Intersection of Sphere and Plane - Lagrange Multiplier

Question

I got a sphere with $x^2+y^2+z^2=1$ and a plane with $x+y+z=1$. I should minimize and maximize the distance from the intersection to (0,0,0).

First thing that threw me off, is that $x^2+y^2+z^2=1$ is centered around $(0,0,0)$. So, every point that is on the sphere, and therefore every point that is on the intersection, has the same distance to $(0,0,0)$, right? Not sure what to minimize?

Since we are so supposed to use Langrage Multipliers, I tried

$$ f:x^2+y^2+z^2=1$$ $$g:x+y+z-1=0 $$ $$ \nabla_f=\nabla_g \lambda, $$ and got $$\left\{\begin{array}{l}2x=\lambda \\2y=\lambda \\2z=\lambda \end{array}\right.$$

Together with g that gave me $x=y=z=1/3$.

Geometrically I have $(1,0,0),(0,1,0),(0,0,1)$ on my intersection, and I concluded that the intersection is the circle with the center $(1/3,1/3,1/3)$ and radius ${\sqrt2/\sqrt3}$.

Why did the Lagrange Multplier Method give me the center of the circle? Shouldn't it give me the circle itself? I am not sure what I did wrong. Also: Any guesses what one could maximize/minimize here?

Every help is much appreciated!

Best regards!

Answer 1

The problem makes no sense for the reason that you have explained: all points of that line are at the same distance from $(0,0,0)$.

Besides, what you did is not the right approach. Suppose that you want to find the minimum and the maximum of a function $\varphi$ restricted to$$\{(x,y,z)\in\Bbb R^3\mid x^2+y^2+z^2=0\wedge x+y+z=1\}.$$In order to do that, you define $f(x,y,z)=x^2+y^2+z^2$ and $g(x,y,z)=x+y+z$, and then you solve the system$$\left\{\begin{array}{l}\nabla\varphi=\lambda\nabla f+\mu\nabla g\\f(x,y,z)=1\\g(x,y,z)=1\end{array}\right.\iff\left\{\begin{array}{l}\varphi_x=2x\lambda+\mu\\\varphi_y=2y\lambda+\mu\\\varphi_z=2z\lambda+\mu\\x^2+y^2+z^2=1\\x+y+z=1.\end{array}\right.$$

For instance, if you want to find the points of that line which minimize and maximize the distance to $(2,0,0)$, then you can take $\varphi(x,y,z)=(x-2)^2+y^2+z^2$, and then you will get the points $(1,0,0)$ ($\longleftrightarrow$ minimal distance) and $\left(-\frac13,\frac23,\frac23\right)$ ($\longleftrightarrow$ maximal distance).

Answer 2

Intersection of a sphere/plane is a small circle, whose points when connected to the origin form the cone base, all whose points are at the same distance from the apex of the cone...so a non-starter.

Either shift the given point away to say $ ( 1,0,2) $ etc. or add non all-unity-coefficients to make the surface as non-spherical conicoid in $ a x^2+b y^2+c z^2=1$ before starting to find extrema.