Let PQRS be a parallelogram whose diagonals PR and QS intersect at O'. IF O is the origin then : `bar(OP) + bar(OQ) + bar(OR) + bar(OS) =`

Options

A. 4 `bar(OO')`
B. 3 `bar(OO')`
C. 2 `bar(OO')`
D. `bar(OO')`

Answer

To solve the problem, we need to find the sum of the vectors \( \bar{OP} + \bar{OQ} + \bar{OR} + \bar{OS} \) in the context of a parallelogram \( PQRS \) with diagonals \( PR \) and \( QS \) intersecting at point \( O' \), where \( O \) is the origin.

### Step-by-Step Solution:

1. **Understanding the Geometry**:
- Let \( O \) be the origin.
- Let \( O' \) be the intersection of the diagonals \( PR \) and \( QS \).
- In a parallelogram, the diagonals bisect each other, so \( O' \) is the midpoint of both diagonals.

2. **Expressing Vectors**:
- We can express the position vectors of points \( P, Q, R, S \) in terms of the origin \( O \) and the midpoint \( O' \).
- Since \( O' \) is the midpoint, we have:
\[ \bar{O'P} = \bar{O'R} \quad \text{and} \quad \bar{O'Q} = \bar{O'S} \]
- This implies:
\[ \bar{OP} = \bar{O'P} + \bar{OO'} \quad \text{and} \quad \bar{OR} = \bar{O'R} + \bar{OO'} \]
\[ \bar{OQ} = \bar{O'Q} + \bar{OO'} \quad \text{and} \quad \bar{OS} = \bar{O'S} + \bar{OO'} \]

3. **Using the Properties of Vectors**:
- Since \( O' \) is the midpoint, we can express:
\[ \bar{OP} + \bar{OR} = 2 \bar{OO'} \]
\[ \bar{OQ} + \bar{OS} = 2 \bar{OO'} \]

4. **Adding the Vectors**:
- Now, we can add the vectors:
\[ \bar{OP} + \bar{OQ} + \bar{OR} + \bar{OS} = (\bar{OP} + \bar{OR}) + (\bar{OQ} + \bar{OS}) \]
- Substituting the previous results:
\[ = 2 \bar{OO'} + 2 \bar{OO'} = 4 \bar{OO'} \]

5. **Final Result**:
- Therefore, we conclude that:
\[ \bar{OP} + \bar{OQ} + \bar{OR} + \bar{OS} = 4 \bar{OO'} \]

### Final Answer:
\[ \bar{OP} + \bar{OQ} + \bar{OR} + \bar{OS} = 4 \bar{OO'} \]