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Math Help - Finding positions at the same distance from two given points

  1. #1
    Myx
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    Solved - 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
    Last edited by Myx; February 8th 2011 at 07:58 AM. Reason: Felt very stupid once I got some fresh air
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  2. #2
    A Plied Mathematician
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    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, \mathbf{r}_{1} and [MATH\mathbf{r}_{2},[/tex] with the given distance being

    a>\sqrt{(\mathbf{r}_{1}-\mathbf{r}_{2})\cdot(\mathbf{r}_{1}-\mathbf{r}_{2})},

    then let \mathbf{n}=\mathbf{r}_{2}-\mathbf{r}_{1}, with length

    n=\|\mathbf{n}\|.

    By the Pythagorean theorem, the circle in question is going to be the intersection of the sphere centered at \mathbf{n}/2 of radius \sqrt{a^{2}-n^{2}/4}, with the plane

    \mathbf{n}\cdot\left(\mathbf{x}-\mathbf{n}/2\right)=0.

    That is, the circle is the set of all points \mathbf{x} such that

    (\mathbf{x}-\mathbf{n}/2)\cdot(\mathbf{x}-\mathbf{n}/2)=a^{2}-n^{2}/4, and

    \mathbf{n}\cdot\left(\mathbf{x}-\mathbf{n}/2\right)=0.
    Last edited by Ackbeet; February 8th 2011 at 09:19 AM. Reason: Refer to Plato's post.
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  3. #3
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    "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.
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  4. #4
    A Plied Mathematician
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    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.
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  5. #5
    Myx
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    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.
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