If P. Q = 0, then| PxQ|is

Options

A. |P||Q|
B. ZERO
C. 1
D. `sqrtPQ`

Answer

To solve the problem, we need to analyze the relationship between the dot product and the cross product of two vectors \( \mathbf{P} \) and \( \mathbf{Q} \).

### Step-by-step Solution:

1. **Understanding the Dot Product**:
We are given that \( \mathbf{P} \cdot \mathbf{Q} = 0 \). The dot product of two vectors can be expressed as:
\[ \mathbf{P} \cdot \mathbf{Q} = |\mathbf{P}| |\mathbf{Q}| \cos \theta \]
where \( \theta \) is the angle between the two vectors.

2. **Finding the Angle**:
Since \( \mathbf{P} \cdot \mathbf{Q} = 0 \), we can conclude that:
\[ |\mathbf{P}| |\mathbf{Q}| \cos \theta = 0 \]
For this equation to hold true, either \( |\mathbf{P}| = 0 \), \( |\mathbf{Q}| = 0 \), or \( \cos \theta = 0 \). The case where \( \cos \theta = 0 \) occurs when \( \theta = 90^\circ \). Therefore, the vectors \( \mathbf{P} \) and \( \mathbf{Q} \) are perpendicular to each other.

3. **Understanding the Cross Product**:
The cross product of two vectors is given by:
\[ \mathbf{P} \times \mathbf{Q} = |\mathbf{P}| |\mathbf{Q}| \sin \theta \, \mathbf{n} \]
where \( \mathbf{n} \) is a unit vector perpendicular to the plane formed by \( \mathbf{P} \) and \( \mathbf{Q} \).

4. **Calculating the Cross Product**:
Since we have established that \( \theta = 90^\circ \), we can substitute this into the equation for the cross product:
\[ \sin 90^\circ = 1 \]
Therefore, we have:
\[ \mathbf{P} \times \mathbf{Q} = |\mathbf{P}| |\mathbf{Q}| \cdot 1 \cdot \mathbf{n} = |\mathbf{P}| |\mathbf{Q}| \mathbf{n} \]

5. **Conclusion**:
The magnitude of the cross product \( |\mathbf{P} \times \mathbf{Q}| \) is:
\[ |\mathbf{P} \times \mathbf{Q}| = |\mathbf{P}| |\mathbf{Q}| \]
Thus, if \( \mathbf{P} \cdot \mathbf{Q} = 0 \), the magnitude of the cross product \( |\mathbf{P} \times \mathbf{Q}| \) is equal to the product of the magnitudes of the two vectors.

### Final Answer:
\[ |\mathbf{P} \times \mathbf{Q}| = |\mathbf{P}| |\mathbf{Q}| \]