( 43.3 \, \text{km} \)
3. **x-component of vector B**: \( 18.8 \, \text{km} \)
4. **y-component of vector B**: \( 6.8 \, \text{km} \)
5. **Overall magnitude of vector C**: \( 66.5 \, \text{km} \)
6. **Direction of vector C**: \( 48.6^\circ \) above the horizontal

Let me know if you'd like further clarification, details, or if you have any questions!

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### Follow-up Questions:
1. How would the resultant change if vector B had a larger magnitude?
2. What if vector A had an angle of 45° instead of 60°?
3. How do you compute the components for a vector in 3D space?
4. Can you explain the significance of vector addition in physics?
5. How would you subtract two vectors geometrically?

**Tip**: Breaking vectors into components is one of the most useful tools for solving problems in physics and engineering, especially when dealing with forces or velocities!

What is the magnitude of the x component of vector A? What is the magnitude of the y component of vector A? What is the magnitude of the x component of vector B? What is the magnitude of the y component of vector B? What is the overall magnitude of vector C? What angle above the horizontal is vector C directed?

Learn how to calculate the components and overall magnitude of vectors A and B, and determine the direction of the resultant vector C. This problem involves vector components, trigonometry, and vector addition, suitable for students in Grades 10-12.

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