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Math Help - Find symmetric equation

  1. #1
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    Find symmetric equation

    Find the symmetric equation of the line passing through the point (-5,-3,2) and perpendicular to both lines:

    (x-1)/3 = (4-y)/1 = z/3 and x/-2 = (y-1)/4, z = 2

    For the first line would the direction vector be (3,1,3) and for the second line would it be (-2,4,1) ?
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  2. #2
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    Quote Originally Posted by Random-Hero- View Post
    Find the symmetric equation of the line passing through the point (-5,-3,2) and perpendicular to both lines:

    (x-1)/3 = (4-y)/1 = z/3 and x/-2 = (y-1)/4, z = 2

    For the first line would the direction vector be (3,1,3) and for the second line would it be (-2,4,1) ?
    1. Re-write the equations of the lines:

    \begin{array}{l}x=1+3t \\ y=4-t \\ z=0+3t\end{array} ...... and ...... \begin{array}{l}x=0-2t \\ y=1+4t \\ z=2+0t\end{array}

    Therefore the direction vectors are \vec u=(3, -1, 3) and \vec v = (-2, 4, 0)

    2. The vector perpendicular to both direction is:

    \vec n = \vec u \times \vec v = (-12, -6, 10)

    3. Now use the point-direction-form of the equation of a line:

    \vec r = (-5, -3, 2) + s \cdot (-12, -6, 10)

    4. Re-write:

    \begin{array}{l}x=-5-12s\\y=-3-6s\\z=2+10s\end{array}~\implies~ \begin{array}{l}\dfrac{x+5}{-12}=s\\ \dfrac{y+3}{-6}=s\\ \dfrac{z-2}{10}=s\end{array}

    5. Symmetric equation:

    \dfrac{x+5}{-12}= \dfrac{y+3}{-6}= \dfrac{z-2}{10}
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  3. #3
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    Wow thanks earboth!!! I totally spaced on the cross product! You're the man! *high fives*
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