To find the resultant vector of the three displacement vectors $\mathbf{A}$, $\mathbf{B}$, and $\mathbf{C}$ using the component method, we need to break down each vector into its components along the x and y axes, then sum up the components to get the resultant vector's components. Finally, we can calculate the magnitude and direction of the resultant vector.

1. Breaking down the vectors into components:

Vector $\mathbf{A}$:

• $\mathbf{A}$ has a magnitude of 5.4 m, and it points in the positive y direction and negative x direction. Thus, we can write:
  • $A_x = -5.4 \, \text{m}$ (negative because it points in the negative x direction)
  • $A_y = 5.4 \, \text{m}$ (positive because it points in the positive y direction)

Vector $\mathbf{B}$:

• $\mathbf{B}$ has a magnitude of 5.14 m, and it points in the positive y direction and positive x direction. Thus, we can write:
  • $B_x = 5.14 \, \text{m}$ (positive because it points in the positive x direction)
  • $B_y = 5.14 \, \text{m}$ (positive because it points in the positive y direction)

Vector $\mathbf{C}$:

• $\mathbf{C}$ has a magnitude of 3.45 m, and it points directly downward along the negative y axis. Thus, we can write:
  • $C_x = 0 \, \text{m}$ (no x component because it points vertically)
  • $C_y = -3.45 \, \text{m}$ (negative because it points in the negative y direction)

2. Summing the components:

Now, we sum the x and y components of all three vectors:

x-components of the resultant vector:

$R_x = A_x + B_x + C_x = -5.4 + 5.14 + 0 = -0.26 \, \text{m}$

y-components of the resultant vector:

$R_y = A_y + B_y + C_y = 5.4 + 5.14 - 3.45 = 7.09 \, \text{m}$

3. Calculating the magnitude of the resultant vector:

The magnitude of the resultant vector $\mathbf{R}$ is given by: $R = \sqrt{R_x^2 + R_y^2} = \sqrt{(-0.26)^2 + (7.09)^2} = \sqrt{0.0676 + 50.4081} = \sqrt{50.4757} \approx 7.11 \, \text{m}$

4. Calculating the directional angle of the resultant vector:

The direction $\theta$ of the resultant vector relative to the positive x-axis is given by: $\theta = \tan^{-1} \left( \frac{R_y}{R_x} \right)$ Since $R_x$ is negative and $R_y$ is positive, the resultant vector is in the second quadrant, and we can compute: $\theta = \tan^{-1} \left( \frac{7.09}{-0.26} \right) = \tan^{-1}(-27.27)$ This gives an angle of approximately: $\theta \approx -88.4^\circ$ Since we need an angle above the positive or negative x-axis, we convert this to a positive angle in the second quadrant: $\theta = 180^\circ - 88.4^\circ = 91.6^\circ$ Thus, the angle is approximately $91.6^\circ$ above the negative x-axis.

Final Answer:

• (a) The magnitude of the resultant vector is approximately 7.11 m.
• (b) The directional angle of the resultant vector is approximately 91.6° above the negative x-axis.

Would you like further clarification or details on any part of this solution?

Here are some related questions you might find interesting:

1. How do you break vectors into components in 3D space?
2. What are some methods to find the direction of a vector in physics?
3. How can the dot product be used to find angles between vectors?
4. What is the difference between vector addition in Cartesian and polar coordinates?
5. Can you solve for the resultant when vectors have both x, y, and z components?

Tip: When dealing with vector components, always ensure you correctly handle signs and directions, especially when the vectors are in different quadrants!