`ABCD` is a parallelogram whose diagonals meet at P. If O is a fixed point, then `bar(OA)+bar(OB)+bar(OC)+bar(OD)` equals :

ABCD is a parallelogram whose diagonals meet at P. If O is a fixed point, then vec(OA)+vec(OB)+vec(OC)+vec(OD) equals :

ABCD is a parallelogram whose diagonals meet at P. If O is a fixed point, then vec(OA)+vec(OB)+vec(OC)+vec(OD) equals :

ABCD is a parallelogram whose diagonals meet at P. If O is a fixed point, then vec(OA)+vec(OB)+vec(OC)+vec(OD) equals :

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) =

If diagonals of a parallelogram ABCD intersect each other in M, then bar(OA) + bar(OB) + bar(OC) + bar(OD) =

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

ABCD is a parallelogram and P is the mid-point of bar(DC) . If Q is a point on bar(AP) , such that bar(AQ)=2/3bar(AP) , show that Q lies on the diagonal bar(BD) " and " bar(BQ)=2/3bar(BD) .

If ABCD is a parallelogram then bar(AC)+bar(BD)=