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Thread: 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 $\displaystyle x = 1+t, y = 2+3t, z = 1-t$
    and the surface $\displaystyle z = x^2 +2y^2$
    Attempt:

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

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

    $\displaystyle 1-t = (1+t)^2 +2(2+3t)^2$

    Solving i get:

    $\displaystyle 19t^2+27t+8=0$


    At this point i need to solve for t

    It appears to be a quadratic equation.

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

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

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

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

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

    $\displaystyle t= \frac{{ - 8 }}{{19}} $ and $\displaystyle 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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