Finding the closest distance between a point and a surface Assuming that I am given a matrix (A) and a vector (b) which represents a surface, say
\begin{bmatrix}2&0&1\\0&1&0\\1&0&2\end{bmatrix} and \begin{bmatrix}4\\0\\2\end{bmatrix}
and the equation representing the surface as:
$r^TAr+b^Tr=1$
If I am asked to find the closest distance between point "a" and the surface, represented by a position vector of
\begin{bmatrix}-1\\0\\0\end{bmatrix}
What am I supposed to do?

Below is what I've attempted:
Using this as an example, I tried to rewrite the equation representing the surface into quadratic form, in this case:
$x^T Ax=3$
where $r=x+a$ (note: $a$ is the point where I am trying to find it's distance from the surface).
I'm thinking about rotating the axis, however I am stuck here (my teacher told me to try not consider a rotation matrix).
 A: My first idea for what to do: First, rewrite your equation $$r^TAr + b^Tr = 1 $$ into $$(r-c)^TA(r-c) = d$$
(You can find a suitable $c$ by multiplying out $(r-c)^TA(r-c)$, rewriting $r^TAc=c^TAr$ and solving $b^T = -2c^TA$. Then $d$ must be $1+c^TAc$ -- though I haven't triple-checked my reasoning for sign errors; caveat lector).
$A$ is real symmetric and therefore orthogonally diagonalizable; it is easy to see that its eigenvalues are $1,1,3$ and therefore your surface is an oblate spheroid centered on $c$.
Now the natural thing to do would be to transform $a-c$ by the orthogonal diagonalizing matrix, so you end up with $a-c$ in a coordinate system where $A=\operatorname{diag}(1,1,3)$. Then by symmetry the shortest line betwen $a$ and the spheroid would be in the plane that contains the transformed point and the $z$-axis, and we've reduced the problem to finding the shortest distance between a point and an ellipse. That seems to have no nice closed-form solution, but at least it is now easy to parameterize the ellipse and find a minimum numerically by setting the derivative of the squared distance to $0$.
However, this works because the diagonalizing matrix for $A$ can be chosen to be orthogonal -- that is, a rotation matrix -- which sounds like what your teacher is asking you not to do.

It is possible that you're supposed to use Lagrange multipliers instead.
A: We are fortunate that the surface is an ellipsoid and (-1,0,0) is the center.
$2x^2 + y^2 + 2z^2 + 2xz + 4x + 2z = 1\\
4x^2 + 2y^2 + 4z^2 + 4xz + 8x + 4z = 2\\
3(x+z+1)^2 + 2y^2 + (x-z+1)^2 = 6$
The shortest distance from (-1,0,0) to the edge of the ellipse is in the (1,0,1) direction.
$x+1 = z, y = 0$
$3(2z)^2 = 6$
the closest point on the surface is $(\frac {\sqrt 2}{2} - 1, 0, \frac {\sqrt 2}{2})$
and the distance from $(-1,0,0)$ is $1$
Suppose you want to do this using linear algebra.
$r = u + x_0 \\
u^T A u +2x_0^TA u + b^t u + x_0^tAx_0 + b^Tx_0  = 1$
$2x_0^TA u = b^t u$ since $x_0$ is the center
$u^T A u = 3\\ 
A = P^TDP\\
v = Pw\\
v^T\begin{bmatrix} 3\\&1\\&&1\end{bmatrix}v = 3$
A: Here is a solution using gradients. I am aware that the OP has said that he has not studied this concept. May this first contact wet his/her apetite for understanding this very fundamental tool, maybe with the assistance of his physics teacher that helps him/her.
Definition: the gradient at a certain point $M(x,y,z)$ of a surface $(S)$ with implicit equation $f(x,y,z)=0$ is the vector of partial derivatives 
$$grad(f):=(\partial f/\partial x, \partial f/\partial y,\partial f/\partial z)$$
with respect to the different variables. 
The fundamental property of $grad(f)$: it gives the orientation of the normal to surface $(S)$ at point $M(x,y,z).$ 
Consider now our surface (S) with equation:
$$f(x,y,z)=2x^2 + y^2 + 2z^2 + 2xz + 4x + 2z -1=0$$
Our formulation will rely on a basic property: the shortest distance of point $B(-1,0,0)$ to surface $(S)$ is realized for a point $M(x,y,z)$ of $(S)$ such that $\vec{BM} \perp (S)$. Otherwise said, such that $\vec{BM}$ is proportional to $grad(f)$. Let $\lambda$ be the proportionality ratio. We thus have a system of 4 equations in 4 unknowns $x,y,z,\lambda$, the first 3 equations express the upsaid proportionality, the last one expresses that $M(x,y,z)$ lies on surface $(S)$:
$$\begin{cases}x + 1 = \lambda*(4x + 2z + 4)\\y = \lambda * 2y\\z = \lambda*(4z + 2x + 2)\\
 2x^2 + y^2 + 2z^2 + 2x z + 4 x + 2 z=1\end{cases}$$
This system can be solved by a Computer Algebra Software (Mathematica), which gives the following set of solutions:
$$\begin{cases}x=-z-1 & y=\pm\sqrt{3-2z^2} & \text{for any} \ z \in [-\sqrt{6}/2,\sqrt{6}/2] & \lambda=1/2 & \\
x=-1\pm\sqrt{2}/2 & y=0 & z=\pm\sqrt{2}/2 & \lambda=1/6 
\end{cases}$$
(the values of $\lambda$ are unimportant). 
The first line means that there is an infinite set of candidate solutions on a certain sectional ellipse.
One verifies that it is the second line which gives an optimal solution with $\pm$ replaced by $+$. This coincides with the result of @Doug M.
Remark: There are many solutions because we are in a very particular case: $B$ is situated at the center of surface $(S)$ (an ellipsoid).
A: Closest distance between a plane and a line goes like this:
If plane is ax+by+cz=d where a,b,c,d are constants and the point is (h,k,l)
Then shortest distance is
(ha+bk+cl-d)/root(a^2+b^2+c^2)
