Vector Question !! ~~

(a)
AB=OB-OA=4i+3j
|AB|=5
The unit vector in the direction of AB
=(4/5)i+(3/5)j
(b)
let P=xi+yj
AP=(x+3)i+(y+2)j
AP=√[(x+3)^2+(y+2)^2]=√(x^2+y^2+6x+4y+13)
Also x^2+y^2=41
So
AP=√(54+6x+4y)=m
Now AP is in the same direction of AB
(x+3):(y+2)=4/5:3/5=4:3
3(x+3)=4(y+2)
3x+9=4y+8
4y=3x+1
Substitution into 16x^2+16y^2=656
16x^2+9x^2+6x+1=656
25x^2+6x-655=0
x=5 or -5.24 (rejected)
y=4
AP=√(54+6x+4y)
m=√(54+30+16)=10