Question

In oab e is the mid point of ab and f is a point on a such that of

Answer 1

in triangle OAB , E is the mid-point of OB and D is a point on AB such that AD:DB=2:1. If OD and AE intrsects at P, determine the ratio OP:PD using vector methods.

sorry i cant construct the triangle and note: A, B,D,E,P represent vectors 

 

D=2(B)+(A)1/2+1

 =2(B)+(A)/3

E=1(O)+1(B)/1+1

  =B/2

 given the point of interesection of OD and AE is P

let the OP:PD is 1:m and the ratio AP:PE is 1:n

therefore P=1(D)+m(O)/1+m

               =D/1+m

               =2(B)+(A)/3(1+m)

and also P=n(A)+1(B/2)/1+n

               =2n(A)+B/2(1+n)

equating values of P

                               2(B)+(A)/3(1+m)=2n(A)+B/2(1+n) 

                             2(B)/3(1+m)+(A)/3(1+m)=2n(A)/2(1+n)+(B)/2(1+n)

comparing values of A and B we get

                               2/3(1+m)=1/2(1+n)---->I

                              1/3(1+m)=2n/2(1+n)----->II

   divide II/I we get

                          1/2=2n

                          n=1/4

  substitute n=1/4 in I we get m=2/3

 OP:PD is 1:2/3

              =3:2