A,B,C ARE THREE NON COLLINEAR PONITS WITH POSITION VECTORS a,b,c respectively given P,Q,R are ponits on BC,CA&AB respectively such that: BP:PC=CQ:QA=AR:RB=1:2 Find the position vectors of the vertices of the triangle XYZ formed lines AB,BQ,CR.hence show that the centroid of tiangle ABC is same as that of triangle XYZ slove with diag.

							
 position vector of p = b +1/3 (b-c )  
                              = 4/3b - 1/3c
 position vector of q =  c + 1/3 ( c-a )
                              = 4/3 c - 1/3 a
                       of r =  4/3 a - 1/3 b
eq. of line AP,
        r = a + n1 ( 4/3b - 1/3c  -a )   = (1-n1 )a  + 4/3 n1 b - 1/3 n1 c   ........................1
line BQ , 
      r = b + n2 ( 4/3 c - 1/3 a  -b )   =   -1/3n2 a + (1- n2 )b  + 4/3 n2 c  ...............................2
 line CR ,
       r = c + n3 (4/3 a - 1/3 b  - c ) = 4/3 n3 a  -1/3 n3 b  + (1-n3 )c   .........................................3
 
position vector of X will be found by solving eq 1 &2 simultaneously , becoz it is the intersecting point of AP & BQ,
equating both eq.
(1-n1 )a  + 4/3 n1 b - 1/3 n1 c  =  -1/3n2 a + (1- n2 )b  + 4/3 n2 c
 comparing coefficients of a, b & c
 1-n1 = -1/3 n2     ,     4/3 n1  = (1- n2 )   ,   - 1/3 n1   =   4/3 n2
gives n1 = 12/13, so the position vector of X will be,              (putting n1 = 12/13 in eq 1 )
  x =  1/13 a + 16/13 b  -4/13 c .
 in the similar way we can solve eq 2 &3 & get Y   &  solve 1 & 3 to get Z
 Y =  1/13 b  + 16/13 c - 4/13 a
 Z = 1/13 c + 16/13 a  - 4/13 b
centroid of the triangle XYZ
           = (x + y +z )1/3
           = (1/13 + 16/13 - 4/13 ) ( a+b +c ) /3
           =  ( a+b +c ) /3
           , which is also the centroid of triangle ABC