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Math Help - Find the point of intersection of the line and surface

  1. #1
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    Find the point of intersection of the line and surface

    I have an odd problem with no solution. I am completely lost on how to solve this.

    Problem:

    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
    Attempt:

    (1) x = 1+t
    (2) y = 2+3t
    (3) z = 1-t

    Subbing in (1), (2), (3) into the surface i have

    1-t = (1+t)^2 +2(2+3t)^2

    Solving i get:

    19t^2+27t+8=0


    At this point i need to solve for t

    It appears to be a quadratic equation.

    \begin{array}{*{20}c}    {t = \frac{{ - b \pm \sqrt {b^2  - 4ac} }}{{2a}}}  \end{array}

    \begin{array}{*{20}c}    {t = \frac{{ - 27 \pm \sqrt {(27)^2  - 4(19)(8)} }}{{2(19)}}}  \end{array}

    \begin{array}{*{20}c}    {t = \frac{{ - 27 \pm \sqrt {121} }}{{38}}}  \end{array}

     \frac{{ - 27 \pm 11 }}{{38}}

     \frac{{ - 27 + 11 }}{{38}} and  \frac{{ - 27 - 11 }}{{38}}

     t= \frac{{ - 8  }}{{19}} and t = -1

    Then just plug these t values into the parametric equations to get the points?
    Did i do this right?
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  2. #2
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    Re: Find the point of intersection of the line and surface

    right
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