Optimization - find point by minimising squared distance

1. Sep 21, 2007

inner08

Find the point in the plane 3x+2y+z=1 that is the closest to the origin by minimising squared distance. (I hope I translated this ok..)

I was thinking I would need to isolate a variable in the equation for the plane above then substitute it into the distance formula then do a partial derivative.

Something like... D=(x^2+y^2+z^2)^(1/2). Since it is squared I could just have (x^2+y^2+z^2).

z=1-2y-3x

D = x^2+y^2+(1-2y-3x)^2

= x^2+y^2+1-4y-6x+12xy+9x^2
=10x^2+y^2+1-4y-6x+12xy

f_x = 20x-6+12y = 0
y = (6-20x)/12

f_y = 2y-4+12x
y = (4-12x)/2

etc....x=1/4, y=1/2, z=-3/4

It seems ok because it works in the given equation(3x+2y+z=1) but I've never done a problem like this so if its wrong, I do hope I'm atleast on the right track. Hope someone can't let me know.

Thanks,

2. Sep 21, 2007

mathman

You made an error in your calculation of D. In squaring the expression for z, you omitted 4y^2. Therefore D should have 5y^2 not y^2. You should be able to get the right answer after that. It'll be (3,2,1)/14.