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Distance in 3D

  1. Apr 13, 2008 #1
    1. The problem statement, all variables and given/known data
    If distance |AB| = root 83, and i also know that |PA| is twice |PB|, then how do you find? |PA| or |PB| ?

    The whole question states: consider the points P such that the distance from P to A (-1,5,3) is twice the distance from P to B (6,2,-2).
    Show that the set of all such points is a sphere, and find its center and radius.
    please help!

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Apr 13, 2008 #2


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    Homework Helper

    Let P=(x,y,z) and then the vector PB would be (6-x,2-y,-2-z)
    Similarly, do with A...then use the fact that |PA|=2|PB|
  4. Apr 14, 2008 #3


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    Science Advisor

    Do you know the distance formula? Let P be the general point (x,y,z). Then the distance from P to (-1, 5, 3) is [itex]\sqrt{(x+1)^2+ (y-5)^2+ (z-3)^2}[/itex] and the distance from P to (6, 2, -2) is [itex]\sqrt{(x- 6)^2+ (y-2)^2+ (z+2)^3}[/itex]. Those are the |PA| and |PB| rock.freak667 is talking about. Put them into the equation he gives and simplify (I would square both sides).
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