The resultant `vec(P) and vec(Q)` is perpendicular to `vec(P)`. What is the angle between `vec(P) and vec(Q)`?

Options

A. `cos^(-1)(P//Q)`
B. `cos^(-1)(-P//Q)`
C. `sin^(-1)(P//Q)`
D. `sin^(-1)(-P//Q)`

Explanation

To solve the problem, we need to find the angle between the vectors \(\vec{P}\) and \(\vec{Q}\) given that their resultant is perpendicular to \(\vec{P}\). Let's denote the angle between \(\vec{P}\) and \(\vec{Q}\) as \(\phi\).

### Step-by-Step Solution:

1. **Understanding the Given Information**:
We know that the resultant vector \(\vec{R} = \vec{P} + \vec{Q}\) is perpendicular to \(\vec{P}\). This means that the dot product of \(\vec{R}\) and \(\vec{P}\) is zero.
\[ \vec{R} \cdot \vec{P} = 0 \]

2. **Expressing the Resultant**:
The resultant vector can be expressed as:
\[ \vec{R} = \vec{P} + \vec{Q} \]

3. **Applying the Dot Product Condition**:
Since \(\vec{R}\) is perpendicular to \(\vec{P}\), we have:
\[ (\vec{P} + \vec{Q}) \cdot \vec{P} = 0 \]
Expanding this using the distributive property of the dot product gives:
\[ \vec{P} \cdot \vec{P} + \vec{Q} \cdot \vec{P} = 0 \]

4. **Using Magnitudes**:
The dot product \(\vec{P} \cdot \vec{P}\) is equal to \(|\vec{P}|^2\) (the magnitude of \(\vec{P}\) squared), and we denote the magnitude of \(\vec{Q}\) as \(q\) and the angle between \(\vec{P}\) and \(\vec{Q}\) as \(\phi\).
Thus, we can rewrite:
\[ |\vec{P}|^2 + |\vec{Q}| |\vec{P}| \cos(\phi) = 0 \]

5. **Rearranging the Equation**:
Rearranging gives:
\[ |\vec{Q}| |\vec{P}| \cos(\phi) = -|\vec{P}|^2 \]
Dividing both sides by \(|\vec{P}|\) (assuming \(|\vec{P}| \neq 0\)):
\[ |\vec{Q}| \cos(\phi) = -|\vec{P}| \]

6. **Finding the Angle**:
Since \(|\vec{Q}| \cos(\phi)\) is negative, \(\cos(\phi)\) must also be negative. This implies that \(\phi\) is in the second or third quadrant. The only angle that satisfies this condition while ensuring that the resultant is perpendicular to \(\vec{P}\) is:
\[ \phi = 90^\circ + \theta \]
where \(\theta\) is the angle between \(\vec{P}\) and the negative direction of \(\vec{Q}\).

7. **Conclusion**:
Therefore, the angle between \(\vec{P}\) and \(\vec{Q}\) is:
\[ \phi = 90^\circ \]