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The problem describes two vectors \( \mathbf{A} \) and \( \mathbf{B} \) as follows:

- **Vector \( \mathbf{A} \)**: 
  - Magnitude: 2 inches.
  - Direction: \( 60^\circ \) above the x-axis (in the first quadrant).
  
- **Vector \( \mathbf{B} \)**: 
  - Magnitude: 2 inches.
  - Direction: \( 60^\circ \) below the x-axis (in the fourth quadrant).

You are asked to find graphically:
1. The vector sum \( \mathbf{A} + \mathbf{B} \).
2. The vector differences \( \mathbf{A} - \mathbf{B} \) and \( \mathbf{B} - \mathbf{A} \).

### Solution Breakdown:
We can break down each vector into its components and then use vector addition/subtraction principles. The graphical method would also involve these components.

#### 1. Components of Vector \( \mathbf{A} \):
Since \( \mathbf{A} \) is 60° above the x-axis in the first quadrant, its components are:
- \( A_x = A \cos(60^\circ) = 2 \times \frac{1}{2} = 1 \) inch (positive x-component).
- \( A_y = A \sin(60^\circ) = 2 \times \frac{\sqrt{3}}{2} = \sqrt{3} \) inches (positive y-component).

Thus, \( \mathbf{A} = (1, \sqrt{3}) \) inches.

#### 2. Components of Vector \( \mathbf{B} \):
Since \( \mathbf{B} \) is 60° below the x-axis in the fourth quadrant, its components are:
- \( B_x = B \cos(60^\circ) = 2 \times \frac{1}{2} = 1 \) inch (positive x-component).
- \( B_y = B \sin(60^\circ) = 2 \times \frac{\sqrt{3}}{2} = \sqrt{3} \) inches (negative y-component).

Thus, \( \mathbf{B} = (1, -\sqrt{3}) \) inches.

### a) Vector Sum \( \mathbf{A} + \mathbf{B} \):
Adding the components:
- \( (\mathbf{A} + \mathbf{B})_x = 1 + 1 = 2 \) inches.
- \( (\mathbf{A} + \mathbf{B})_y = \sqrt{3} + (-\sqrt{3}) = 0 \) inches.

Thus, \( \mathbf{A} + \mathbf{B} = (2, 0) \) inches, which is a vector of magnitude 2 inches along the x-axis.

### b) Vector Differences \( \mathbf{A} - \mathbf{B} \) and \( \mathbf{B} - \mathbf{A} \):

1. **\( \mathbf{A} - \mathbf{B} \)**:
   Subtracting the components:
   - \( (\mathbf{A} - \mathbf{B})_x = 1 - 1 = 0 \) inches.
   - \( (\mathbf{A} - \mathbf{B})_y = \sqrt{3} - (-\sqrt{3}) = 2\sqrt{3} \) inches.

   Thus, \( \mathbf{A} - \mathbf{B} = (0, 2\sqrt{3}) \) inches, which is a vertical vector of magnitude \( 2\sqrt{3} \) inches.

2. **\( \mathbf{B} - \mathbf{A} \)**:
   Subtracting the components:
   - \( (\mathbf{B} - \mathbf{A})_x = 1 - 1 = 0 \) inches.
   - \( (\mathbf{B} - \mathbf{A})_y = -\sqrt{3} - \sqrt{3} = -2\sqrt{3} \) inches.

   Thus, \( \mathbf{B} - \mathbf{A} = (0, -2\sqrt{3}) \) inches, which is a vertical vector in the negative y-direction with magnitude \( 2\sqrt{3} \) inches.

### Summary:
- \( \mathbf{A} + \mathbf{B} = (2, 0) \) inches.
- \( \mathbf{A} - \mathbf{B} = (0, 2\sqrt{3}) \) inches.
- \( \mathbf{B} - \mathbf{A} = (0, -2\sqrt{3}) \) inches.

Would you like further details or have any questions?

### Related Questions:
1. What is the general process to find the magnitude and direction of a vector?
2. How does vector subtraction differ from vector addition graphically?
3. What are the differences between first quadrant and fourth quadrant vectors?
4. How would changing the angle of vectors \( \mathbf{A} \) or \( \mathbf{B} \) affect the resultant sum?
5. Can you solve vector addition problems using trigonometry instead of graphical methods?

### Tip:
Always ensure that you correctly account for the signs of vector components based on their direction in the coordinate plane.

Vector A is 2 in. long and is 60° above the x-axis in the first quadrant. Vector B is 2 in. long and is 60° below the x-axis in the fourth quadrant. Find graphically: a) the vector sum A + B, and b) the vector differences A - B and B - A.

This problem involves two vectors, A and B, with equal magnitudes but different directions. Vector A is in the first quadrant, and Vector B is in the fourth quadrant. Learn how to find the vector sum and differences graphically using trigonometry and vector components.

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