Two vectors A and B are shown in the figure (not drawn to scale). Vector A has a magnitude rA = 15 units and an angle θA = 37o. Vector B has a magnitude rB = 40 units and an angle θB of 53o. 

Express these vectors in ijk unit vector notation, that is in the form A = Axi + Ayj and B = Bxi + Byj, where Ax, Ay, Bx, and By are the x and y components of vectors A and B, respectively;  and        find C = A -B. (Magnitude and direction)

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.

Learn how to express vectors A and B in ijk unit vector notation and calculate their magnitudes and direction. Vector A has a magnitude of 15 units at an angle of 37 degrees, while vector B has a magnitude of 40 units at an angle of 53 degrees. Find vector C = A - B, including its magnitude and direction. Suitable for Grades 11-12.

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