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Find the closest point on the plane $x+2y+3z=4$ to the point $(0,1,0)$ and the minimum distance.

How can this be solved with another method (not Lagrange method)?

I had no problem to solve it using the Lagrange multipliers. My question is, we know that planes (and lines) are not compact. So how we can explain the existion of an absolute extrema (absolute minimum). And in general what we are supposed to do in similar cases where tha constraints set is not compact?

Many thanks

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2 Answers 2

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The method of Lagrange multipliers, without further ado, only states that if you have a constraint extremum, then at a solution of some system of equations. So it' just a necessary condition.

To prove that you indeed have a local constained extremum or even a global extremum you must come up with something else. If the constraint is fulfilled only on a compact set, then you mentioned how to solve it. If you have an unbounded plain the usual approach would be to show that outside some ball around your special point, the distance to that point is bounded from below by something bigger than the suspected minimum value. Then you know that the value you have found is indeed the global minimum.

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  • $\begingroup$ Yes, so I took a ball with a large enough radius around my point/s, (which I got using the Lagrange method), such that the ball intersect my plane so this intersect will be a compact set so by weirestrass we'll have absolute points, because points that are outside the ball their distance from (0,1,0) is bigger than those points that points in the ball. @nicrot000 $\endgroup$
    – user864806
    Commented Jun 9, 2021 at 15:25
  • $\begingroup$ I don't know. I can only tell, that if I had to grade your exercises, I would definitely expect you argue why your result is a minimum. Otherwise it's just a false/incomplete conclusion of the theorem. $\endgroup$
    – nicrot000
    Commented Jun 9, 2021 at 21:34
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You were asking for an alternative method. One is to eliminate the constraint. Here we can recast the problem by rotating space so that the given plane becomes $w=0$, while the target points becomes $(a,b,c)$.

Now we solve the unconstrained problem $$\min((u-a)^2+(v-b)^2+(w-c)^2)=\min((u-a)^2+(v-b)^2+c^2),$$ having the obvious solution $(a,b,0)$.

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  • $\begingroup$ Hi @Yves Daoust thanks!. Can I ask, when solving it with Lagrange method, how can I justify that there's actually a minimum point (here the plane is not compact so we cannot use the weirestrass theorem!)? $\endgroup$
    – user864806
    Commented Jun 10, 2021 at 7:04
  • $\begingroup$ @Bestmat: I answered for an alternative method. For Lagrange, see the one you accepted. $\endgroup$
    – user65203
    Commented Jun 10, 2021 at 7:16

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