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?
 A: 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.
A: Define,
$$F(x,y,z)=z-x^2-y^2$$
$$G(x,y,z)=x+y-2z$$
Now because gradients are perpendicular to level curves we wish to find when $\nabla F=\lambda \nabla G$. 
So we have,
$$\langle -2x,-2y,1 \rangle = \lambda \langle 1,1,-2 \rangle$$
$$z-x^2-y^2=0$$
And we proceed. Finding the closest point on $z=x^2+y^2$ to the plane. And then using the formula for the distance from a point to the plane to proceed.
The closest point as you found is $(1/4,1/4,z)$. As we are on the surface $z=x^2+y^2$ then we have $z=(1/4)^2+(1/4)^2=1/8$. Then the distance is:
$$d=\frac{|1(1/4)+1(1/4)-2(1/8)-8|}{\sqrt{1^2+1^2+2^2}}$$
A: By visualizing the shape of the surfaces you should realize that the shortest distance will be between some point on the surface when $x=y$ (a parabola) and the plane when $x=y$ (a line).
There are at least two methods for solving this.  First, there is a general fact that if $L$ is a line with vector equation $r(t)=(a,b,c)+t(\alpha,\beta,\gamma)$ where $(\alpha,\beta,\gamma)$ is a unit vector and $P=(x_0,y_0,z_0)$ a point in $\mathbb{R}^3$
and if $d$ is the distance from $P$ to $L$, then 
$$d^2=(x_0-a)^2+(y_0-b)^2+(z_0-c)^2-[\alpha(x_0-a)+\beta(y_0-b)+\gamma(z_0-c)]^2$$
Under this fact we can simply solve the problem by fixing the line $x+x-2z=8$ and varying $x$ along the curve $z=x^2+x^2$.
To this end, let $(a,b,c)=(0,0,-4)$ and the unit direction vector is $\frac{1}{\sqrt{3}}(1,1,1)$.  Then our problem becomes minimizing the function $f(x)=d^2(P(x),L)$ where $P(x)$ is any point along the parabola.
$$f(x)= 2x^2+(2x^2+4)^2-\frac{1}{3}(2x+2x^2+4)$$
$$f'(x)=\frac{1}{3}(32x^3 -24x^2 + 68x-16)=0$$
now use whatever method you want to solve this (the cubic formula, rational root test, or bisection method starting at $0$ and $1$).
You'll find that $f'(\frac{1}{4})=0$ and consequently $d=\frac{31}{4\sqrt{6}}$.
Alternatively, you could do the more obvious but less elegant method of simply using the normal distance function between the surfaces and minimizing it.
To this end, let $(x,y,z)$ describe the surface of the parabola and $(u,v,w)$ describe the plane.
As before, $x=y$ and $u=v$ and $w=\frac{1}{2}(u+u-8)$ and $z=x^2+x^2$.
Then the function we want to minimize is 
$$f((x,y,z),(u,v,w))=f(x,u)=2(x-u)^2 + (u-4-2x^2)^2$$
The minimum occurs at 
$$\nabla f(x,u)=(-8ux-4u+16x^3+36x,6u-4x^2-4x-8)=(0,0)$$
solve for $u$ in terms8 of $x$ from the second equation. That is $u=\frac{1}{6}(4x^2+4x+8)$ then plug this into the other equation and you'll end up with a multiple of the cubic equation we found in the first method.
The solution for this is as before.
A: In general the cross-product method does not work. In this special case, it does, however. Suppose $A\in \Sigma_U = U^{-1}(0)$ and $B\in \Sigma_V=V^{-1}(0)$ verify:
$$ (A-B,A-B)=|A-B|^2 = \inf \{|P-Q|^2: P\in \Sigma_U, Q\in \Sigma_V\}>0$$
Then $|A-B|^2$ is stationary under tangential variations of $A$ and $B$. So let $$U'_A(h)=(\nabla U(A),h)=0 \ \ \mbox{and} \ \ V'_B(k)=(\nabla V(B),k)=0$$ 
describe such tangential variations. Then  $(h,A-B)=(k,A-B)=0$ meaning that $h$ and $k$ are any vectors orthogonal to $A-B$. But as they are also orthogonal to the gradients we must have:
$$ \nabla U(A)\  \parallel \nabla V(B) \parallel A-B .$$
That the first two are parallel is equivalent to their cross product being zero. But you also need that they should be parallel to $A-B$. In general for two curved surfaces, there is a phletora of (irrelevant) couples of points for which this  cross product vanishes, just imagine e.g. two non-intersecting spheres. For any point on one of them there are two points on the other for which this cross product vanishes. 
Now in your case the situation is very special as one surface is a plane for which the gradient $n=(1,1,-2)$ is constant. As there is only one point on $U$ for which the gradient is parallel to $n$ the method just happens to work in this case and gives the point $A$. To find the point $B$ you look at the line passing through $A$ in the direction $n$ and find its intersection with the plane. I omit the explicit solution which is given in other answers to this post.
In order to see that you really get a minimum for the couple $(A,B)$ you should also show that the surfaces don't intersect. Furthermore, that when $P\in\Sigma_U$ and $Q\in \Sigma_V$ go to infinity then so does their distance (not so difficult, simply show that $\Sigma_U$ does not intersect a ball $B(Q,r)$, $Q\in \Sigma_V$ with $r$ going to infinity as $Q$ goes to infinity). A compactness argument then shows that the (strictly positive) minimum is attained in some bounded region.
In the general (curved surface) case, a part from the compactness problem which may be difficult to show you may look at two Lagrange multipliers and look for extremal values of the function $$F(A,B,\lambda,\mu)=(A-B,A-B)+\lambda U(A) + \lambda V(B).$$ This gives rise to the two equations: 
$$A-B = \lambda \nabla U(A) \ \ , \ \   A-B = \mu \nabla V(B) $$
which should be solved under the constraints $U(A)=0$ and $U(B)=0$. As explained above this happens (because of $\Sigma_V$ being a plane) to be the same as finding the zero of the cross-product etc...
A: 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.
$$\begin{eqnarray}
\text{d}|PQ|^2
  &=& 2(x-\frac14)\text{d}x + 2(y-\frac14)\text{d}y + 2(\frac12(x+y-8)-\frac18)(\text{d}x + \text{d}y) \\
  &=& (2x-\frac12+x+y-8+\frac14)\text{d}x + (2x-\frac12+x+y-8+\frac14)\text{d}y \\
  &=& (3x+y-\frac{33}4)\text{d}x + (x+3y-\frac{33}4)\text{d}y
\end{eqnarray}$$
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:
$$\begin{eqnarray}
|PQ| &=& \sqrt{(\frac{33}{16}-\frac14)^2 + (\frac{33}{16}-\frac14)^2 + (\frac12(\frac{33}{16}+\frac{33}{16}-8)-\frac18)^2} \\
  &=& \sqrt{\frac{29^2}{16^2} + \frac{29^2}{16^2} + \frac{33^2}{16^2}} \\
  &=& \frac{\sqrt{2771}}{16} \approx 3.290018047062964
\end{eqnarray}$$
