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Show that two lines(in 3D) are parallel, skew or intercept

  1. Jan 21, 2010 #1
    http://img204.imageshack.us/img204/7130/screenshotxe.png [Broken]
    Hi,
    I was given this online assignment and this was one question I could not get.
    I know that two parallel lines parallel if their normal vectors' are scalar multiples of each other. And that if solving for t and s with two equations gives a t and s not satisfying the third equation, they do not intercept and are thus skew.
    I've spent about 4 hours on this now without any luck. From top to bottom, my answers are B, B, A, A, B, B. I know that there must be an intercept somewhere in there but I can't seem to find it even after repeating the questions over and over.
    If anyone could help me out that would be great,
    Thanks
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Jan 21, 2010 #2

    Matterwave

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    Well, you are only given 2 possible points of intersection. The easiest way to find which ones intersect at those point is to plug those numbers into the equations to see if the both lines go through either point.
     
  4. Jan 21, 2010 #3
    Well according to my calculator the first one intersects at (7,5,-3). But I couldn't get this on paper! When I solve for t and s, they end up as t=-3/2 and s=-4/3. It works fine for solving the x and y equations, but once I plug those into the z equations, it doesn't satisfy.
     
  5. Jan 21, 2010 #4

    HallsofIvy

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    The first two lines are given by
    [tex]\frac{x- 11}{4}= y- 6= \frac{z+5}{-2}[/tex]
    and
    [tex]\frac{x- 13}{6}= \frac{y- 2}{-3}= \frac{z- 5}{8}[/tex]

    One way to do this:
    [tex]\frac{x- 11}{4}= y- 6= \frac{z+5}{-2}= t[/tex]
    so x- 11= 4t, y- 6= t, z+ 5= -2t
    x= 4t+ 11, y= t+ 6, z= -2t- 5

    [tex]\frac{x- 13}{6}= \frac{y- 2}{-3}= \frac{z- 5}{8}= s[/tex]
    so x- 13= 6s, y- 2= -3s z- 5= 8s
    x= 6s+ 13, y= -3s+ 2, z= 8s+ 5

    The lines are not parallel sin <4, 1, -2> is not a multiple of <6, -3, 8>

    At a point of intersection, if any
    x= 4t+ 11=6s+ 13 and y= t+ 6= -3s+ 2. Then t= -3s+ 2- 6= -3s- 4. Putting that into the x-equation, 4(-3s-4)+ 11= -12s- 16+ 11= -12s- 5= 6s+ 13. 18s= -18 and s= -1 and t= 3- 4= -1.

    x= 6s+13= -6+13= 7, y= -3s+2= 3+ 2= 5.

    Now, check that in the z-equations. z= -2t- 5= 2- 5= -3 and z= 8s+ 5= -8+ 5= -3. Since those are the same the lines intersect at (7, 5, -3).

    Another way:
    [tex]\frac{x- 11}{4}= \frac{x- 13}{6}[/tex]
    so 6(x- 11)= 4(x- 13) or 6x- 66= 4x- 52. 2x= 14, x= 7.

    [tex]y- 6= \frac{y- 2}{-3}[/tex]
    so -3(y- 6)= -3y+ 18= y- 2 or 4y= 20, y= 5.

    [tex]\frac{z+ 5}{-2}= \frac{z- 5}{8}[/tex]
    so 8(z+ 5)= -2(z- 5) or 10z= -30, z= -3.

    Of course, we still have to check those answers:
    [tex]\frac{7- 11}{4}= 5- 6= \frac{-3+5}{-2}[/tex]
    [tex]\frac{-4}{4}= -1= \frac{2}{-2}[/tex]\
    -1= -1 = -1 so that equation checks.

    [tex]\frac{7- 13}{6}= \frac{5- 2}{-3}= \frac{-3- 5}{8}[/tex]
    [tex]\frac{-6}{6}= \frac{-3}{3}= \frac{-8}{8}
    -1= -1 = -1 so that equation checks.

    Again, the two lines intersect at (7, 5, -3).

    Now, you try the others.
     
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