Three vectors `vecP, vecQ, vecR` obey`P^(2) + Q^(2) =R^(2)` angle between `vecP` & `vecQ` . is

Options

A. `0^(@)`
B. `30^(@)`
C. `60^(@)`
D. `90^(@)`

Answer

To solve the problem, we need to find the angle between the vectors \(\vec{P}\) and \(\vec{Q}\) given that they satisfy the equation \(P^2 + Q^2 = R^2\).

### Step-by-Step Solution:

1. **Understanding the Given Equation**:
The equation \(P^2 + Q^2 = R^2\) resembles the Pythagorean theorem, which suggests that the vectors \(\vec{P}\) and \(\vec{Q}\) are perpendicular to each other.

2. **Defining the Angle**:
Let the angle between the vectors \(\vec{P}\) and \(\vec{Q}\) be \(\theta\).

3. **Using the Law of Cosines**:
According to the law of cosines, the square of the magnitude of the resultant vector \(\vec{R}\) can be expressed as:
\[ R^2 = P^2 + Q^2 + 2PQ \cos(\theta) \]
Here, \(P\) and \(Q\) are the magnitudes of vectors \(\vec{P}\) and \(\vec{Q}\), respectively.

4. **Equating the Two Expressions for \(R^2\)**:
From the given equation, we have:
\[ R^2 = P^2 + Q^2 \]
We can set the two expressions for \(R^2\) equal to each other:
\[ P^2 + Q^2 = P^2 + Q^2 + 2PQ \cos(\theta) \]

5. **Simplifying the Equation**:
By subtracting \(P^2 + Q^2\) from both sides, we get:
\[ 0 = 2PQ \cos(\theta) \]

6. **Analyzing the Result**:
Since \(P\) and \(Q\) are not equal to zero (as they are magnitudes of vectors), we can divide both sides by \(2PQ\):
\[ \cos(\theta) = 0 \]

7. **Finding the Angle**:
The cosine of an angle is zero at \(\theta = \frac{\pi}{2}\) radians, which corresponds to \(90^\circ\).

### Conclusion:
The angle between the vectors \(\vec{P}\) and \(\vec{Q}\) is \(90^\circ\).