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=`

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) =

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) =

A parallelogram ABCD. Prove that vec(AC)+ vec (BD) = 2 vec(BC) '

If ABCD is a parallelogram, then vec(AC) - vec(BD) =

ABCD is a quadrilateral and E is the point of intersection of the lines joining the middle points of opposite side. Show that the resultant of vec (OA) , vec(OB) , vec(OC) and vec(OD) = 4 vec(OE) , where O is any point.

E and F are the interior points on the sides BC and CD of a parallelogram ABCD. Let vec(BE)=4vec(EC) and vec(CF)=4vec(FD) . If the line EF meets the diagonal AC in G, then vec(AG)=lambda vec(AC) , where lambda is equal to :

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

ABCD is a parallelogram . If vec(AB)=vec(a), vec(BC)=vec(b) , then what vec(BD) equal to ?

ABCD is a quadrilateral.E is the point of intersection of the line joining the midpoints of the opposite sides.If O is any point and vec OA+vec OB+vec OC+vec OD=xvec OE, then x is equal to a.3 b.9 c.7 d.4