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Math Help - Line-Surface intersection

  1. #1
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    Line-Surface intersection

    Find the coordinates of the point(s) of intersection of the line x = 1 - t, y = 2 + 3t, z = 1 - t and the surface z = x^2 + 2y^2.

    Im completely lost. My book has not example of this and I dont remember seeing a problem like this.
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by kenshinofkin View Post
    Find the coordinates of the point(s) of intersection of the line x = 1 - t, y = 2 + 3t, z = 1 - t and the surface z = x^2 + 2y^2.

    Im completely lost. My book has not example of this and I dont remember seeing a problem like this.
    are you sure taht you have this copied correctly?
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  3. #3
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    Hello, kenshinofkin!

    I agree with Mathstud28 . . . The problem has no solution.


    Find the coordinates of the point(s) of intersection of the line: . \begin{array}{ccc}x &= &1 - t \\ y &= &2 + 3t \\ z &=& 1 - t\end{array}
    . . and the surface: . z \:= \:x^2 + 2y^2

    The usual way is to substitute the parametric equations into the surface's equation.

    Then we have: . z \:=\:x^2 + 2y^2

    . . . . . . . . . 1-t \;=\;(1-t)^2 + 2(2+3t)^2


    But this simplifies to: . 19t^2 + 23t + 8 \:=\:0 . . . which has no real solutions.

    . . The line does not intersect the elliptic paraboloid.

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  4. #4
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    hmm my teacher gave it to us. Maybe its wrong.
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