If vector `P, Q and R` have magnitude 5,12,and 13 units and `vec(P)+vec(Q)=vec(R )`, the angle between Q and R is

Options

A. `cos^(-1)(5)/(12)`
B. `cos^(-1)(5)/(13)`
C. `cos^(-1)(12)/(13)`
D. `cos^(-1)(7)/(13)`

Explanation

To solve the problem, we need to find the angle between vectors Q and R given the magnitudes of vectors P, Q, and R, and the relationship \( \vec{P} + \vec{Q} = \vec{R} \).

### Step-by-Step Solution:

1. **Identify the Magnitudes:**
- Given magnitudes are:
- \( |\vec{P}| = 5 \)
- \( |\vec{Q}| = 12 \)
- \( |\vec{R}| = 13 \)

2. **Use the Vector Addition Relationship:**
- From the problem, we know that:
\[ \vec{P} + \vec{Q} = \vec{R} \]

3. **Visualize the Vectors:**
- We can visualize the vectors as forming a triangle. Here, \( \vec{P} \) and \( \vec{Q} \) are two sides of the triangle, and \( \vec{R} \) is the resultant side.

4. **Check for Right Triangle:**
- To confirm if it forms a right triangle, we can use the Pythagorean theorem:
\[ |\vec{R}|^2 = |\vec{P}|^2 + |\vec{Q}|^2 \]
- Substitute the values:
\[ 13^2 = 5^2 + 12^2 \]
\[ 169 = 25 + 144 \]
\[ 169 = 169 \]
- Since the equation holds true, \( \vec{P}, \vec{Q}, \) and \( \vec{R} \) form a right triangle.

5. **Determine the Angle Between Q and R:**
- In a right triangle, the angle opposite the side of length 5 (which is \( |\vec{P}| \)) is the angle we are interested in. We can use cosine to find the angle \( \theta \) between \( \vec{Q} \) and \( \vec{R} \):
\[ \cos(\theta) = \frac{|\vec{Q}|}{|\vec{R}|} \]
- Substitute the magnitudes:
\[ \cos(\theta) = \frac{12}{13} \]

6. **Calculate the Angle:**
- To find \( \theta \), take the inverse cosine:
\[ \theta = \cos^{-1}\left(\frac{12}{13}\right) \]

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