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 vector \(\vec{Q}\) and the resultant of vectors \(\vec{2P}\) and \(\vec{Q}\). Given that the magnitude of the resultant of vectors \(\vec{P}\) and \(\vec{Q}\) is equal to the magnitude of vector \(\vec{P}\), we can proceed with the following steps:

### Step-by-Step Solution:

1. **Understanding the Given Information:**
We are given that the magnitude of the resultant of vectors \(\vec{P}\) and \(\vec{Q}\) is equal to the magnitude of vector \(\vec{P}\). Mathematically, this can be expressed as:
\[ |\vec{R}| = |\vec{P}| \]
where \(\vec{R} = \vec{P} + \vec{Q}\).

2. **Expressing the Magnitude of the Resultant:**
The magnitude of the resultant vector \(\vec{R}\) can be calculated 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 vectors \(\vec{P}\) and \(\vec{Q}\).

3. **Setting 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. **Squaring Both Sides:**
Squaring both sides of the equation gives:
\[ |\vec{P}|^2 = |\vec{P}|^2 + |\vec{Q}|^2 + 2 |\vec{P}| |\vec{Q}| \cos \theta \]

5. **Simplifying the Equation:**
By simplifying the equation, we can subtract \(|\vec{P}|^2\) from both sides:
\[ 0 = |\vec{Q}|^2 + 2 |\vec{P}| |\vec{Q}| \cos \theta \]

6. **Rearranging the Equation:**
Rearranging gives:
\[ |\vec{Q}|^2 = -2 |\vec{P}| |\vec{Q}| \cos \theta \]
Since \(|\vec{Q}|^2\) cannot be negative, this implies:
\[ \cos \theta = -\frac{|\vec{Q}|}{2 |\vec{P}|} \]

7. **Finding the Angle Between \(\vec{Q}\) and the Resultant of \(\vec{2P}\) and \(\vec{Q}\):**
Now we need to find the angle \(\alpha\) between vector \(\vec{Q}\) and the resultant of \(\vec{2P}\) and \(\vec{Q}\). The resultant \(\vec{R'}\) can be expressed as:
\[ \vec{R'} = \vec{2P} + \vec{Q} \]
The angle \(\alpha\) can be found using the tangent function:
\[ \tan \alpha = \frac{|\vec{2P}| \sin \theta}{|\vec{Q}| + |\vec{2P}| \cos \theta} \]

8. **Substituting Values:**
Since we established that \(\cos \theta = -\frac{|\vec{Q}|}{2 |\vec{P}|}\), we can substitute this into the tangent equation. After simplification, we find that:
\[ \tan \alpha \to \infty \]
This implies that:
\[ \alpha = 90^\circ \]

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