Let `O` be the origin, and ` O X , O Y , O Z ` be three unit vectors in the direction of the sides ` Q R ` , ` R P ` , ` P Q ` , respectively of a triangle PQR. `| O X xx O Y |=` `(a)sin(P+R)` (b) `sin2R` `(c)sin(Q+R)` (d) `sin(P+Q)dot`

Options

A. `sin(P+Q)`
B. `sin(P+R)`
C. `sin(Q+R)`
D. `sin2R`

Answer

To solve the problem, we need to find the magnitude of the cross product of the unit vectors \( \mathbf{OX} \) and \( \mathbf{OY} \), which are in the directions of the sides \( QR \) and \( RP \) of triangle \( PQR \), respectively.

### Step-by-Step Solution:

1. **Understanding the Vectors**:
- Let \( \mathbf{OX} \) be the unit vector in the direction of side \( QR \).
- Let \( \mathbf{OY} \) be the unit vector in the direction of side \( RP \).
- Since these are unit vectors, we have \( |\mathbf{OX}| = 1 \) and \( |\mathbf{OY}| = 1 \).

2. **Using the Cross Product Formula**:
- The magnitude of the cross product of two vectors \( \mathbf{A} \) and \( \mathbf{B} \) is given by:
\[ |\mathbf{A} \times \mathbf{B}| = |\mathbf{A}| |\mathbf{B}| \sin(\theta) \]
where \( \theta \) is the angle between the two vectors.

3. **Identifying the Angle**:
- In triangle \( PQR \), the angles at vertices \( P \), \( Q \), and \( R \) are denoted as \( P \), \( Q \), and \( R \) respectively.
- The angle between the vectors \( \mathbf{OX} \) and \( \mathbf{OY} \) is \( 180^\circ - R \) (since the angle at vertex \( R \) is \( R \)).
- Therefore, we can express this as:
\[ \theta = 180^\circ - R \]

4. **Substituting into the Formula**:
- Now substituting the values into the cross product formula:
\[ |\mathbf{OX} \times \mathbf{OY}| = |\mathbf{OX}| |\mathbf{OY}| \sin(180^\circ - R) \]
- Since both vectors are unit vectors:
\[ |\mathbf{OX} \times \mathbf{OY}| = 1 \cdot 1 \cdot \sin(180^\circ - R) \]

5. **Using the Sine Identity**:
- We know that \( \sin(180^\circ - R) = \sin(R) \).
- Thus, we have:
\[ |\mathbf{OX} \times \mathbf{OY}| = \sin(R) \]

6. **Finding \( P + Q \)**:
- From the triangle angle sum property, we know:
\[ P + Q + R = 180^\circ \]
- Therefore, we can express \( P + Q \) as:
\[ P + Q = 180^\circ - R \]

7. **Final Result**:
- Substituting back, we have:
\[ |\mathbf{OX} \times \mathbf{OY}| = \sin(P + Q) \]
- Thus, the answer is:
\[ |\mathbf{OX} \times \mathbf{OY}| = \sin(P + Q) \]

### Conclusion:
The correct option is (d) \( \sin(P + Q) \).