Question

If G is the intersection of diagonals of a parallelogram ABCD and O is any point, then $O\overrightarrow{A}+O\overrightarrow{B}+O\overrightarrow{C}+O\overrightarrow{D}=$
(a) $2\overrightarrow{OG}$

(b) $4\overrightarrow{OG}$

(c) $5\overrightarrow{OG}$

(d) $3\overrightarrow{OG}$

Answer 1

(b) $4\overrightarrow{OG}$

Let us consider the point O as origin.
G is the mid point of AC.




$$\begin{gathered} \therefore \overrightarrow{OG}=\frac{\overrightarrow{OA}+\overrightarrow{OC}}{2} \\ 2\overrightarrow{OG}=\overrightarrow{OA}+\overrightarrow{OC}.....\left(1\right) \end{gathered}$$

Also, G is the mid point BD

$$\begin{gathered} \therefore \overrightarrow{OG}=\frac{\overrightarrow{OB}+\overrightarrow{OD}}{2} \\ 2\overrightarrow{OG}=\overrightarrow{OB}+\overrightarrow{OD}.....\left(2\right) \end{gathered}$$

On adding (1) and (2) we get,

$$\begin{gathered} 2\overrightarrow{OG}+2\overrightarrow{OG}=\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}+\overrightarrow{OD} \\ 4\overrightarrow{OG}=\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}+\overrightarrow{OD} \\ \therefore \overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}+\overrightarrow{OD}=4\overrightarrow{OG} \end{gathered}$$