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

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) =

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 bar(OA)+bar(OB)+bar(OC)+bar(OD) equals :

If E is the intersection point of diagonals of parallelogram ABCD and vec( OA) + vec (OB) + vec (OC) + vec (OD) = xvec (OE) where O is origin then x=

Let G be the centroid of a triangle ABC and O be any other point, then bar(OA)+bar(OB)+bar(OC) is equal to

Let G be the centroid of a triangle ABC and O be any other point, then bar(OA)+bar(OB)+bar(OC) is equal to

Unit vectors along the diagonals of the parallelogram whose adjacent sides are bar(i)+bar(j)+bar(k) and bar(i)-bar(j)+bar(k) are

If ABCD is a parallelogram such that bar(AB)=bar(a), bar(BC)=bar(b)" then "bar(AC), bar(BD) are

The resultant of the three vectors bar(OA),bar(OB) and bar(OC) shown in figure :-