Let ABCD be a parallelogram whose diagonals intersect at P and let O be the origin. What is `vec(OA) + vec(OB) + vec(OC ) + vec(OD)` equal to

Options

A. `2 vec(OP)`
B. `4 vec(OP)`
C. `6 vec(OP)`
D. `8 vec(OP)`

Answer

To solve the problem, we need to find the sum of the vectors from the origin O to the vertices A, B, C, and D of the parallelogram ABCD.

### Step-by-Step Solution:

1. **Understanding the Position Vectors**:
- Let \( \vec{OA} \), \( \vec{OB} \), \( \vec{OC} \), and \( \vec{OD} \) be the position vectors of points A, B, C, and D respectively from the origin O.

2. **Using the Property of Midpoints**:
- The diagonals of a parallelogram bisect each other. Therefore, the point P, where the diagonals AC and BD intersect, is the midpoint of both diagonals.
- For diagonal AC, we can express this as:
\[ \vec{OP} = \frac{\vec{OA} + \vec{OC}}{2} \]
- Rearranging this gives:
\[ \vec{OA} + \vec{OC} = 2\vec{OP} \quad \text{(Equation 1)} \]

3. **Applying the Same Concept to the Other Diagonal**:
- For diagonal BD, we have:
\[ \vec{OP} = \frac{\vec{OB} + \vec{OD}}{2} \]
- Rearranging this gives:
\[ \vec{OB} + \vec{OD} = 2\vec{OP} \quad \text{(Equation 2)} \]

4. **Adding the Two Equations**:
- Now, we add Equation 1 and Equation 2:
\[ (\vec{OA} + \vec{OC}) + (\vec{OB} + \vec{OD}) = 2\vec{OP} + 2\vec{OP} \]
- This simplifies to:
\[ \vec{OA} + \vec{OB} + \vec{OC} + \vec{OD} = 4\vec{OP} \]

5. **Conclusion**:
- Therefore, the sum of the vectors from the origin O to the vertices A, B, C, and D is:
\[ \vec{OA} + \vec{OB} + \vec{OC} + \vec{OD} = 4\vec{OP} \]

### Final Answer:
\[ \vec{OA} + \vec{OB} + \vec{OC} + \vec{OD} = 4\vec{OP} \]