To solve this problem using the component method, we need to break each of the displacement vectors into its x- and y-components. Then, we can sum the components to find the components of the resultant vector and its magnitude and direction.

Step 1: Break the vectors into components.

Vector A (magnitude $A = 5.4 \, \text{m}$)

• The vector $A$ is going to the $+y$ axis and $-x$ axis. So, its components will be:
  • $A_x = -A \cdot \cos(45^\circ) = -5.4 \cdot \frac{\sqrt{2}}{2} = -3.82 \, \text{m}$
  • $A_y = A \cdot \sin(45^\circ) = 5.4 \cdot \frac{\sqrt{2}}{2} = 3.82 \, \text{m}$

Vector B (magnitude $B = 5.14 \, \text{m}$)

• The vector $B$ is going to the $+y$ axis and $+x$ axis. So, its components will be:
  • $B_x = B \cdot \cos(45^\circ) = 5.14 \cdot \frac{\sqrt{2}}{2} = 3.63 \, \text{m}$
  • $B_y = B \cdot \sin(45^\circ) = 5.14 \cdot \frac{\sqrt{2}}{2} = 3.63 \, \text{m}$

Vector C (magnitude $C = 3.45 \, \text{m}$)

• The vector $C$ is going straight down the $-y$ axis. So, its components are:
  • $C_x = 0 \, \text{m}$ (no displacement in the x-direction)
  • $C_y = -C = -3.45 \, \text{m}$

Step 2: Sum the components.

Now, sum the components in both the x- and y-directions to get the resultant vector components.

Resultant in the x-direction:

$R_x = A_x + B_x + C_x = -3.82 + 3.63 + 0 = -0.19 \, \text{m}$

Resultant in the y-direction:

$R_y = A_y + B_y + C_y = 3.82 + 3.63 - 3.45 = 4.00 \, \text{m}$

Step 3: Find the magnitude of the resultant vector.

The magnitude of the resultant vector $R$ is given by: $R = \sqrt{R_x^2 + R_y^2} = \sqrt{(-0.19)^2 + (4.00)^2} = \sqrt{0.0361 + 16.00} = \sqrt{16.0361} \approx 4.00 \, \text{m}$

Step 4: Find the directional angle.

The directional angle $\theta$ of the resultant vector is the angle above the negative x-axis (since $R_x$ is negative and $R_y$ is positive). We can find it using the tangent function: $\theta = \tan^{-1}\left(\frac{|R_y|}{|R_x|}\right) = \tan^{-1}\left(\frac{4.00}{0.19}\right) = \tan^{-1}(21.05) \approx 87.5^\circ$ This angle is above the negative x-axis, so it is measured from the negative x-axis to the positive y-direction.

Final Answer:

(a) The magnitude of the resultant vector is approximately 4.00 m.

(b) The directional angle of the resultant vector is approximately 87.5° above the negative x-axis.

Would you like more detailed steps or explanations on any part of this problem?

Here are 5 related questions you might find interesting:

1. How do you break a vector into components when it is not aligned along the coordinate axes?
2. What happens to the magnitude of the resultant vector if the vectors are collinear?
3. How can you calculate the angle of a resultant vector if you have the components along the x- and y-axes?
4. How do you handle vectors in three-dimensional space when using the component method?
5. What are the applications of the component method in real-world scenarios?

Tip: Always ensure that your angles are measured relative to the correct axis when finding directional angles.