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Parametrize plane-sphere intersection

  1. Aug 20, 2013 #1
    1. The problem statement, all variables and given/known data.
    Parametrize a circumference contained in the plane [itex]x+y+z=1[/itex], centered at [itex](2,-2,1)[/itex], and of radius [itex]40[/itex].


    2. The attempt at a solution.
    At first I thought I could intersect the plane [itex]x+y+z=1[/itex] with the sphere [itex](x-2)^2+(y+2)^2+(z-1)^2=40^2[/itex], but then I realized that this is wrong: the circumference can be obtained intersecting the given plane with some sphere, but this sphere doesn't necessarily have a radius of [itex]40[/itex]. So, what can I do know? Could I describe the circumference using polar coordinates? I had in mind a parametrization of the form: [itex]σ(t)=(40cos(t)+2, 40sin(t)-2, 1)[/itex] [itex]0≤t≤2π[/itex] but I'm not sure if this circumference lies on the plane.
     
  2. jcsd
  3. Aug 20, 2013 #2

    ehild

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    Your first equation is correct, but you have to add that x,y,z have to be points of the plane x+y+z=1. You can eliminate z .

    ehild
     
  4. Aug 20, 2013 #3

    HallsofIvy

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    What sphere that "doesn't necessarily have radius 40" are you talking about? There is only one sphere in this problem and it is the one with center at (2, -2, 1) with radius 40!

    It certainly doesn't because 40cos(t)+ 2+ 40sin(t)- 2+ 1= 40(cos(t)+ sin(t))+ 1, not 1!

    I see two ways to approach it:
    1) Write the sphere in spherical (not "polar") coordinates as
    [tex]x= 40 cos(\theta)sin(\phi)+ 2[/tex]
    [tex]y= 40 sin(\theta)sin(\phi)- 2[/tex]
    [tex]z= 40 cos(\phi)+ 1[/tex]
    and replace x, y, z in the equation of the plane, x+ y+ z= 1, with those. Since this "circumference" (I would say "circle"- circumference is a number) is one dimensional, you would want to use that equation to eliminate either [itex]\theta[/itex] or [itex]\phi[/itex]

    2) Start with x+ y+ z= 1 and write, say, z= 1- x- y. Replace z in [itex](x- 2)^2+ (y+ 2)^2+ (z- 1)^2= 1600[/itex] with that to get [itex](x- 2)^2+ (y+2)^2+ (x+y)^2=16[/itex]. Again solve the that equation for x as a function of y or y as a function of x to get a single parameter.
     
  5. Aug 20, 2013 #4
    You really should learn how to plot these things to get an intuitive feel for them. Since the radius of the sphere is greater than the distance from the x-y plane, it will intersect the x-y plane in a circle. Contrast a small sphere far away from the x-y plane which does not. Anyway, since the intersection is a circle, then we can parameterize that circle in terms of sines and cosines like you did but not exactly. So we have [itex]x+y+z=1[/itex] and [itex](x-2)^2+(y+2)^2+(z-1)^2=1600[/itex]. So what happens when we let z=0? Well, we have the equation:

    [tex](x-2)^2+(y+2)^2=1599[/tex]

    and we know [itex]z=1-(x+y)[/itex]

    Ok then, bingo-bango. Please provide a nice-looking plot of all this.

    Edit: Now that I plotted it, I believe this is wrong. Sorry about that.
     
    Last edited: Aug 20, 2013
  6. Aug 20, 2013 #5

    LCKurtz

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    I think there is a lot of misunderstanding in this problem because of the way it is stated. Since the point (2,-2,1) is on the plane ##x+y+z = 1## I believe the OP wants the circle in that plane which has radius ##40## and is centered at that point. He has used the term "circumference" instead of circle and has led things astray by deciding that that circle must have come from the intersection of the plane with some larger sphere.

    Assuming I am correct in this interpretation, here's what I would suggest.

    1. Start with the normal vector to the plane ##\vec n = \langle 1,1,1\rangle##.
    2. Calculate ##\vec u = \vec n \times \langle 1,0,0\rangle##. This will be a vector parallel to the plane.
    3. Calculate ##\vec v = \vec u \times \vec n##. This will be a vector perpendicular to both ##\vec u## and ##\vec n##.

    Now make unit vectors ##\hat u,~ \hat v## from ##\vec u## and ##\vec v##. This will be a pair of orthogonal unit vectors in the plane. Then parameterize your circle as ##\vec r(\theta) =
    \langle 2,-2,1\rangle +40\hat u\cos\theta + 40\hat v \sin\theta##.
     
  7. Aug 21, 2013 #6
    You're 100% correct in your interpretation, in spanish we use "circumference" (circunferencia) instead of "circle", that may have confused the others. The problem requests the circle to be of radius 40, the part about the sphere is something I've added to solve the problem. I'll use the information you gave me to do the parametrization. Thanks.
     
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