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Math Help - Vector Calculus equation for planes

  1. #1
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    Vector Calculus equation for planes

    I am having real trouble understanding this concept and I am hoping that maybe someone would be willing to help me out.

    I need to find the equation for a plane that just touches the sphere (x - 2)^2 + (y - 3)^2 = (z - 3)^2 = 16 and is parallel to 1) xy plane 2) yz plane and 3) the xz plane. If someone can walk me through one of these I should be able to apply the info to the other two. Thanks very much!!!

    Also, if the function (x,y,z) = x^2 + y^2 + z^2 - 2x + 3y + z I need to find a point (a,b,c) such that the function can be expressed in terms of distance from (a,b,c)

    I know I need to complete the square but I am lost........

    Can someone recommend a good online 3D graphing program? I need to really work on understanding what different equations look like in 3D. Thanks again, this site is WONDERFUL!
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  2. #2
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    Bad answer.
    Last edited by lukaszh; January 13th 2009 at 04:07 AM.
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  3. #3
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    Quote Originally Posted by Frostking View Post
    I need to find the equation for a plane that just touches the sphere (x - 2)^2 + (y - 3)^2 + (?) (z - 3)^2 = 16 and is parallel to 1) xy plane 2) yz plane and 3) the xz plane.
    A plane parallel to the x-y plane has equation of the form z=k for some constant k.

    The sphere (x - 2)^2 + (y  - 3)^2 + (z - 3)^2 = 16 has centre (2,3,3) and radius 4. So at its highest point z will be 3+4, and at its lowest point z will be 34. There are therefore two planes parallel to the x-y plane that touch the sphere, namely z=7 and z=-1.

    Quote Originally Posted by Frostking View Post
    Also, if the function (x,y,z) = x^2 + y^2 + z^2 - 2x + 3y + z I need to find a point (a,b,c) such that the function can be expressed in terms of distance from (a,b,c)

    I know I need to complete the square
    That is correct. Write the function as f(x,y,z) = x^2 2x + y^2 + 3y + z^2 + z. Complete the squares in the x, y and z terms, and you will get f(x,y,z) = (xa)^2 + (yb)^2 + (zc)^2 k (for some constants a, b, c, k). That tells you that f(x,y,z) is the square of the distance from (x,y,z) to (a,b,c), less k.
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