If `vecP -vecQ - vecR=0` and the magnitudes of `vecP, vecQ and vecR` are 5 , 4 and 3 units respectively, the angle between `vecP and vecR` is

Options

A. `cos ^(-1) (3/5)`
B. `cos^(-1) (4/5)`
C. `pi/2`
D. `sin^(-1) (3/4)`

Answer

To solve the problem where we have the equation \( \vec{P} - \vec{Q} - \vec{R} = 0 \) and the magnitudes of \( \vec{P}, \vec{Q}, \) and \( \vec{R} \) are given as 5, 4, and 3 units respectively, we can follow these steps:

### Step-by-Step Solution:

1. **Rearranging the Equation**:
From the equation \( \vec{P} - \vec{Q} - \vec{R} = 0 \), we can rearrange it to:
\[ \vec{P} = \vec{Q} + \vec{R} \]
This means that vector \( \vec{P} \) is the resultant of vectors \( \vec{Q} \) and \( \vec{R} \).

2. **Using the Law of Cosines**:
The magnitude of the resultant vector \( \vec{P} \) can be calculated using the law of cosines:
\[ |\vec{P}|^2 = |\vec{Q}|^2 + |\vec{R}|^2 + 2 |\vec{Q}| |\vec{R}| \cos(\theta) \]
where \( \theta \) is the angle between vectors \( \vec{Q} \) and \( \vec{R} \).

3. **Substituting the Known Values**:
We know:
- \( |\vec{P}| = 5 \)
- \( |\vec{Q}| = 4 \)
- \( |\vec{R}| = 3 \)

Substituting these values into the equation gives:
\[ 5^2 = 4^2 + 3^2 + 2 \cdot 4 \cdot 3 \cdot \cos(\theta) \]
Simplifying this, we have:
\[ 25 = 16 + 9 + 24 \cos(\theta) \]
\[ 25 = 25 + 24 \cos(\theta) \]

4. **Solving for \( \cos(\theta) \)**:
Rearranging the equation:
\[ 0 = 24 \cos(\theta) \]
This implies:
\[ \cos(\theta) = 0 \]
Therefore, \( \theta = 90^\circ \).

5. **Finding the Angle Between \( \vec{P} \) and \( \vec{R} \)**:
Since we need to find the angle \( \alpha \) between \( \vec{P} \) and \( \vec{R} \), we can use the sine rule in the triangle formed by \( \vec{P}, \vec{Q}, \) and \( \vec{R} \).

From the triangle:
\[ \sin(\alpha) = \frac{|\vec{Q}|}{|\vec{P}|} = \frac{4}{5} \]
Thus, we can find \( \alpha \):
\[ \alpha = \sin^{-1}\left(\frac{4}{5}\right) \]

6. **Calculating \( \alpha \)**:
We can also find \( \cos(\alpha) \):
\[ \cos(\alpha) = \frac{|\vec{R}|}{|\vec{P}|} = \frac{3}{5} \]
Therefore, \( \alpha \) can also be expressed as:
\[ \alpha = \cos^{-1}\left(\frac{3}{5}\right) \]

### Final Answer:
The angle between \( \vec{P} \) and \( \vec{R} \) is:
\[ \alpha = \sin^{-1}\left(\frac{4}{5}\right) \text{ or } \alpha = \cos^{-1}\left(\frac{3}{5}\right) \]