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Checking if line is orthogonal to two skew lines

  1. Sep 20, 2016 #1
    1. The problem statement, all variables and given/known data
    So, I'm doing a long exercise, you can check here the first part: https://www.physicsforums.com/threads/checking-if-the-following-lines-are-coplanar.885948/
    The second part asks me to find, if one of the couple of lines are skew, the orthogonal line to two skew lines.

    2. Relevant equations


    3. The attempt at a solution
    Here is how I did it:
    I first found the directional vector of the orthogonal line, which is ##\vec v_o = \vec v_s \times \vec v_t##.
    ##\begin{vmatrix}
    \hat i & \hat j & \hat k \\
    3 & 0 & -3 \\
    -12 & -4 & 8
    \end{vmatrix} = (-12, 12, -12)##

    After this, I'll need the parametric form of one of the two lines. In this case I'll take ##s##.
    ##\begin{cases}
    x = 4 - \tau \\
    y = 2 \\
    z = \tau
    \end{cases}##

    Now we will need a line bundle and use the equations of the parametric form of ##s## and the vector ##\vec v_o##.
    ##\begin{cases}
    x' = 4 - \tau - 12m\\
    y' = 2 + 12m\\
    z' = \tau - 12m
    \end{cases}##

    At this point, we will substitute the unknowns to the Cartesian form of ##t## .
    ##\begin{cases}
    4(4 - \tau - 12m) - 2(2 + 12m) + 5(\tau - 12m) - 3 = 0 \\
    2(2 + 12m) + \tau - 12m + 1 = 0
    \end{cases}##
    and I end up with two unique solutions for ##\tau## and ##m##.
    ##\begin{cases}
    \tau = \frac{22}{7} \\
    m = \frac{1}{42}
    \end{cases}##

    Now I have to substitute these in the line bundle equations so I'll have the point where the intersection between the line bundle, which is orthogonal to the line ##s## because we used the parametric form of ##s## in order to make it, and the line ##t## happens.

    ##\begin{cases}
    x' = 4 - \frac{22}{7} - \frac{12}{42} = \frac{4}{7} \\
    y' = 2 + \frac{12}{42} = \frac{16}{7} \\
    z' = \frac{22}{7} - \frac{12}{42} = \frac{20}{7}
    \end{cases}##

    Now we know that the line we are searching for has to pass in that point and that the directional vector is ##\vec v_o##, so:
    ##\begin{cases}
    x' = \frac{4}{7} - 12\tau\\
    y' =\frac{16}{7} + 12\tau\\
    z' = \frac{20}{7} - 12\tau
    \end{cases}##

    Is this correct? Because every time I see fractions in this kind of exercises makes me think I did something wrong.
     
  2. jcsd
  3. Sep 20, 2016 #2

    Mark44

    Staff: Mentor

    Is the line you found perpendicular to the two given skew lines? This is simple to check, and is something that you should do.

    Also, just because you end up with fractions doesn't mean that your answer is wrong.

    One other thing: the vector you found, <-12, 12, -12> has the same direction as <-1, 1, -1> and the opposite direction as <1, -1, 1>. You can use any of the three as the direction for your orthogonal line. Note that I didn't check your work in getting this vector.
     
  4. Sep 20, 2016 #3
    Yes, because the scalar product of the directional vector of the line I found and the other two skew lines is ##0##. And yeah, you're right, I should do it.

    Yeah, I know. It's that usually my professor makes the exercises so that the results look... simple?

    Anyway, thank you for the help!
     
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