ABCD is a parallelogram If M and N are the midpoints of BC and DC respectively,If `bar(AM)+bar(AN)=lambda(bar(AC))` then value of `lambda`

ABCD is parallelogram.If L and M are the middle points of BC and CD, then bar(AL)+bar(AM) equals

ABCD is a parallelogram. P and R are the midpoints of DC and BC respectively. The line PR intersects the diagonal AC at Q . The distance CQ will be equal to

Let ABC be a triangle. Three points D, E and F are on the sides AB, AC and BC respectively so that BDEF is a parallelogram. M and N are mid points of the sides DE and EF. If vector bar(MN)=(1)/(4)bar(AC) . Then find the ratio in which D and F divides the sides AB and BC.

If D is the mid -point of side AB of DeltaABC , then bar(AB) + bar(BC) + bar(AC) =

M and N are mid-point of the diagnols AC and BD respectivley of quadrilateral ABCD, then bar(AB) + bar(AD)+bar(CB)+bar(CD) =

If bar(DA)=bar(a);bar(AB)=bar(b) and bar(CB)=kbar(a) where k>0 and X,Y are the midpoint of DB and AC respectively such that |bar(a)|=17 and |XY|=4, then k=

ABC is an equilateral triangle. P and Q are two points on bar(AB) and bar(AC) respectively such that bar(PQ) || bar(BC) . If bar(PQ) = 5cm , then area of Delta APQ is :

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 :

Let bar(a) , bar(b) and bar(c) be vectors with magnitudes 3,4 and 5 respectively and bar(a)+bar(b)+bar(c)=0 then the value of bar(a).bar(b)+bar(b).bar(c)+bar(c).bar(a) is

If ABCD is a square then, 2bar(AB)+4bar(BC)+5bar(CD)+7bar(DA)=