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Intuition behind the distance between two lines

  1. Aug 9, 2010 #1
    RE-EDITING POST: I had misread some of my notes. :redface:
     

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    Last edited: Aug 9, 2010
  2. jcsd
  3. Aug 10, 2010 #2

    HallsofIvy

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    What is your question?
     
  4. Aug 10, 2010 #3
    I've been working on trying to solve the following question. (See figure)

    (NOTE FOR FIGURE: Ignore salshed out part)

    The first thing I did is write the two lines in vector format with different parameters (s&t). After that, looking at the direction vectors of each line we can determine that one is not a multiple of the other, therefore they musn't be parallel. I double checked this with the cross product between the two direction vectors, and I did not obtain the zero vector.

    From here I established a system of equations, equating the x, y and z coordinates and solving for t and s.

    Since I obtained t = -12 and s = 7, the two lines should intersect at some point, and the distance should simply be 0, but I'm not entirely sure this is true.

    Am I interpreting the question wrong?

    Thanks again.

    EDIT: Is it possible that both the lines are defined by the same parameter, t? If so then they lines would be skew lines, correct?

    As soon as I can clarify the first few questions I have, then I may have a question regarding the intution behind solving for the distance between two lines.
     

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  5. Aug 11, 2010 #4
    I think I've made some progressive after doing some more reading/review.

    Since there are no values of t and s that satisfy the three equations, [tex]l_1[/tex] and [tex]l_2[/tex] do not intersect. Thus [tex]l_1[/tex] and [tex]l_2[/tex] are skew lines.

    Since the two lines are skew, we can think of them as two parallel planes [tex]P_1[/tex] and [tex]P_2[/tex]. The normal vector to both planes will be orthogonal to the direction vectors of [tex]l_1[/tex] and [tex]l_2[/tex]. ([tex] \vec{v_{1}}\times \vec{v_{2}}[/tex])

    [tex]\vec{n} = 8\hat{i} - 5\hat{j} -1\hat{k}[/tex]

    Now if we simply use a point on either [tex]l_1[/tex] or [tex]l_2[/tex] we should be able to describe either of the planes [tex]P_1[/tex] or [tex]P_2[/tex].

    If we choose [tex]P_1[/tex],

    [tex]8(x +1) -5(y -3) -(z+4) = 0[/tex] or [tex]8x -5y -z + 19 = 0[/tex]

    Now if we set t = 0 in the equations for [tex]l_2[/tex], we get the point P(1,0,-3) on [tex]P_2[/tex].

    That being said the distance between the two lines is now reduced to the distance between a plane and a given point.

    This will distance will be equal to the absolute value of the scalar projection of any vector that lies in the plane and travels to the point P(1,0,-3) projected onto the normal vector of that plane.

    [tex]D = \frac{|8(1)-5(0)-1(3)+19|}{|\sqrt{8^{2} + (-5)^{2} + (-1)^{2}}|} = \frac{24}{\sqrt{90}} = \frac{4\sqrt{10}}{5} \approx 2.530[/tex]

    Is this correct?
     
    Last edited: Aug 11, 2010
  6. Aug 12, 2010 #5
    Bump, still looking for a quick check on this one!
     
  7. Aug 12, 2010 #6

    vela

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    Your reasoning is correct, but you flipped a sign when calculating the dot product in your last step.
     
  8. Aug 12, 2010 #7
    Thank you for pointing that out. I'll edit this post later when I have the time to redo the calculations!

    EDIT:

    [tex]D = \frac{|8(1)-5(0)-1(-3)+19|}{|\sqrt{8^{2} + (-5)^{2} + (-1)^{2}}|} = \frac{30}{\sqrt{90}} = \sqrt{10} \approx 3.162[/tex]
     
    Last edited: Aug 12, 2010
  9. Aug 12, 2010 #8
    Hopefully that's correct now!
     
  10. Aug 12, 2010 #9

    vela

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    Yup, that's it.
     
  11. Aug 13, 2010 #10

    ehild

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    And it is equal to the minimum distance between points on the two lines.

    ehild
     
  12. Aug 13, 2010 #11
    I have one more quick question.

    When I preformed the dot product with the normal vector of the plane, it seems as though I didn't dot together two vectors, but rather a plane and a point.

    Can someone try and clarfiy this for me, it's the only part left that has me rattled!

    Thanks again!
     
  13. Aug 13, 2010 #12

    vela

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    You're choosing one point on P1. It's just that regardless of which point on the plane you choose, you always have [itex]\vec{n}\cdot\vec{x} = -19[/itex].
     
  14. Aug 13, 2010 #13

    ehild

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    Quite a lot of way to attack a problem... The distance between the skew lines is the length of the straight line which intersects both lines and is perpendicular to both of them. The tangent vector of this line is the vector product of

    v1 and v2.

    [tex]\vec n = \vec v_1\times \vec v_2[/tex]

    The length of the perpendicular line segment is equal to the projection of any vector which connects a point on l1 to a point on l2, for example r12 that points from P1 to P2. The distance is

    [tex]D=\frac{\vec n\cdot\vec r_{12}}{|\vec n|}[/tex]

    ehild
     
  15. Aug 13, 2010 #14
    Thank you, this cleared things up for me!
     
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