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Find 2 lines in R3 which satisfy some conditions, and find the point on the lines whe

  1. Jul 9, 2012 #1
    1. The problem statement, all variables and given/known data

    Find two lines in R3 in parametric form which satisfy the following conditions. Also, find the points on the lines which achieve the closest distance.

    Conditions:
    1. They are not parallel to any of the coordinate planes
    2. They do not intersect and are not parallel
    3. They are a distance of 2 apart at the points which have parameter values t = 0 and u = 0
    4. The lines are NOT closest when the parameter values are t = 0 and u = 0.


    2. Relevant equations


    3. The attempt at a solution

    I know how to find the shortest distance between two skew lines, but how do I find where that distance occurs in terms of points on the lines if all I have are parameters? I'm thinking that I need to use projections from the lines onto a plane then find the intersection of the projections, but when I try to do this I either get an unsolvable equation or infinite number of answers, since I'm dealing with parameters rather than exact coordinates.

    This problem would be easier if the closest distance was 2 and occurred when t = 0 and u = 0, since I could start with two lines distance 2 apart parallel to the xz plane, then rotate them around the z axis 45 degrees using a rotation matrix. If the closest distance was 2 and occurred when t = 0 and u = 0, my lines could be (-t*sqrt(2)/2, -t*sqrt(2)/2, t) and (u*sqrt(2)/2 - sqrt(2), u*sqrt(2)/2 + sqrt(2), u) but unfortunately that is not the case here.

    In terms of the distances, for 2 lines
    L1 = (a,b,c) + t[v1,v2,v3]
    L2 = (d,e,f) + u[w1,w2,w3]

    distance between them must be 2:

    2 = sqrt((a-d)^2 + (b-e)2 + (c-f)^2)
    so 4 = (a-d)^2 + (b-e)2 + (c-f)^2

    2 sets of lines I tried:

    (0,0,2) + t[3,1,-1]
    (0,0,0) + u[1,2,3]

    for these the shortest distance seems to be 2/sqrt(6), which is indeed less than 2, but where does this distance occur? and

    (0,0,3) + t[1,2,-1]
    (0,0,1) + u[1,-1,1]

    while for these its 6/sqrt(14), but I still don't see how I can find where this distance occurs.
     
  2. jcsd
  3. Jul 9, 2012 #2

    LCKurtz

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    Re: Find 2 lines in R3 which satisfy some conditions, and find the point on the lines



    Calculate the distance squared (to avoid square roots) between those points as a function of t and u. Then use calculus to minimize d2. That should give you the values of t and u and you can use the points they give to check your ##\frac 6 {\sqrt{14}}##. ( I didn't check it so I'm not saying it is wrong, just that it gives a nice way to check it.)
     
  4. Jul 11, 2012 #3
    Re: Find 2 lines in R3 which satisfy some conditions, and find the point on the lines

    im sorry, but I don't understand. What do you mean "use calculus to minimize d2?
    you mean take derivative of d and set it to 0? Then in this case ut/√6 would just be 0, and that doesn't make sense. When I'm finding the shortest distance between two lines I'm taking the cross product of their direction vectors. If I try including u and t in that equation for say these lines:

    (0,0,2) + t[3,1,-3]
    (0,0,0) + u[1,2,3]

    you get 5uti - 10utj + 5utk

    turn this into a unit vector and scale it down:

    1/(ut√6) * (1*ut, -2*ut, 1*ut)

    then multiplying by position vector

    1/(ut√6) * (1*ut, -2*ut, 1*ut) dot (0,0,2) = d1
    1/(ut√6) * (1*ut, -2*ut, 1*ut) dot (0,0,0) = d2 (obviously this is 0)
    d1 - d2 = d
    d = 2/√6

    ...

    :confused:

    if you mean minimizing the original equation for distance, wouldn't that require solving differential equations? I think this problem is just supposed to use basic linear algebra
     
    Last edited: Jul 11, 2012
  5. Jul 12, 2012 #4

    LCKurtz

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    Re: Find 2 lines in R3 which satisfy some conditions, and find the point on the lines

    The distance squared between those two lines is, calling it ##D##:$$
    D = (3t-u)^2+(t-2u)^2+(2-3t-3u)^2$$
    You can find the critical values for ##t## and ##u## by setting$$
    \frac{\partial D}{\partial u}=0,\ \frac {\partial D}{\partial t}=0$$
    You will get two simple equations in ##t## and ##u## which you can solve simultaneously.

    [Edit] Added the next morning, it was getting late.

    If you want a purely non-calculus method you could do this. Cross [3,1,-3] and [1,2,3] to get a vector N in the mutually perpendicular direction. You don't have to normalize it. Call R1(t) = (0,0,2) + t[3,1,-3] and R2(u) = (0,0,0) + u[1,2,3]. Now you need to find t and u making R1(t) - R2(u) parallel to N. So set ##R_1(t)-R_2(u) = \lambda N##. Write that out and you will have three equations in the unknowns ##t,\ u,\ \lambda##. Solve for ##t## and ##u## that way.
     
    Last edited: Jul 12, 2012
  6. Jul 12, 2012 #5
    Re: Find 2 lines in R3 which satisfy some conditions, and find the point on the lines

    hm yeah I just tried the calculus way, it works great. I got t = .2666666 and u = .46666, then checked by plugging them back in and I get the 2/sqrt(6) so yeah it works.

    thanks a lot!

    I'll try the non calculus way right now.

    k just tried vector way and got same answer, thx.

    oh and btw I accidently wrote the vector wrong on my second post, for the 1st line its [3,1,-1] not [3,1,-3]

    Hm yeah on my sheet I write the exact values, in my brain though sometimes I'm checking things I don't like fractions, fortunately my ti 89 calc gives me both exact vals and decimals for everything
     
    Last edited: Jul 12, 2012
  7. Jul 12, 2012 #6

    LCKurtz

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    Re: Find 2 lines in R3 which satisfy some conditions, and find the point on the lines

    I'm curious why you use decimal approximations instead of the simple exact values of 4/15 and 7/15.
     
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