If O is the origin and `P(x_(1),y_(1)), Q(x_(2),y_(2))` are two points then `OPxOQ sin angle POQ=`

Options

A. `x_(1)x_(2)+y_(1)+y_(2)`
B. `x_(1)y_(2)+x_(2)+y_(1)`
C. `|x_(1)y_(2)-x_(2)y_(1)|`
D. none of these

Answer

To solve the problem, we need to find the expression for \( OP \cdot OQ \cdot \sin(\angle POQ) \) where \( O \) is the origin, and \( P(x_1, y_1) \) and \( Q(x_2, y_2) \) are two points.

### Step-by-Step Solution:

1. **Define the Vectors**:
- The vector \( \vec{OP} \) from the origin \( O(0, 0) \) to the point \( P(x_1, y_1) \) can be expressed as:
\[ \vec{OP} = x_1 \hat{i} + y_1 \hat{j} \]
- The vector \( \vec{OQ} \) from the origin \( O(0, 0) \) to the point \( Q(x_2, y_2) \) can be expressed as:
\[ \vec{OQ} = x_2 \hat{i} + y_2 \hat{j} \]

2. **Cross Product of the Vectors**:
- The magnitude of the cross product \( \vec{OP} \times \vec{OQ} \) gives us the area of the parallelogram formed by the two vectors, which is also equal to \( OP \cdot OQ \cdot \sin(\angle POQ) \).
- The cross product in two dimensions can be calculated using the determinant:
\[ \vec{OP} \times \vec{OQ} = \begin{vmatrix} \hat{i} & \hat{j} \\ x_1 & y_1 \\ x_2 & y_2 \end{vmatrix} \]
- This determinant simplifies to:
\[ \vec{OP} \times \vec{OQ} = x_1y_2 - x_2y_1 \]

3. **Magnitude of the Cross Product**:
- The magnitude of the cross product is:
\[ |\vec{OP} \times \vec{OQ}| = |x_1y_2 - x_2y_1| \]

4. **Final Expression**:
- Therefore, we have:
\[ OP \cdot OQ \cdot \sin(\angle POQ) = |x_1y_2 - x_2y_1| \]

### Conclusion:
The value of \( OP \cdot OQ \cdot \sin(\angle POQ) \) is given by:
\[ |x_1y_2 - x_2y_1| \]