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..)(adsbygoogle = window.adsbygoogle || []).push({});

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,

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# Optimization - find point by minimising squared distance

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