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Math Help - maximum distance from point to sphere

  1. #1
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    maximum distance from point to sphere

    I have to calculate the maximum distance from the point T(1, -1, 3) to the sphere x^2+y^2+z^2-6x+4y-10z-62=0.

    Here's what I did.
    The equation for the sphere is (x-3)^2+(y+2)^2+(z-5)^2=100.
    That means the center of the sphere is C(3, -2, 5), and the radius is r=10.
    I think the maximum distance d_m is the distance from the point T to the center of the sphere plus the radius:
    d_m=d(T, C)+r.

    Is this okay?
    Please advise.
    Thank you!
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  2. #2
    A Plied Mathematician
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    Everything looks good so far. What do you get?
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  3. #3
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    I get
    d((1, -1, 3), (3, -2, 5))=\sqrt((3-1)^2+(-2+1)^2+(5-2)^2 )=\sqrt(9)=3 so d_m=3+10=13.

    It seems too simple, are you sure I'm not thinking something terribly wrong?

    Thank you for your quick reply!
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  4. #4
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    You're not computing the distance formula correctly. It should be more like

    \sqrt{(3-1)^2+(-2+1)^2+(5-2)^2}=\sqrt{4+1+9}=\sqrt{14}.

    [EDIT]: Ignore this post. See below for correction.
    Last edited by Ackbeet; January 31st 2011 at 03:57 PM. Reason: Incorrect.
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  5. #5
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    Why do we have (5-2)^2 instead of (5-3)^2? I'm afraid I don't understand.
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  6. #6
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    I'm sorry. You're correct. It should be 3, like you got.
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  7. #7
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    Thank you so much! I'm quite insecure when I'm doing the exercises on my own, it really helps when someone checks your work.
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  8. #8
    A Plied Mathematician
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    If you feel insecure when you're doing exercises, I would come up with a system of checks you can run against your answers to justify them to yourself. It's good problem-solving strategy anyway.

    You're very welcome. Have a good one!
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