Distance between surfaces I'm trying to find the minimal distance between the surfaces described by $z=x^2+y^2$ and $x+y-2z=8$. I would imagine there are several approaches including the use of Lagrange multipliers. I attempted to find the spot when their normal vectors are parallel ( since i believe if these vectors are not parallel there is a direction that will bring the distance lower). So if $U(x,y,z)=x^2+y^2$ and $V(x,y,z)=x+y-2z$ are functions describing these surfaces then their normal vectors are $(2x,2y,-1)$ and $(1,1-2)$ respectively. Then I want their cross product to be zero; this is at $(1/4,1/4,z)$. But when i solve for $z$ and I cannot simultaneously satisfy both equations. What gives?! Am I along the right line of thinking?
You won't be able to solve for $z $ on both equations  if the surfaces dont intersect. You have to find the $z $ coordinate of the point of each surface with $x = y = \frac14$ and then find the distance between the two points.
You have correctly identified that the normal to the bottom surface is $(1, 1, -2)$. You have correctly identified that the only point on the top surface with the same normal (up to scaling) is $(\frac14, \frac14, \frac18)$, which I will call $P$. It remains to find the point on the bottom surface that is closest to $P$.

We can rewrite the formula for the bottom surface as $z = \frac12(x+y-8)$. As we vary $x$ and $y$, the point $Q = (x, y, \frac12(x+y-8))$ explores the whole surface. The distance we need to minimise is

$$|PQ| =
\sqrt{(x-\frac14)^2 + (y-\frac14)^2 + (\frac12(x+y-8)-\frac18)^2}$$

For the sake of an easy life, let's minimise $|PQ|^2$ instead.

$$\begin{eqnarray}
\text{d}|PQ|^2
  &amp;=&amp; 2(x-\frac14)\text{d}x + 2(y-\frac14)\text{d}y + 2(\frac12(x+y-8)-\frac18)(\text{d}x + \text{d}y) \\
  &amp;=&amp; (2x-\frac12+x+y-8+\frac14)\text{d}x + (2x-\frac12+x+y-8+\frac14)\text{d}y \\
  &amp;=&amp; (3x+y-\frac{33}4)\text{d}x + (x+3y-\frac{33}4)\text{d}y
\end{eqnarray}$$

This is zero when $3x+y = x+3y = \frac{33}4$, i.e. when $x = y = \frac{33}{16}$. Plugging this into the formula above gives:

$$\begin{eqnarray}
|PQ| &amp;=&amp; \sqrt{(\frac{33}{16}-\frac14)^2 + (\frac{33}{16}-\frac14)^2 + (\frac12(\frac{33}{16}+\frac{33}{16}-8)-\frac18)^2} \\
  &amp;=&amp; \sqrt{\frac{29^2}{16^2} + \frac{29^2}{16^2} + \frac{33^2}{16^2}} \\
  &amp;=&amp; \frac{\sqrt{2771}}{16} \approx 3.290018047062964
\end{eqnarray}$$