The moment I left the office I realised it. Excuse me for my sudden loss of intelligence.

// Myx

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- Feb 8th 2011, 07:33 AMMyxSolved - Finding positions at the same distance from two given points
The moment I left the office I realised it. Excuse me for my sudden loss of intelligence.

// Myx - Feb 8th 2011, 08:08 AMAckbeet
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

$\displaystyle \mathbf{n}\cdot\left(\mathbf{x}-\mathbf{n}/2\right)=0.$ - Feb 8th 2011, 09:15 AMPlato
"

**The set of all points equidistant from two points in 3D space**" is a**plane**that is perpendicular to the line segment determined by the two points at its midpoint. - Feb 8th 2011, 09:19 AMAckbeet
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.

- Feb 8th 2011, 10:37 AMMyx
Thank you for your answer, that is indeed pretty much the solution I came up with.

The moment I took a breath of fresh air I felt very silly as Pythagora came to mind and solved the problem.