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.