Given `vecP. vecQ=|vecPxxvecQ|` and `vecR=vecP+vecQ` then `|vecR|` is

Options

A. `sqrt(P^2+Q^2)`
B. P+Q
C. `sqrt(P^2+Q^2 + "PQ"/sqrt2)`
D. `[P^2+Q^2+sqrt2PQ ]^(1/2)`

Explanation

To solve the problem, we need to find the magnitude of the vector \(\vec{R} = \vec{P} + \vec{Q}\) given that \(\vec{P} \cdot \vec{Q} = |\vec{P} \times \vec{Q}|\).

### Step-by-Step Solution:

1. **Understanding the Given Condition**:
We know that:
\[ \vec{P} \cdot \vec{Q} = |\vec{P} \times \vec{Q}| \]
This implies that the angle \(\theta\) between vectors \(\vec{P}\) and \(\vec{Q}\) is \(45^\circ\). This is because:
\[ \vec{P} \cdot \vec{Q} = |\vec{P}| |\vec{Q}| \cos \theta \]
and
\[ |\vec{P} \times \vec{Q}| = |\vec{P}| |\vec{Q}| \sin \theta \]
Setting these equal gives us:
\[ |\vec{P}| |\vec{Q}| \cos \theta = |\vec{P}| |\vec{Q}| \sin \theta \]
Dividing both sides by \(|\vec{P}| |\vec{Q}|\) (assuming they are non-zero) leads to:
\[ \cos \theta = \sin \theta \]
This holds true when \(\theta = 45^\circ\).

2. **Finding the Magnitude of \(\vec{R}\)**:
The magnitude of \(\vec{R}\) can be calculated using the formula:
\[ |\vec{R}| = |\vec{P} + \vec{Q}| = \sqrt{|\vec{P}|^2 + |\vec{Q}|^2 + 2 |\vec{P}| |\vec{Q}| \cos \theta} \]
Substituting \(\theta = 45^\circ\) (where \(\cos 45^\circ = \frac{1}{\sqrt{2}}\)):
\[ |\vec{R}| = \sqrt{|\vec{P}|^2 + |\vec{Q}|^2 + 2 |\vec{P}| |\vec{Q}| \cdot \frac{1}{\sqrt{2}}} \]

3. **Simplifying the Expression**:
Let \(|\vec{P}| = p\) and \(|\vec{Q}| = q\):
\[ |\vec{R}| = \sqrt{p^2 + q^2 + \frac{2pq}{\sqrt{2}}} \]
This simplifies to:
\[ |\vec{R}| = \sqrt{p^2 + q^2 + \sqrt{2}pq} \]

4. **Conclusion**:
Therefore, the magnitude of \(\vec{R}\) is:
\[ |\vec{R}| = \sqrt{p^2 + q^2 + \sqrt{2}pq} \]