`DeltaABC` be an equilateral triangle whose orthocentre is the origin 'O' . If `bar (OA)=bar a.bar (OB)=bar b` then `bar (OC)` is

ABC is an equilateral triangle of side ‘a'. Then bar(AB).bar(BC) +bar(BC).bar(CA)+ bar(CA).bar(AB) =

OACB is a parallelogram with bar(OC)=bar(a), bar(AB)=bar(b)" then "bar(OA)=

Let ABC be a triangle whose circumcentre is at P. If the position vectors of A,B,C and P are bar(a),bar(b),bar(c) and (bar(a)+bar(b)+bar(c))/(4) respectively then the positive vector of the orthocentre of the triangle is

Let ABC be a triangle whose circumcentre is at P. If the position vectors of A, B, C and P are bar(a), bar(b), bar(c) and (bar(a)+bar(b)+bar(c))/4 respectively, then the position vector of the orthocentre of this triangle is

In DeltaOAB , L is the midpoint of OA and M is a point on OB such that (OM)/(MB)=2 . P is the mid point of LM and the line AP is produced to meet OB at Q. If bar(OA)=bar(a), bar(OB)=bar(b) then find vectors bar(OP) and bar(AP) interms of bar(a) and bar(b) .

ABCDEF is a regualar hexagon whose centre is O. Then bar(AB)+bar(AC)+bar(AD)+bar(AE)+bar(AF) is

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

From the given figure (bar) OA , (bar) OB are called