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
On
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 $(a,b,0)$.
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.