To solve this problem, we'll break it down into a few steps:

Step 1: Express Vectors A and B in Component Form

Given:

• Vector $\mathbf{A}$:
  • Magnitude $r_A = 15$ units
  • Angle $\theta_A = 37^\circ$
• Vector $\mathbf{B}$:
  • Magnitude $r_B = 40$ units
  • Angle $\theta_B = 53^\circ$

For a vector $\mathbf{V}$ with magnitude $r_V$ and angle $\theta_V$, the components are given by: $V_x = r_V \cos(\theta_V)$ $V_y = r_V \sin(\theta_V)$

Vector A Components:

$A_x = 15 \cos(37^\circ)$ $A_y = 15 \sin(37^\circ)$

Vector B Components:

$B_x = 40 \cos(53^\circ)$ $B_y = 40 \sin(53^\circ)$

Let's calculate these components.

Step 2: Calculate the Components

For Vector A:

$A_x = 15 \cos(37^\circ) \approx 15 \times 0.7986 \approx 11.979$ $A_y = 15 \sin(37^\circ) \approx 15 \times 0.6018 \approx 9.027$

So, $\mathbf{A}$ in component form is: $\mathbf{A} = 11.979\hat{i} + 9.027\hat{j}$

For Vector B:

$B_x = 40 \cos(53^\circ) \approx 40 \times 0.6018 \approx 24.072$ $B_y = 40 \sin(53^\circ) \approx 40 \times 0.7986 \approx 31.944$

So, $\mathbf{B}$ in component form is: $\mathbf{B} = 24.072\hat{i} + 31.944\hat{j}$

Step 3: Find Vector $\mathbf{C} = \mathbf{A} - \mathbf{B}$

Subtract the components of vector $\mathbf{B}$ from vector $\mathbf{A}$: $C_x = A_x - B_x = 11.979 - 24.072 \approx -12.093$ $C_y = A_y - B_y = 9.027 - 31.944 \approx -22.917$

So, vector $\mathbf{C}$ in component form is: $\mathbf{C} = -12.093\hat{i} - 22.917\hat{j}$

Step 4: Calculate the Magnitude and Direction of Vector $\mathbf{C}$

Magnitude of $\mathbf{C}$:

$|\mathbf{C}| = \sqrt{C_x^2 + C_y^2} = \sqrt{(-12.093)^2 + (-22.917)^2} \approx \sqrt{146.24 + 525.17} \approx \sqrt{671.41} \approx 25.9 \text{ units}$

Direction of $\mathbf{C}$ (Angle with the positive x-axis):

$\theta_C = \tan^{-1}\left(\frac{C_y}{C_x}\right) = \tan^{-1}\left(\frac{-22.917}{-12.093}\right) \approx \tan^{-1}(1.895) \approx 62.1^\circ$

Since both $C_x$ and $C_y$ are negative, $\theta_C$ is in the third quadrant. The angle in the third quadrant relative to the positive x-axis is: $\theta_C \approx 180^\circ + 62.1^\circ = 242.1^\circ$

Final Answer:

• The vectors in component form are: $\mathbf{A} = 11.979\hat{i} + 9.027\hat{j}$ $\mathbf{B} = 24.072\hat{i} + 31.944\hat{j}$
• The vector $\mathbf{C} = \mathbf{A} - \mathbf{B}$ has a magnitude of approximately 25.9 units and a direction of $242.1^\circ$ relative to the positive x-axis.

Would you like more details, or do you have any questions?

Related Questions:

1. How would you find the vector sum $\mathbf{A} + \mathbf{B}$?
2. What is the physical significance of vector subtraction in real-world applications?
3. How can you express the direction of vector $\mathbf{C}$ using other trigonometric functions?
4. How does the change in the angle $\theta_A$ or $\theta_B$ affect the resultant vector?
5. Can vector components be negative, and what does that imply about the direction of the vector?

Tip:

Always visualize the vectors and their components on the Cartesian plane to better understand their resultant magnitudes and directions.