If G is the intersection of diagonals of a parallelogram ABCD and O is any point, then ` O vecA + O vec B + O vec C + vec (OD) = `

Options

A. `2vec(OG)`
B. `4vec(OG)`
C. `5vec(OG)`
D. `3vec(OG)`

Answer

To solve the problem, we need to find the expression \( O \vec{A} + O \vec{B} + O \vec{C} + O \vec{D} \) where \( G \) is the intersection of the diagonals of the parallelogram \( ABCD \) and \( O \) is any point.

### Step-by-Step Solution:

1. **Define the Position Vectors**:
Let the position vectors of points \( A, B, C, \) and \( D \) be represented as:
- \( \vec{A} \) for point \( A \)
- \( \vec{B} \) for point \( B \)
- \( \vec{C} \) for point \( C \)
- \( \vec{D} \) for point \( D \)

2. **Express Vectors with Reference Point \( O \)**:
The vectors from point \( O \) to points \( A, B, C, \) and \( D \) can be expressed as:
- \( \vec{OA} = \vec{A} - \vec{O} \)
- \( \vec{OB} = \vec{B} - \vec{O} \)
- \( \vec{OC} = \vec{C} - \vec{O} \)
- \( \vec{OD} = \vec{D} - \vec{O} \)

3. **Sum the Vectors**:
Now, we can sum these vectors:
\[ \vec{OA} + \vec{OB} + \vec{OC} + \vec{OD} = (\vec{A} - \vec{O}) + (\vec{B} - \vec{O}) + (\vec{C} - \vec{O}) + (\vec{D} - \vec{O}) \]
Simplifying this gives:
\[ = (\vec{A} + \vec{B} + \vec{C} + \vec{D}) - 4\vec{O} \]

4. **Finding the Midpoint \( G \)**:
Since \( G \) is the intersection of the diagonals of the parallelogram, we know that:
\[ \vec{G} = \frac{\vec{A} + \vec{C}}{2} = \frac{\vec{B} + \vec{D}}{2} \]

5. **Substituting \( G \) in the Expression**:
We can express \( \vec{A} + \vec{B} + \vec{C} + \vec{D} \) in terms of \( \vec{G} \):
\[ \vec{A} + \vec{B} + \vec{C} + \vec{D} = 2\vec{G} + 2\vec{G} = 4\vec{G} \]

6. **Final Expression**:
Thus, substituting back into our earlier expression:
\[ \vec{OA} + \vec{OB} + \vec{OC} + \vec{OD} = 4\vec{G} - 4\vec{O} \]
Therefore, we can conclude:
\[ O \vec{A} + O \vec{B} + O \vec{C} + O \vec{D} = 4\vec{G} - 4\vec{O} \]

### Conclusion:
The final expression is \( O \vec{A} + O \vec{B} + O \vec{C} + O \vec{D} = 4\vec{G} - 4\vec{O} \).