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Find the equation of a sphere.

  1. Jan 25, 2007 #1
    1. The problem statement, all variables and given/known data
    find an equation of a sphere if one of its diameters has endpoints (2,1,4)and
    (4,3,10)


    2. Relevant equations



    3. The attempt at a solution

    it should be in the form of (x-h)^2+(y-k)^2+(z-l)^2=r^2
    i think i found the radius by:
    r= sqrt((4-2)^2+(3-1)^2+(10-4)^2)
    r= sqrt(44)/2
    r= sqrt(11)
    im not sure about the coordinates for the center i think i should just take the difference between the two endpoints of the diameter?
    if someone could plz give me some info thank you
     
  2. jcsd
  3. Jan 25, 2007 #2
    nvm i figured it i need to use midpt formula ty anyway
     
  4. Jan 25, 2007 #3
    Your equation for the radius certainly seems to be fine, but the centre is not calculated by the difference of the points.

    Consider something simple like the real line. If you want to find the point between 3 and 1, you take the average of these two points. That is

    [tex]\frac{x_1+x_2}{2}[/tex]

    [tex]\frac{1+3}{2}=2[/tex]

    This applies in higher dimensional space [tex]\mathbf{R^3}[/tex] as well, and can be worked out co-ordinate wise. This will give you the centre of the circle.
     
  5. Jan 26, 2007 #4

    Gib Z

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    Yes the midpoint formula would have been alot easier huh? Just sub those points into the spheres equation >.<
     
  6. Jan 26, 2007 #5

    HallsofIvy

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    The formula for the radius is not quite "fine"!

    You have
    Your final result is good by that first equation does not have the "/2".
     
  7. Jan 26, 2007 #6
    Yes, I noticed this mistake as well, though I didn't think that it warranted comment. Seeing as how bobbarkernar did indeed derive the final result properly, I think it's fair to assume that the /2 was accidently ommited.
     
  8. Jan 27, 2007 #7

    HallsofIvy

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    Yes, but I'm just dang picky!!
     
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