The resultant of `vecP` and `vecQ` is perpendicular to `vecP`. What is the angle between `vecP` and `vecQ`

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 vectors \(\vec{P}\) and \(\vec{Q}\) given that their resultant \(\vec{R}\) is perpendicular to \(\vec{P}\). Here’s a step-by-step solution:

### Step 1: Understand the relationship between the vectors
Given that the resultant vector \(\vec{R}\) of \(\vec{P}\) and \(\vec{Q}\) is perpendicular to \(\vec{P}\), we can write:
\[ \vec{R} = \vec{P} + \vec{Q} \]
and since \(\vec{R}\) is perpendicular to \(\vec{P}\), we have:
\[ \vec{R} \cdot \vec{P} = 0 \]

### Step 2: Use the dot product condition
From the dot product condition, we can expand it:
\[ (\vec{P} + \vec{Q}) \cdot \vec{P} = 0 \]
This simplifies to:
\[ \vec{P} \cdot \vec{P} + \vec{Q} \cdot \vec{P} = 0 \]
Let \(P = |\vec{P}|\) and \(Q = |\vec{Q}|\). Then:
\[ P^2 + Q \cdot P \cos(\theta) = 0 \]
where \(\theta\) is the angle between \(\vec{P}\) and \(\vec{Q}\).

### Step 3: Rearranging the equation
Rearranging the equation gives:
\[ Q \cdot P \cos(\theta) = -P^2 \]
This leads to:
\[ \cos(\theta) = -\frac{P}{Q} \]

### Step 4: Finding the angle
Now, we can express the angle \(\theta\) as:
\[ \theta = \cos^{-1}\left(-\frac{P}{Q}\right) \]

### Conclusion
Thus, the angle between \(\vec{P}\) and \(\vec{Q}\) is:
\[ \theta = \cos^{-1}\left(-\frac{P}{Q}\right) \]

### Final Answer
The angle between \(\vec{P}\) and \(\vec{Q}\) is \(\cos^{-1}\left(-\frac{P}{Q}\right)\).

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