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/
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