How can one find the point of tangency between a plane and a sphere in $\mathbb{R}^3$? The equations of the plane and the sphere are $x + y + z - 5 = 0$ and $(x-1)^2 + (y-2)^2 + (z+1)^2 = 3$ respectively.
I realized that the distance between the point of tangency and the center of the sphere must equal its radius, but setting this constraint up just yielded the equation $x + y + z - 5 = 0$. Am I doing something wrong?
Intersect the line (1/2/-1) + t (1/1/1) with the plane.
The point (1/2/-1) is the mid-point of the sphere. The normal vector of the plane has the same direction as the line between the mid-point of the sphere and the point of tangency.
Note that the distance of the sphere-mid-point from the plane is $\sqrt{3}$, so the plane indeed is a tangency plane.
I guess you need a point on both the sphere and the plane where the gradient $\nabla f$ of $$f(x,y,z)=(x-1)^2+(y-2)^2+(z+1)^2$$ is parallel to (ie. a constant multiple of) $\langle 1,1,1 \rangle$. This leads to the following set of 5 equations in 4 unknowns. $$ \begin{array}{l} 2 (x-1)=\lambda \\ 2 (y-2)=\lambda \\ 2 (z+1)=\lambda \\ (x-1)^2+(y-2)^2+(z+1)^2=3 \\ x+y+z=5 \\ \end{array} $$
This is an overdetermined system because it's unusual for there to be a solution. In this particular case, there is a solution, namely $x=2$, $y=3$, $z=0$, and $\lambda=2$, as you can easily verify.
