There vectors `vecP, vecQ and vecR` are such that `vecP+vecQ+vecR=0` Vectors `vecP` and `vecQ` are equal in , magnitude . The magnitude of vector `vecR` is `sqrt2` times the magnitude of either `vecP or vecQ` . Calculate the angle between these vectors .

Answer

To solve the problem, we need to analyze the given vectors \( \vec{P}, \vec{Q}, \) and \( \vec{R} \) under the conditions provided:

1. **Understanding the Given Information**:
- We have three vectors: \( \vec{P} + \vec{Q} + \vec{R} = 0 \).
- Vectors \( \vec{P} \) and \( \vec{Q} \) are equal in magnitude.
- The magnitude of vector \( \vec{R} \) is \( \sqrt{2} \) times the magnitude of either \( \vec{P} \) or \( \vec{Q} \).

2. **Assigning Magnitudes**:
- Let the magnitude of \( \vec{P} \) and \( \vec{Q} \) be \( p \).
- Therefore, \( |\vec{R}| = \sqrt{2} p \).

3. **Using the Vector Addition Condition**:
- From the equation \( \vec{P} + \vec{Q} + \vec{R} = 0 \), we can rearrange it to find \( \vec{R} = -(\vec{P} + \vec{Q}) \).
- This means that \( \vec{R} \) is equal in magnitude to the resultant of \( \vec{P} \) and \( \vec{Q} \).

4. **Finding the Magnitude of the Resultant**:
- Since \( \vec{P} \) and \( \vec{Q} \) have equal magnitudes and we denote their angle as \( \theta \), the magnitude of the resultant \( \vec{R} \) can be calculated using the formula:
\[ |\vec{R}| = \sqrt{|\vec{P}|^2 + |\vec{Q}|^2 + 2 |\vec{P}| |\vec{Q}| \cos(\theta)} \]
- Substituting the magnitudes:
\[ \sqrt{2} p = \sqrt{p^2 + p^2 + 2p^2 \cos(\theta)} \]
- Simplifying gives:
\[ \sqrt{2} p = \sqrt{2p^2(1 + \cos(\theta))} \]
- Squaring both sides:
\[ 2p^2 = 2p^2(1 + \cos(\theta)) \]
- Dividing by \( 2p^2 \) (assuming \( p \neq 0 \)):
\[ 1 = 1 + \cos(\theta) \]
- This leads to:
\[ \cos(\theta) = 0 \]

5. **Finding the Angle**:
- The angle \( \theta \) for which \( \cos(\theta) = 0 \) is \( \theta = 90^\circ \).

6. **Conclusion**:
- Therefore, the angle between vectors \( \vec{P} \) and \( \vec{Q} \) is \( 90^\circ \).

### Summary of Angles:
- The angle between \( \vec{P} \) and \( \vec{Q} \) is \( 90^\circ \).
- The angle between \( \vec{Q} \) and \( \vec{R} \) is \( 135^\circ \).
- The angle between \( \vec{R} \) and \( \vec{P} \) is \( 135^\circ \).