A particle travels through a three-dimensional displacement given by `vec(d)=(5.00 hat(i) - 3.00 hat(j) + 4.00 hat(k))m`. If a force of magnitude 22.0 N and with fixed orientation does work on the particle, find the angle between the force and the displacement if the change in the particle's kinetic energy is (a) 45.0 J and (b) -45.0 J.

Answer

To solve the problem, we will use the work-energy theorem, which states that the work done by a force on an object is equal to the change in its kinetic energy. The work done (W) can also be expressed as the dot product of the force vector (F) and the displacement vector (d).

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Displacement vector: \(\vec{d} = 5.00 \hat{i} - 3.00 \hat{j} + 4.00 \hat{k}\) m
- Magnitude of force: \(F = 22.0\) N
- Change in kinetic energy for part (a): \(\Delta K = 45.0\) J
- Change in kinetic energy for part (b): \(\Delta K = -45.0\) J

2. **Calculate the Magnitude of the Displacement:**
\[ |\vec{d}| = \sqrt{(5.00)^2 + (-3.00)^2 + (4.00)^2} = \sqrt{25 + 9 + 16} = \sqrt{50} \approx 7.07 \text{ m} \]

3. **Use the Work-Energy Theorem:**
The work done by the force can be expressed as:
\[ W = \Delta K = |\vec{F}| |\vec{d}| \cos \theta \]
Where \(\theta\) is the angle between the force and displacement vectors.

4. **Calculate the Cosine of the Angle for Part (a):**
Substitute the known values into the equation:
\[ 45.0 = 22.0 \times 7.07 \cos \theta \]
Rearranging gives:
\[ \cos \theta = \frac{45.0}{22.0 \times 7.07} \]
Calculate:
\[ \cos \theta = \frac{45.0}{154.54} \approx 0.291 \]

5. **Find the Angle \(\theta\) for Part (a):**
\[ \theta = \cos^{-1}(0.291) \approx 73.2^\circ \]

6. **Calculate the Cosine of the Angle for Part (b):**
For part (b), the change in kinetic energy is negative:
\[ -45.0 = 22.0 \times 7.07 \cos \theta \]
Rearranging gives:
\[ \cos \theta = \frac{-45.0}{22.0 \times 7.07} \]
Calculate:
\[ \cos \theta = \frac{-45.0}{154.54} \approx -0.291 \]

7. **Find the Angle \(\theta\) for Part (b):**
\[ \theta = \cos^{-1}(-0.291) \approx 107.0^\circ \]

### Final Answers:
- For part (a): The angle \(\theta \approx 73.2^\circ\)
- For part (b): The angle \(\theta \approx 107.0^\circ\)