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Angle between two skew lines

  1. Oct 11, 2016 #1
    1. The problem statement, all variables and given/known data
    The problem asks me to evaluate the angle between these two lines:
    ##r : \begin{cases}
    x - 2y - 3 = 0 \\
    3y + z = 0
    \end{cases} s : \begin{cases}
    x = 1 + 4t \\
    y = 2 - 3t \\
    z = 3
    \end{cases}##
    both oriented to the decreasing ##y##.

    2. Relevant equations


    3. The attempt at a solution
    Having found ##\vec v_r = (-2, -1, 3)##, ##\vec v_s = (4, -3, 0)##, ##P_r (1, -1, 3)##, and ##P_s (1, 2, 3)##
    I already know that the lines are askew. I then found out that in order to find the angle between the two lines, I have to first find a plane containing one of the two lines(for example ##r##) that is at the same time parallel to the other one(in this example ##s##). In a few words I have to find a line parallel to ##s## that meets the line ##r## in a point that belongs to ##r##.
    The thing is that I don't know how to find that parallel line to ##s## that at the same time passes into a point ##P_r## belonging to the line ##r##.
    Should I take one of the Cartesian equations of ##r## and see the projection of ##s## on it so to have the parallel line? And then see the interjection between this parallel line and ##r##? Or I should proceed in another way?
    By the way, this is the Cartesian form of the ##s## line I found:
    ##s : \begin{cases}
    \frac{3}{4}x + y - \frac{11}{4} = 0 \\
    z - 3 = 0
    \end{cases}##
     
  2. jcsd
  3. Oct 11, 2016 #2

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    Do you know the dot product = scalar product? You don't have to construct new planes and whatever, once you have vr and vs the angle can be found in a single line on paper.
     
  4. Oct 11, 2016 #3
    Ohw... I was so concentrated on the parallel line I didn't thought of that...
    Thank you and sorry.
     
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