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Homework Help: Calc III: sphere & plane intersection.

  1. Jan 22, 2010 #1
    Consider the intersection of the sphere (x-3)^2+(y+2)^2+(z-1)^2=13 with the plane x+y=0

    a) This intersection should be a familiar curve. Describe the curve.

    b) Substituing x=-y or y=-x into the equation of the sphere does not give this curve. Explain the difference.


    I know that the intersectionis a circle but I am having trouble proving that. I rearranged the above equation and set it equal to x+y:

    (x-3)^2+(y+2)^2+(z-1)^2-13=x+y

    The next thing I did was put the like terms together to give the following equation:

    x^2-6x+9-x+y^2+4y+4+y+z^2-2z+1-13=0

    I then simplified this equation:

    X^2-7x+9+y^2+3y+4+(z-1)^2-13=0

    Now I am stuck. This just looks like a smaller sphere.
    Also, I would plug in x=-y or vice versa but the second part of the question says that will not give me the result that I want.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jan 22, 2010 #2

    Mark44

    Staff: Mentor

    The intersection of a sphere and a plane will be one of the following.
    1. The empty set (the two don't actually intersect).
    2. A single point.
    3. A circle.
     
    Last edited: Jan 22, 2010
  4. Jan 22, 2010 #3

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi hq1330! :smile:

    (try using the X2 tag just above the Reply box :wink:)
    I think this question is unnecessarily obscure. :frown:

    A curve, in three dimensions, requires two equations to describe it, so of course substituing x=-y or y=-x into one equation, which will only give one equation, cannot give the curve.

    (in fact, it will give the two-dimensional projection of the curve onto the xz or yz plane)

    Sooo … plug away! :smile:

    (also, the method you used gives a whole surface, including eg the intersection of the sphere (x-3)2+(y+2)2+(z-1)2=14 with the plane x+y=1, doesn't it? :wink:)
     
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