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Finding the distance

  1. Mar 1, 2014 #1
    1. The problem statement, all variables and given/known

    Find the distance from the line

    L1: t
    2t
    3t-1

    And
    L2: -t+1
    t+2
    t





    2. Relevant equations

    Cross product








    3. The attempt at a solution

    well I ASSUME that my L1 and L2 were given in parametric equation form Thus L1 would be : x=t y=2t z=3t-1
    L2 : x=-t+1 y=t+2 z=t

    If I am correct in assuming that,

    I then found my vector for L1 u= <1,2,3>
    L2 v=<-1,1,1>

    Used cross product : uXv
    Which I then got n= <-1,4,3>
    if I am correct so far... I am stuck on my next step ImageUploadedByPhysics Forums1393661649.302359.jpg
     
  2. jcsd
  3. Mar 1, 2014 #2

    mfb

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    Staff: Mentor

    You don't have to assume that, they are given in parametric form.

    What is your next step, and where did you get stuck there?
     
  4. Mar 1, 2014 #3
    I actually don't know what my next step would be?
    Maybe find an equation of a plane for L1 and L2 and then find the distance between those 2 planes?
     
  5. Mar 1, 2014 #4

    ehild

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    Find the distance of what???


    ehild
     
  6. Mar 1, 2014 #5
    Of line_1 and line _2. I posted a photo of the question, that was all the info given to me
     
  7. Mar 1, 2014 #6

    Mark44

    Staff: Mentor

    The problem statement as posted seems confusing to me. It's possible the problem is asking for the distance between the two lines. "Find the distance from L1 and L2 ..." implies that there is a "to" part that is missing from the description. It's similar to asking "Find the distance from San Francisco." This is impossible to answer because it doesn't give you the other location.

    If all that is needed is the distance between the two lines, the problem could have been stated in a clearer manner. My guess is that the problem writer is not a native speaker of English.
     
  8. Mar 1, 2014 #7

    Yes my professor makes up his own test, so I can't find anything similar to it from the examples on our book. Which is why I get confused.
     
  9. Mar 1, 2014 #8

    Mark44

    Staff: Mentor

    It's probably safe to assume that the problem is asking for the distance between the two lines (or the distance from one line to the other). By "distance" I mean the shortest distance between any point on the first line and any point on the second line.

    What you have about the cross product seems irrelevant to me. This is what I would do.
    1. Determine whether the lines intersect. If they do, the distance is zero.
    2. If they don't intersect, find an arbitrary point on the first line, and an arbitrary point on the second line.
    3. Calculate the length of the line segment between these two points. This will be the hypotenuse of a right triangle.
    4. Find the angle that this line segment makes with one of the two lines. From the hypotenuse and the angle you found, you can use trig to find the length of the side opposite to this angle. That will be the distance between the two lines.
     
  10. Mar 1, 2014 #9

    mfb

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    I don't see how this would work. In general both ends of the arbitrary line segment will not be at the point of closest approach. You would have to repeat this triangle step with the other line.

    The method with the cross-product is fine and it is the quickest method, you just have to know (or figure out, or look up) the steps to take.
     
  11. Mar 1, 2014 #10

    Mark44

    Staff: Mentor

    You're right. I didn't think this all the way through. In my diagram, one of the points on one line was the closest to the the other line, and this is an incorrect assumption.
     
  12. Mar 1, 2014 #11

    ehild

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    The problem asks the distance between the two lines.
    First you have to show that the lines do not intersect: they are skew lines. The distance between skew lines is the length of a the connecting line which is perpendicular to both lines.
    You have found the vector that is normal to both lines: The projection of any connecting vector onto the direction of the normal vector is equal to the distance between the lines.

    http://en.wikipedia.org/wiki/Skew_lines

    ehild
     
  13. Mar 1, 2014 #12

    So ImageUploadedByPhysics Forums1393731302.344252.jpg

    So my last step I used the distance
    . D = projection of PQ onto n would this be correct?
     
  14. Mar 1, 2014 #13

    ehild

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    Yes, but the distance can not be negative. Take the absolute value.

    ehild
     
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