Minimum Distance

**billym** · October 18th 2009, 11:01 AM

Can somebody tell me if I've started this off right?

I need to use the Lagrange method to find the minimum distance from the point (1,2,3) to any point of the surface: $x + 2y - z = 2$

Let: $D^2=f(x,y,z)=(x-1)^2+(y-2)^2+(z-3)^2$

and

$g(x,y,z)=x+2y-z-2=0$

and then just carry on in the usual manner?

**tonio** · October 18th 2009, 03:00 PM

Originally Posted by billym

Can somebody tell me if I've started this off right?

I need to use the Lagrange method to find the minimum distance from the point (1,2,3) to any point of the surface: $x + 2y - z = 2$

Let: $D^2=f(x,y,z)=(x-1)^2+(y-2)^2+(z-3)^2$

and

$g(x,y,z)=x+2y-z-2=0$

and then just carry on in the usual manner?

If by "the usual manner" you mean Lagrange multipliers and stuff, the answer is, I think, yes.

Anyway, from geometry you already know what the answer must be:

$D=\frac{|1+4-3-2|}{\sqrt{1+4+1}} = \frac{0}{\sqrt{6}}=0$

which doesn't surprise since the point (1,2,3) is on the surface.

Tonio

**billym** · October 18th 2009, 04:44 PM

I can see that... so what is the problem with my method?

$f_x+ \lambda g_x=0 \Rightarrow 2(x-1)+ \lambda =0 \Rightarrow x=1-\frac{\lambda}{2}$ ...(1)

$f_y+ \lambda g_y=0 \Rightarrow 2(y-2)+ 2\lambda =0 \Rightarrow y=2-\lambda$ ...(2)

$f_z+ \lambda g_z=0 \Rightarrow 2(z-3)- \lambda =0 \Rightarrow z=3+\frac{\lambda}{2}$ ...(3)

Plug (1) to (3) into the constraint:

$(1-\frac{\lambda}{2})+2(2-\lambda)-(3+\frac{\lambda}{2})-2=0$

$\lambda=-\frac{8}{3}$

Which gives $x = 7/3, y = 14/3, z = 5/3$

So:

$D=\sqrt{(7/3-1)^2+(14/3-2)^2+(5/3-1)^2}=3.06$

**tonio** · October 18th 2009, 05:01 PM

Originally Posted by billym

I can see that... so what is the problem with my method?

$f_x+ \lambda g_x=0 \Rightarrow 2(x-1)+ \lambda =0 \Rightarrow x=1-\frac{\lambda}{2}$ ...(1)

$f_y+ \lambda g_y=0 \Rightarrow 2(y-2)+ 2\lambda =0 \Rightarrow y=2-\lambda$ ...(2)

$f_z+ \lambda g_z=0 \Rightarrow 2(z-3)- \lambda =0 \Rightarrow z=3+\frac{\lambda}{2}$ ...(3)

Plug (1) to (3) into the constraint:

$(1-\frac{\lambda}{2})+2(2-\lambda)-(3+\frac{\lambda}{2})-2=0$

$\lambda=-\frac{8}{3}$

Which gives $x = 7/3, y = 14/3, z = 5/3$

So:

$D=\sqrt{(7/3-1)^2+(14/3-2)^2+(5/3-1)^2}=3.06$

Simply you must form the function $H = f -\lambda g$ , with a minus and not with a plus! With this simple change you get $\lambda = 0$ , as you should.

Tonio

**billym** · October 18th 2009, 05:11 PM

I see. I suppose this will explain my other wrong answers. I have about ten example questions from my notes and they all involve plus signs. How do you determine if its plus or minus? (obviously I've stopped visualizing it at this point)

**billym** · October 19th 2009, 11:43 AM

I've looked at a bunch of Lagrange tutorials, but they're all so different. I'm sure these basic questions aren't too difficult, but I seem to be stuck.

The question i am trying solve is:

Using the Lagrange multiplier method, find the minimum distance from the point (1,2,3) to any point on the surface defined by:

$x+2y-z=2$

Every example that I have tells me that I should solve like this:

We have:

$D^2=f(x,y,z)=(x-1)^2+(y-2)^2+(z-3)^2, g(x,y,z)=x+2y-z-2=0$

$f_x+\lambda g_x = 0 \Rightarrow$ ...(and so on)

$f_y+\lambda g_y = 0 \Rightarrow$ ...(and so on)

$f_z+\lambda g_z = 0 \Rightarrow$ ...(and so on)

Why do I not get the right answer unless I use $-\lambda$ in these equations?

**billym** · October 19th 2009, 11:59 AM

Actually that doesn't give me the correct answer either. In my journeys I have got:

$D=3.06,3,\sqrt 6$ , and now $6.22$

**tonio** · October 19th 2009, 12:11 PM

Originally Posted by billym

Actually that doesn't give me the correct answer either. In my journeys I have got:

$D=3.06,3,\sqrt 6$ , and now $6.22$

Well, I keep on getting zero....Let's see:

$H(x,y,z,\lambda)=f(x,y,z)-\lambda g(x,y,z)$ , and from here:

$H_{x}=f_{x}-\lambda g_{x}=2(x-1)-\lambda=0 \Longrightarrow x=\frac{\lambda}{2}+1$

$H_{y}=f_{y}-\lambda g_{y}=2(y-2)-2\lambda=0 \Longrightarrow y=\lambda +2$

$H_{z}=f_{z}-\lambda g_{z}=2(z-3)+\lambda=0 \Longrightarrow z=-\frac{\lambda}{2}+3$

Now just input in $g(x,y,z)$ this and you must get $3\lambda =0$

Tonio

**billym** · October 19th 2009, 12:12 PM

Ok so the the original equation should have been:

$D^2=f(x,y,z)=(1-x)^2+(2-y)^2+(3-z)^2$

Got it. Thanks. Goodnight.

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