The moment I left the office I realised it. Excuse me for my sudden loss of intelligence.
// Myx
The moment I left the office I realised it. Excuse me for my sudden loss of intelligence.
// Myx
I see you've found the answer to your problem. Here's my answer, if you find it helpful:
Sounds like a circle to me. You're looking for the set of all points equidistant from two points in 3D space, right? [EDIT]: See Plato's post below for a clarification of this phraseology. That's going to be a circle. If you have two points, $\displaystyle \mathbf{r}_{1}$ and [MATH\mathbf{r}_{2},[/tex] with the given distance being
$\displaystyle a>\sqrt{(\mathbf{r}_{1}-\mathbf{r}_{2})\cdot(\mathbf{r}_{1}-\mathbf{r}_{2})},$
then let $\displaystyle \mathbf{n}=\mathbf{r}_{2}-\mathbf{r}_{1},$ with length
$\displaystyle n=\|\mathbf{n}\|.$
By the Pythagorean theorem, the circle in question is going to be the intersection of the sphere centered at $\displaystyle \mathbf{n}/2$ of radius $\displaystyle \sqrt{a^{2}-n^{2}/4}$, with the plane
$\displaystyle \mathbf{n}\cdot\left(\mathbf{x}-\mathbf{n}/2\right)=0.$
That is, the circle is the set of all points $\displaystyle \mathbf{x}$ such that
$\displaystyle (\mathbf{x}-\mathbf{n}/2)\cdot(\mathbf{x}-\mathbf{n}/2)=a^{2}-n^{2}/4,$ and
$\displaystyle \mathbf{n}\cdot\left(\mathbf{x}-\mathbf{n}/2\right)=0.$
Sorry. My phraseology was unfortunate. The OP originally stated that he wanted all the points that are the same, single, distance from two points. That is, you have two points A and B, and you want the locus of all points that are a distance a away from A, and a distance a away from B, where a is greater than the distance from A to B.