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, and [MATH\mathbf{r}_{2},[/tex] with the given distance being
then let with length
By the Pythagorean theorem, the circle in question is going to be the intersection of the sphere centered at of radius , with the plane
That is, the circle is the set of all points such that
and
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.