# Point on surface closest to origin.

by mathman44
Tags: closest, origin, point, surface
 Mentor P: 21,216 You want the points on this surface whose distance is smallest. This means you want to minimize $$d~=~\sqrt{x^2 + y^2 + z^2}~=~\sqrt{x^2 + y^2 + xy + 1}$$ In the second square root, I replaced z2 with an expression it is equal to, because of the definition of the surface. Equivalently, you can minimize the distance squared, which is d2 = f(x, y) = x2 + y2 + xy + 1 You need to keep in mind that there is a domain here, {(x, y) | xy + 1 >= 0}. It seems very likely to me that you'll have a minimum point on the boundary of this domain.
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,316 Point on surface closest to origin. Or you could use Lagrange multipliers. Since the same point that minimizes distance will minimize distance square you can take $F(x,y,z)= x^2+ y^2+ z^2$ as the function to be minimized subject to the constraint that $G(x,y,z)= z^2- xy= 1$. The max or min will occur where $\nabla F$ and $\nabla G$ are parallel- that is, that $\nabla F= \lambda\nabla G$ for some number $\lambda$, the "Lagrange multiplier". Here that becomes $2x\vec{i}+ 2y\vec{j}+ 2z\vec{k}= \lambda(-y\vec{i}- x\vec{j}+ 2z\vec{k})$
 HW Helper P: 3,307 thats interesting geomtrically as well, as $\nabla G$ is perpindicular to the level surfaces of G(x,y,z). And $\nabla F$ will always point in the radial direction of the position vector so i think you could probably use this to show that the vector connecting a point with the closest point on a given surface, will be normal to the tangent plane of the surface...