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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).

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  • $\begingroup$ @Peter: Look closer at the definition -- the surface is not a plane but has degree $2$. $\endgroup$ – Henning Makholm Dec 22 '16 at 18:46
  • $\begingroup$ Sorry, I did not notice that $\endgroup$ – Peter Dec 22 '16 at 18:47
  • $\begingroup$ The surface is an ellipsoid (E). Do you know how to compute the normal to any point $(x,y,z)$ ? $\endgroup$ – Jean Marie Dec 22 '16 at 18:58
  • $\begingroup$ @JeanMarie: No I don't know how to compute the normal to any point (x,y,z). $\endgroup$ – The First StyleBender Dec 22 '16 at 18:59
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    $\begingroup$ @JeanMarie: No, I have not learnt anything about Lagrange multipliers nor about maxi/minimization under constraints. This question should all be about Linear Algebra. My Physics professor suggested I should consider the rotation of axis in this question (but also explicitly mentioned that using/computing a rotational matrix is unnecessary and shouldn't be used). $\endgroup$ – The First StyleBender Dec 22 '16 at 19:13
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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.

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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

$u^T A u = 3\\ A = P^TDP\\ v = Pw\\ v^T\begin{bmatrix} 3\\&1\\&&1\end{bmatrix}v = 3$

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  • $\begingroup$ Caution: the coefficient of $z^2$ is $2$ in the first equation. $\endgroup$ – Jean Marie Dec 22 '16 at 22:37
  • $\begingroup$ Indeed, transcription error. Thanks. $\endgroup$ – Doug M Dec 22 '16 at 22:40
  • $\begingroup$ @DougM: by the way, are you dougal main from oxford university? $\endgroup$ – The First StyleBender Dec 23 '16 at 6:13
  • $\begingroup$ @foxielmao do I look like dougal main? sorry no. $\endgroup$ – Doug M Dec 23 '16 at 20:01
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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).

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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)

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  • $\begingroup$ Not relevant, surface not a plane. $\endgroup$ – coffeemath Dec 22 '16 at 18:54
  • $\begingroup$ Its ok but for plane it goes like what i mentioned above $\endgroup$ – H4K3R Dec 22 '16 at 18:55
  • $\begingroup$ Something is strange: you mention the distance between a plane and a line, and afterwards, from a plane to a point. $\endgroup$ – Jean Marie Dec 22 '16 at 19:01
  • $\begingroup$ #jeanmaire just figure out my answer is correct $\endgroup$ – H4K3R Dec 22 '16 at 19:08
  • $\begingroup$ The fact that the last sentence is correct is not all. As you are new in math SE, you will improve ! But take care when answering that you have coherent writing, and coherence also with the question. A last remark, you need an absolute value sign on the numerator of your formula. $\endgroup$ – Jean Marie Dec 22 '16 at 19:14

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