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Closest Point to Origin in a Plane

Date: 11/04/2004 at 06:46:38
From: Ali
Subject: The closest point to origin

Let x + y + 3z = 7 represent a plane.  We find the closest point to
the origin in this plane by [d/[n]^2] * n.  In this case n = (1,1,3);
d = 7; [n]^2 = 1^2 + 1^2 + 3^2 = 11; then the vector that gives us the
closest point is (7/11, 7/11, 21/11).

I don't understand WHY this operation gives us the closest point and 
Strang's book doesn't really explain.  I'd appreciate it if you can help.



Date: 11/05/2004 at 23:28:31
From: Doctor Jordan
Subject: Re: The closest point to origin

Hi Ali,

This problem can be easily solved using Lagrange's method.  If you're
not familiar with it, look in the back of your book.  It should be in
there; I haven't seen Strang's book, but I think most multivariable
calculus courses should cover it.

We take f(x,y,z) = sqrt(x^2 + y^2 + z^2), and u = f + lambda(x+y+3z).
For Lagrange's method, the function we want to minimize/maximize is
taken as f.  In this case we want to find the minimum distance to the
origin for the point (x,y,z), and sqrt(x^2+y^2+z^2) is precisely the
distance of (x,y,z) to the origin.  The other term lambda(x+y+3z) is
the product of a constant lambda multiplying the nonconstant part of
the constraint.  In this case the constraint is that we must always
have x + y + 3z = 7.

I express partial derivatives with d here, but these are all partials.
Then we set df/dz = 0, df/dy = 0, df/dz = 0 (we can check that the
partials can take the value 0), and take this along with the 
constraint x + y + 3z = 7 as a system of equations.  In general, this 
is not a system of linear equations, so solving it can be difficult. 
For us, we note that the df/dx and df/dy both have lambda in them (so 
does df/dz, but it has 3lambda).  Thus we can move lambda to the right 
side in both which was 0 thus making both have -lambda on the right, 
and so they are equal to each other.  We divide away the 
sqrt(x^2+y^2+z^2) which is the denominator of each which can never be 
0 because of the constraint.  Then we have x = y.

We can then do the same with df/dx (or df/dy) and df/dz.  We note that
three of df/dx is equal to df/dz because of df/dz has three lambdas,
and with some algebra we see that z = 3x.

This is our solution.  This must be a minimum, because this function
clearly has no maximum (we can verify this if we want).  Thus the
minimum is found for x = x, y = x, z = 3x. 

But x + y + 3z = 7, so x + x + 9x = 7 so 11x = 7 so x = 7/11.  Thus 
the minimum of this function is at (7/11, 7/11, 21/11), which agrees 
with the answer from your book.

If your book doesn't cover Lagrange's method, you can probably find
one at your university library that does.  A good book is _Advanced
Calculus_ by Taylor and Mann, and you should find Lagrange's method in
the index.

Does this answer your question?  If you have any questions about this
solution or any other problems, please write me back!

- Doctor Jordan, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
College Calculus

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