ABCD is a parallelogram and P is the point of intersection of the diagonals. If O is the origin, then $\overrightarrow {{\rm{OA}}} + \overrightarrow {{\rm{OB}}} + \overrightarrow {{\rm{OC}}} + \overrightarrow {{\rm{OD}}} $ is equal to

Question

ABCD is a parallelogram and P is the point of intersection of the diagonals. If O is the origin, then $\overrightarrow {{\rm{OA}}} + \overrightarrow {{\rm{OB}}} + \overrightarrow {{\rm{OC}}} + \overrightarrow {{\rm{OD}}} $ is equal to

TuteeHUB · Accepted Answer

$2{\rm{\;}}\overrightarrow {{\rm{OP}}} $

TuteeHUB · Answer

$4{\rm{\;}}\overrightarrow {{\rm{OP}}} $

TuteeHUB · Answer

$\overrightarrow {{\rm{OP}}} $

TuteeHUB · Answer

Null vector

1.	ABCD is a parallelogram and P is the point of intersection of the diagonals. If O is the origin, then \(\overrightarrow {{\rm{OA}}} + \overrightarrow {{\rm{OB}}} + \overrightarrow {{\rm{OC}}} + \overrightarrow {{\rm{OD}}} \) is equal to
A.	\(4{\rm{\;}}\overrightarrow {{\rm{OP}}} \)
B.	\(2{\rm{\;}}\overrightarrow {{\rm{OP}}} \)
C.	\(\overrightarrow {{\rm{OP}}} \)
D.	Null vector
Answer» B. \(2{\rm{\;}}\overrightarrow {{\rm{OP}}} \)

ABCD is a parallelogram and P is the point of intersection of the diagonals. If O is the origin, then \(\overrightarrow {{\rm{OA}}} + \overrightarrow {{\rm{OB}}} + \overrightarrow {{\rm{OC}}} + \overrightarrow {{\rm{OD}}} \) is equal to

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