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Checking if the following lines are coplanar

  1. Sep 20, 2016 #1
    1. The problem statement, all variables and given/known data
    Set an orthonormal reference system in the euclidean space ##E^3## and consider the following lines:
    $$r: \begin{cases}
    x = 3 \tau \\
    y = 1 + \tau \\
    z = 3 - 2 \tau
    \end{cases}s: \begin{cases}
    x + z - 4 = 0 \\
    x - 3y + z + 2 = 0
    \end{cases}t: \begin{cases}
    4x - 2y + 5z - 3 = 0 \\
    2y + z + 1 = 0 .
    \end{cases}$$
    Find the mutual position of each possible couple of lines.

    2. Relevant equations


    3. The attempt at a solution
    I've calculated the mutual positions of the three couple of lines(##r## and ##s##, ##s## and ##t##, and ##r## and ##t##) but I'm not sure about it so here it's how I proceeded:

    Before starting, I transformed $r$ from the parametric form into the Cartesian form, so:

    $$r: \begin{cases}
    x - 3y + 3 = 0 \\
    2y + z - 5 = 0 .
    \end{cases}$$

    Now, to know if the lines are parallel, orthogonal, etc, I first find out about the directional vector of each line by finding the determinant of the matrices made of the components of the lines:
    ##\begin{vmatrix}
    \hat i & \hat j & \hat k \\
    1 & -3 & 0 \\
    0 & -2 & 1
    \end{vmatrix} = (-3, -1, -2)

    \begin{vmatrix}
    \hat i & \hat j & \hat k \\
    1 & 0 & 1 \\
    1 & -3 & 1
    \end{vmatrix} = (3, 0, -3)

    \begin{vmatrix}
    \hat i & \hat j & \hat k \\
    4 & -2 & 5 \\
    0 & 2 & 1
    \end{vmatrix} = (-12, -4, 8)##

    Then I find a point for each line:
    ##P_r (0, 1, 3), P_s (2, 2, 2), P_t (-1, -1, 1)##

    After that I do the determinant with the first row as the difference between the two points of the two lines, and on the second and third raw I put the directional vectors of the two lines.
    For ##r## and ##s##:
    ##\begin{vmatrix}
    0-2 & 1-2 & 3-2 \\
    -3 & -1 & -2 \\
    3 & 0 & -3
    \end{vmatrix} = 12##
    Non-coplanar.

    ##\begin{vmatrix}
    2-(-1) & 2-(-1) & 2-1 \\
    3 & 0 & -3 \\
    -12 & -4 & 8 \\
    \end{vmatrix} = -10##
    Non-coplanar.

    ##\begin{vmatrix}
    0-(-1) & 1-(-1) & 3-1 \\
    -3 & -1 & -2 \\
    -12 & -4 & 8 \\
    \end{vmatrix} = 80##
    Non-coplanar.

    The thing is that all three of them are askew (non-coplanar). Could someone check it out and tell me if they are really all non-coplanar or maybe I did it wrong?
     
  2. jcsd
  3. Sep 20, 2016 #2

    Ssnow

    User Avatar
    Gold Member

    Hi, finding the directional vector of the first line ##r##, in the determinant the third line is ## 0 \ \ 2 \ \ 1## and not ##0 \ \ -2 \ \ 1##, this can change something later ... the procedure seems reasonable ...

    Ssnow
     
  4. Sep 20, 2016 #3
    Wow... that sign changed everything...
    ##r## and ##s## are intersecting lines.
    ##s## and ##t## are still non-coplanar. (because the directional vector wasn't needed here)
    ##r## and ##t## are parallel.
    Thank you very much for noticing that mistake.
     
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