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Math Help - Intersection of line and surface

  1. #1
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    Intersection of line and surface

    Find the coordinates of the points where the line

    x = t, y = 1 + t, z = 5t

    intersects the surface

    z = x^2 + y^2
    Last edited by mr fantastic; March 10th 2010 at 04:07 PM. Reason: Changed post title
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  2. #2
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    Hello, looseenz2!

    I have a strange answer . . . Did I mess up somewhere?


    Find the coordinates of the points where the line: . \begin{Bmatrix}x &=& t \\ y &=& 1 + t \\ z &=& 5t\end{Bmatrix} .[1]
    . . intersects the surface: . z \:=\: x^2 + y^2 .[2]

    Substitute [1] into [2]: . 5t \:=\:t^2 + (1+t)^2 \quad\Rightarrow\quad 3t^2 - 2t - 1 \:=\:0

    . . . . . . . . . . . . . . . . (t-1)(3t+1) \:=\:0 \quad\Rightarrow\quad t \:=\:1,\:-\tfrac{1}{3}


    \text{The points of intersection are: }\;(1,2,5),\;{\color{red}\rlap{//////////}}\left(\text{-}\tfrac{1}{3},\:\tfrac{2}{3},\:\text{-}\tfrac{5}{3}\right)


    The second point is extraneous. .
    (I don't know why it appeared.)

    z \,=\,x^2+y^2 is a paraboloid completely above the xy-plane.

    There can be no intersection below the xy-plane.

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