Given : `vecP=vecA+vecBandvecQ=vecA-vecB`. If the magnitudes of vectors `vecP and vecQ` are equal, what is the angle between vectors `vecAandvecB`?

Answer

To solve the problem, we need to find the angle between vectors \(\vec{A}\) and \(\vec{B}\) given that the magnitudes of vectors \(\vec{P}\) and \(\vec{Q}\) are equal. We start with the definitions of the vectors:

1. **Define the vectors**:
\[ \vec{P} = \vec{A} + \vec{B} \]
\[ \vec{Q} = \vec{A} - \vec{B} \]

2. **Set up the equation for magnitudes**:
Since the magnitudes of \(\vec{P}\) and \(\vec{Q}\) are equal, we can write:
\[ |\vec{P}| = |\vec{Q}| \]

3. **Express the magnitudes**:
The magnitude of \(\vec{P}\) is given by:
\[ |\vec{P}| = \sqrt{|\vec{A}|^2 + |\vec{B}|^2 + 2 |\vec{A}| |\vec{B}| \cos \theta} \]
where \(\theta\) is the angle between \(\vec{A}\) and \(\vec{B}\).

Similarly, the magnitude of \(\vec{Q}\) is:
\[ |\vec{Q}| = \sqrt{|\vec{A}|^2 + |\vec{B}|^2 - 2 |\vec{A}| |\vec{B}| \cos \theta} \]

4. **Set the magnitudes equal**:
Since \(|\vec{P}| = |\vec{Q}|\), we have:
\[ \sqrt{|\vec{A}|^2 + |\vec{B}|^2 + 2 |\vec{A}| |\vec{B}| \cos \theta} = \sqrt{|\vec{A}|^2 + |\vec{B}|^2 - 2 |\vec{A}| |\vec{B}| \cos \theta} \]

5. **Square both sides to eliminate the square roots**:
\[ |\vec{A}|^2 + |\vec{B}|^2 + 2 |\vec{A}| |\vec{B}| \cos \theta = |\vec{A}|^2 + |\vec{B}|^2 - 2 |\vec{A}| |\vec{B}| \cos \theta \]

6. **Simplify the equation**:
Cancel out \(|\vec{A}|^2 + |\vec{B}|^2\) from both sides:
\[ 2 |\vec{A}| |\vec{B}| \cos \theta = -2 |\vec{A}| |\vec{B}| \cos \theta \]

7. **Combine like terms**:
\[ 2 |\vec{A}| |\vec{B}| \cos \theta + 2 |\vec{A}| |\vec{B}| \cos \theta = 0 \]
This simplifies to:
\[ 4 |\vec{A}| |\vec{B}| \cos \theta = 0 \]

8. **Solve for \(\cos \theta\)**:
Since \( |\vec{A}| \) and \( |\vec{B}| \) are not zero, we can divide by \(4 |\vec{A}| |\vec{B}|\):
\[ \cos \theta = 0 \]

9. **Determine the angle**:
The angle \(\theta\) for which \(\cos \theta = 0\) is:
\[ \theta = 90^\circ \]

Thus, the angle between vectors \(\vec{A}\) and \(\vec{B}\) is \(90^\circ\).