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Intersection of a Quadric Surface and a Plane in 3-D

  1. Jan 28, 2014 #1
    1. "Find the equation that describes the intersection of the quadric given by [itex]x^2 + y^2 = 4[/itex] with the plane [itex]x + y + z = 1[/itex]."


    2. Parametric equations for elliptic curve: [itex]x = a cos(t)[/itex] , [itex]y = b sin(t)[/itex] , z = ?


    3. Surface is an [EDIT: right circular] cylinder. Plane is not parallel to xy plane, has some increasing z and will not cut a circle in the cylinder. Curve of intersection of plane and elliptic cylinder will result in an elliptic curve.

    3a. My first approach would be to plug in [itex]a cos(t)[/itex] and [itex]b sin(t)[/itex] for x and y respectively in the [itex]x^2 + y^2 = 4[/itex] equation.


    [itex](a cos(t))^2 + (b sin(t))^2 = 4[/itex] ---> [itex]a^2(cost)^2 + b^2(sint)^2 = 4[/itex] ---> [itex]\frac{a^2(cost)^2}{4} + \frac{b^2(sint)^2}{4} = 1[/itex] ---> a = b = 2 --> [itex]x = 2cost[/itex] and [itex]y = sint[/itex]

    This step confuses me since I would think that because the curve is elliptic, the a and b coefficients would differ.


    3b. My second step would be to rearrange the equation of the plane to be some z = x + y + c to find the parametric equation for z.

    [itex]x + y + z + (- x - y) = 1 + ( - x - y)[/itex] ---> [itex]z = - x - y + 1[/itex] ---> [itex]z = - (2cost) - (2sint) + 1[/itex]

    3c. Last step would be to combine everything into parametric equations.

    [itex]x = 2cos(t)[/itex] , [itex]y = 2sin(t)[/itex] , [itex]z = - 2sin(t) - 2cos(t) + 1[/itex].

    I think I have the right idea, but any suggestions or guidance would be appreciated.
     
    Last edited: Jan 28, 2014
  2. jcsd
  3. Jan 28, 2014 #2

    HallsofIvy

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    This is wrong. [tex]x^2+ y^2= 4[/itex] is the equation of a circular cylinder. It can be written as x= 2cos(t), y= 2 sin(t). Yes, its intersection with the tilted plane will be an ellipse but that is not relevant here.

    The plane, x+ y+ z= 1 can be written 2cos(t)+ 2sin(t)+z= 1 so that z= 1- 2cos(t)- 2sin(t).
    And that gives the parametric equations describing the intersection:
    x= 2cos(t)
    y= 2sin(t)
    z= 1- 2cos(t)- 2sin(t).
     
  4. Jan 28, 2014 #3
    Thank you. Don't know why I typed elliptic cylinder. I think I understand why it doesn't matter that a doesn't equal b. Since the curve of intersection is tilted by the z component anyways, it's going to be an ellipse. If the z parametric equation wasn't in effect, the curve would be a circle. I think this makes sense. Is my overall logic and procedure sound?
     
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