Magnitude of resultant of two vectors `vecP` and `vecQ` is equal to magnitude of `vecP` . Find the angle between `vecQ` and resultant of `vec2P` and `vecQ` .

Explanation

To solve the problem, we need to find the angle between the vector \(\vec{Q}\) and the resultant of the vectors \(\vec{2P}\) and \(\vec{Q}\), given that the magnitude of the resultant of the vectors \(\vec{P}\) and \(\vec{Q}\) is equal to the magnitude of \(\vec{P}\).

### Step-by-Step Solution:

1. **Understand the Given Information**:
We have two vectors \(\vec{P}\) and \(\vec{Q}\). The magnitude of the resultant vector \(\vec{R}\) formed by \(\vec{P}\) and \(\vec{Q}\) is given as:
\[ |\vec{R}| = |\vec{P}| \]

2. **Use the Formula for Resultant**:
The magnitude of the resultant vector \(\vec{R}\) can be expressed using the formula:
\[ |\vec{R}| = \sqrt{|\vec{P}|^2 + |\vec{Q}|^2 + 2 |\vec{P}| |\vec{Q}| \cos \theta} \]
where \(\theta\) is the angle between the vectors \(\vec{P}\) and \(\vec{Q}\).

3. **Set Up the Equation**:
Since we know that \(|\vec{R}| = |\vec{P}|\), we can set up the equation:
\[ |\vec{P}| = \sqrt{|\vec{P}|^2 + |\vec{Q}|^2 + 2 |\vec{P}| |\vec{Q}| \cos \theta} \]

4. **Square Both Sides**:
Squaring both sides gives:
\[ |\vec{P}|^2 = |\vec{P}|^2 + |\vec{Q}|^2 + 2 |\vec{P}| |\vec{Q}| \cos \theta \]

5. **Simplify the Equation**:
By canceling \(|\vec{P}|^2\) from both sides, we have:
\[ 0 = |\vec{Q}|^2 + 2 |\vec{P}| |\vec{Q}| \cos \theta \]

6. **Rearranging the Equation**:
Rearranging gives:
\[ |\vec{Q}|^2 = -2 |\vec{P}| |\vec{Q}| \cos \theta \]
This implies:
\[ |\vec{Q}| + 2 |\vec{P}| \cos \theta = 0 \]

7. **Finding the Angle**:
From the equation above, we can express \(\cos \theta\):
\[ \cos \theta = -\frac{|\vec{Q}|}{2 |\vec{P}|} \]
Since \(|\vec{Q}|\) must be positive, this indicates that \(\theta\) must be such that \(\cos \theta\) is negative, which means \(\theta\) is in the second quadrant.

8. **Consider the Resultant of \(\vec{2P}\) and \(\vec{Q}\)**:
Now, we need to find the angle \(\alpha\) between \(\vec{Q}\) and the resultant \(\vec{R'}\) of \(\vec{2P}\) and \(\vec{Q}\):
\[ |\vec{R'}| = \sqrt{|\vec{2P}|^2 + |\vec{Q}|^2 + 2 |\vec{2P}| |\vec{Q}| \cos \phi} \]
where \(\phi\) is the angle between \(\vec{2P}\) and \(\vec{Q}\).

9. **Using the Resultant Formula**:
Since \(\vec{2P}\) is twice the magnitude of \(\vec{P}\), we can substitute:
\[ |\vec{R'}| = \sqrt{(2|\vec{P}|)^2 + |\vec{Q}|^2 + 2(2|\vec{P}|)(|\vec{Q}|) \cos \phi} \]

10. **Finding the Angle \(\alpha\)**:
We can use the tangent function to find the angle \(\alpha\):
\[ \tan \alpha = \frac{2 |\vec{P}| \sin \phi}{|\vec{Q}| + 2 |\vec{P}| \cos \phi} \]
Since we found that \( |\vec{Q}| + 2 |\vec{P}| \cos \phi = 0 \), we have:
\[ \tan \alpha = \frac{2 |\vec{P}| \sin \phi}{0} \]
This indicates that \(\alpha\) is \(90^\circ\).

### Final Answer:
The angle between \(\vec{Q}\) and the resultant of \(\vec{2P}\) and \(\vec{Q}\) is \(90^\circ\).