OA=(3,1),OB=(-1,2)
OC⊥OB,BC//OA
OD+OA=OC
To find:OD
let OD(x,y)
OC= OA+OD
=(3+x,1+y)
OC⊥OB
=> OC.OB=0
(3+x,1+y).(-1,2)=0
-3-x+2+2y=0
-x+2y = 1 (1)
BC= OC-OB = (4+x,-1+y)
BC//OA
=> (4+x)/(-1+y) = 3
4+x= -3+3y
x-3y= -7 (2)
(1)+(2)
-y=-6
y = 6
x= 11
OD=(11,6)