If vectors `vecP, vecQ and vecR` have magnitudes 5, 12 and 13 units and `vecP + vecQ = vecR`, the angle between `vecQ and vecR` is :

Options

A. `MLT^(-1)C^(-1)`
B. `MT^(2)C^(-2)`
C. `MT^(-1)C^(-1)`
D. `MT^(-2)C^(-1)`

Explanation

To solve the problem, we need to find the angle between the vectors \(\vec{Q}\) and \(\vec{R}\) given that \(\vec{P} + \vec{Q} = \vec{R}\) and their magnitudes are as follows:

- Magnitude of \(\vec{P} = 5\) units
- Magnitude of \(\vec{Q} = 12\) units
- Magnitude of \(\vec{R} = 13\) units

### Step-by-step Solution:

1. **Express \(\vec{P}\) in terms of \(\vec{Q}\) and \(\vec{R}\)**:
\[ \vec{P} = \vec{R} - \vec{Q} \]

2. **Use the magnitude relation**:
We take the magnitude of both sides:
\[ |\vec{P}| = |\vec{R} - \vec{Q}| \]
This gives us:
\[ 5 = |\vec{R} - \vec{Q}| \]

3. **Apply the formula for the magnitude of the difference of two vectors**:
The magnitude of the difference of two vectors can be expressed as:
\[ |\vec{R} - \vec{Q}| = \sqrt{|\vec{R}|^2 + |\vec{Q}|^2 - 2 |\vec{R}| |\vec{Q}| \cos(\theta)} \]
where \(\theta\) is the angle between \(\vec{R}\) and \(\vec{Q}\).

4. **Substitute the known magnitudes**:
Substitute \( |\vec{R}| = 13 \), \( |\vec{Q}| = 12 \), and \( |\vec{P}| = 5 \):
\[ 5 = \sqrt{13^2 + 12^2 - 2 \cdot 13 \cdot 12 \cdot \cos(\theta)} \]

5. **Square both sides**:
\[ 25 = 169 + 144 - 2 \cdot 13 \cdot 12 \cdot \cos(\theta) \]

6. **Simplify the equation**:
\[ 25 = 313 - 312 \cos(\theta) \]
Rearranging gives:
\[ 312 \cos(\theta) = 313 - 25 \]
\[ 312 \cos(\theta) = 288 \]

7. **Solve for \(\cos(\theta)\)**:
\[ \cos(\theta) = \frac{288}{312} \]
Simplifying this fraction:
\[ \cos(\theta) = \frac{12}{13} \]

8. **Find the angle \(\theta\)**:
\[ \theta = \cos^{-1}\left(\frac{12}{13}\right) \]

### Final Answer:
The angle between \(\vec{Q}\) and \(\vec{R}\) is \(\theta = \cos^{-1}\left(\frac{12}{13}\right)\).