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Math Help - Shortest distance plane, sphere [difficult]

  1. #1
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    Shortest distance plane, sphere [difficult]

    Given the sphere
    (x-3)^2 + (y-4)^2 + z^2 = 36

    And the plane
    3x-2y+4z=8

    Find the shorterst distance from the center of the sphere to the plane.
    ___________
    This is what I've tried:

    Center of the sphere C(3, 4, 0)

    The normal vector of the plane \vec n =[3, -2, 4]

    I chose an arbitrary point on the plane P(x, y, z)

    Which gives us;
    [x, y, z]=[3, 4, 0]+[3, -2, 4]t

    So the distance should be
    \rm{D} = \sqrt{(3+3t)^2 + (4-2t)^2 + (4t)^2}

    Where;
    f(t) = 29t^2 + 2t + 25

    Which gives;
    f\prime(t) = 58t+2

    And
    f\prime(t) = 0

    t = -\frac{2}{58}=\underline{-\frac{1}{29}}

    So
     f(-\frac{1}{29}) = 29(-\frac{1}{29})^2 + 2(-\frac{1}{29}) + 25 = \underline{\frac{724}{29}}

    \rm{D} = \sqrt{\frac{724}{29}}

    But this is not correct according to my book. What am I doing wrong?
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  2. #2
    Super Member wingless's Avatar
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    Quote Originally Posted by MatteNoob View Post
    Given the sphere
    (x-3)^2 + (y-4)^2 + z^2 = 36

    And the plane
    3x-2y+4z=8

    Find the shorterst distance from the center of the sphere to the plane.
    ___________
    This is what I've tried:

    Center of the sphere C(3, 4, 0)

    The normal vector of the plane \vec n =[3, -2, 4]

    I chose an arbitrary point on the plane P(x, y, z)

    Which gives us;
    [x, y, z]=[3, 4, 0]+[3, -2, 4]t
    From there, follow like this:
    - Write the line as a vector function in parametric form.
    - Find the limits of t so the line will be limited between (3,4,0) and the plane.
    - The length is \int_a^b |V|~dt where a and b are the limits of t, and V is the velocity of the line (derivative of the line function).
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  3. #3
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    Find the shorterst distance from the center of the sphere to the plane.
    Don't make it hard.
    If A is a point not on the plane \Pi and P \in \Pi then the distance from A to \Pi is D(\Pi ,A) = \frac{{\left| {\overrightarrow {PA}  \cdot n} \right|}}{{\left\| n \right\|}}.

    You can find a point on the plane. You already know the center and normal.
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  4. #4
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    Before there was calculus or vector notation, there was geometry and algebra.

    Plane: 3x - 2y + 4z - 8 = 0
    Point: (3,4,0)

    \frac{|3(3) - 2(4) + 4(0) - 8|}{\sqrt{3^{2}+(-2)^{2}+4^{2}}}

    You will pardon me if I updated the notation from the ancient days.
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  5. #5
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    Hello, MatteNoob!

    You started off great . . . I'd take a different direction.


    Given the sphere: . (x-3)^2 + (y-4)^2 + z^2 \:= \:36\;\;{\color{blue}[1]}

    and the plane: . 3x-2y+4z\:=\:8\;\;{\color{blue}[2]}

    Find the shortest distance from the center of the sphere to the plane.

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    This is what I've tried:

    Center of the sphere: . C(3, 4, 0)

    The normal vector of the plane: . \vec n \:=\:[3, -2, 4] . . . . Good!
    The equation of the line through C, perpendicular to the plane: . \begin{array}{ccc}x &=& 3+3t \\ y &=& 4 - 2t \\ z &=&0 + 4t \end{array}\;\;{\color{blue}[3]}

    Now where does this line intersect the plane?

    . . Substitute [3] into [2]: . 3(3+3t) - 2(4-2t) + 4(4t) \:=\:0

    . . Solve for t\!:\;\;t \:=\:-\frac{1}{29}

    Substitute into [3]: . \begin{array}{ccccc}x &=& 3 + 3\left(\text{-}\frac{1}{29}\right) &=& \frac{84}{29} \\ \\[-3mm]<br />
y &=& 4 - 2\left(\text{-}\frac{1}{29}\right) &=& \frac{118}{29} \\ \\[-3mm]<br />
z &=& 4\left(\text{-}\frac{1}{29}\right) &=& \text{-}\frac{4}{29} \end{array}

    Hence, P\left(\frac{84}{29},\:\frac{118}{20},\:\text{-}\frac{4}{29}\right) is the nearest point on the plane to center C.


    Now, find the distance: . \overline{CP}.

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  6. #6
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    I chose an arbitrary point on the plane P(x, y, z)

    Which gives us;
    [x, y, z]=[3, 4, 0]+[3, -2, 4]t
    Is this really a description of points in the plane? It appears to me that only one point in the plane fits this description.
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