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Math Help - Finding point on sphere furthest from point in space

  1. #1
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    Finding point on sphere furthest from point in space

    Find the point on the sphere (x - 2)^2 + (y - 2)^2 + (z-7)^2 = 99 which is furthest from the point C(1,3,4).

    Now I know that you suppose to find an equation of a line relating the centre of the sphere (2,2,7) and C(1,3,4) e.g. x = (1,3,4) + t(1,-1,3) where t can be any real number. But I dont understand what you would do to find the furthest point on the sphere?
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  2. #2
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    Quote Originally Posted by SyNtHeSiS View Post
    Find the point on the sphere (x - 2)^2 + (y - 2)^2 + (z-7)^2 = 99 which is furthest from the point C(1,3,4).

    Now I know that you suppose to find an equation of a line relating the centre of the sphere (2,2,7) and C(1,3,4) e.g. x = (1,3,4) + t(1,-1,3) where t can be any real number. But I dont understand what you would do to find the furthest point on the sphere?
    Think about it geometrically. To get from C to the furthest point on the sphere, you have to go from C to the centre of the sphere and then continue in the same direction for a distance equal to the radius of the sphere, which is \sqrt{99} = 3\sqrt{11}. You have already identified the direction of travel as the vector (1,1,3), which has length \sqrt{11}.

    Those \sqrt{11}s can't be coincidental, can they?
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  3. #3
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    Hello, SyNtHeSiS!

    \text{Find the point on the sphere }\,(x - 2)^2 + (y - 2)^2 + (z-7)^2 \:= \:99

    \text{which is furthest from the point }\,C(1,3,4).
    Code:
                  * * *         o C
              *           *   *
            *               ♥ P
           *              *  *
                        *
          *           *       *
          *         o         *
          *       * A         *
                *
           *  *              *
          Q ♥               *
          *   *           *
        *         * * *

    Construct a line through \,C and the center of the sphere \,A.

    It will intersect the sphere in two points, \,P and Q.
    . . We want Q, the point further from \,C.

    The line from C(1,3,4) to A(2,2,7) has direction \langle 1,-1,3\rangle

    The line has parametric equations: . \begin{Bmatrix}x &=& 1 + t \\ y &=& 3 - t \\ z &=& 4+3t \end{Bmatrix}


    Substitute into the equation of the sphere:

    . . \left[(1-t) - 2\right]^2 + \left[(3+t)-2\right]^2 + \left[([4-3t)-7\right]^2 \:=\:99

    . . 11t^2 + 22t + 11 \:=\:99 \quad\Rightarrow\quad t^2 + 2t + 1 \:=\:9

    . . t^2 + 2t - 8 \:=\:0 \quad\Rightarrow\quad (t-2)(t+4)\:=\:0 \quad\Rightarrow\quad t \:=\:2,\:-4


    \begin{array}{ccccccc}t = 2\!: & (x,y,z) &=& (\text{-}1,5,\text{-}2) & \text{point J} \\<br />
t = \text{-}4\!: & (x,y,z) &=& (5,\text{-}1,16) & \text{point K}\end{array}


    I suspect that point \,K is further from \,C, but let's make sure.


    Distance from C(1,3,4) to J(\text{-}1,5,\text{-}2)\!:

    . . \overline{CJ} \;=\;\sqrt{(\text{-}1-1)^2 + (5-3)^3 + (\text{-}2-4)^2} \;=\;\sqrt{44}


    Distance from C(1,3,4) to K(5,\text{-}1,16)\!:

    . . CK \;=\;\sqrt{(5-1)^2 + (\text{-}1-3)^2 + (16-4)^2} \;=\;\sqrt{176}

    Hence, K is the point we are seeking.


    Therefore: . Q(5,\text{-}1,16)
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  4. #4
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    Quote Originally Posted by Soroban View Post
    The line from C(1,3,4) to A(2,2,7) has direction \langle 1,-1,3\rangle
    Thanks for the great post. I was just curious why is it that choosing a line from A(2,2,7) to C(1,3,4) with direction <-1,1,-3> equivalent to the quote above? I thought this vector be the same magnitude but opposite, since points A and C is switched around.
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  5. #5
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    Just to point out that the makings of a much simpler solution were already in the original post, where the equation of the line from C to the centre of the sphere is given as x = (1,3,4) + t(1,-1,3). If you put t=1 in that equation then you get the point (2,2,7) (the centre of the sphere). To get to the furthest point of the sphere you have to travel a further distance along that same line. The extra distance that you need to travel is \sqrt{99} units, which is three times the length of the vector (1,–1,3).

    So to get to the furthest point of the sphere you need to take t = 4 (one unit of t to take you to the centre of the sphere, and another three units of t to take you to the furthest point). The furthest point is therefore at (1,3,4) + 4(1,–1,3) = (5,–1,16).
    Last edited by Opalg; September 8th 2010 at 12:45 PM. Reason: corrected typo
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