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Inquiry: Find two lines in R^3 that are not parallel but do not intersect

  1. Feb 16, 2008 #1
    Question: Find two lines in R[tex]^{3}[/tex] that are not parallel but do not intersect.

    My Thoughts:

    I have never seen this type of question before and the material in my text is unfortunately lacking. However, I was able to piece these thoughts together.

    The cross product of three vectors should not equal zero. (Otherwise they are orthogonal, and hence intersect).

    None of the vectors should be multiples of the other vectors.

    So my question is, I can draw to lines in R^3 space that don't intersect. I suppose I
    could find two lines by taking z2 -z1 where z = f(x,y) but this feels sloppy. Can someone
    help me construct something more formally with logic and method?

    Thank You,
    HFO8
     
  2. jcsd
  3. Feb 16, 2008 #2

    Hurkyl

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    Huh? :confused: What vectors?

    None of the vectors should be multiples of the other vectors.


    So do it, and tell me what they are.
     
  4. Feb 16, 2008 #3
    To Mr. Hurkly

    For line 1: let P1(0,0,4) and P2(4,0,0)
    For line 2: let B1(0,0,3) and B2(0,3,0)

    These two lines appear not to intersect and are not parallel.

    Even if my above example is wrong. Consider a cube. The front horizontals and the vertical lines on the back of the cube are not intersecting, not parallel, and hence
    examples of lines that will answer my question.

    So basically, my question, sir, is asking how to formally "do it" and show what my
    intuition and work has shown me thus far. I apologize sir, if my vector teminology
    has confused you. Please disregard and consider only the question and my progress
    thus far.

    Thank You,
    HF08

    PS - Anyone is welcome to post here. I thank those who have viewed my post. Especially Mr. Hurkyl
     
  5. Feb 16, 2008 #4

    Hurkyl

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    What is P1, P2, B1, B2? What are those numbers?

    I'm guessing that P1 and P2 are supposed to be the names of points on the first line (and the following numbers are the coordinates of those points), and similarly for B1 and B2. You really should say such things -- you shouldn't introduce a new symbol without stating what it's supposed to be. For example, you could say something like:

    Let P1=(0,0,4) and P2=(4,0,0) be points in R^3, and Let line 1 be the line through them​

    Anyways, if you prove:
    (1) line 1 and line 2 do not intersect,
    (2) line 1 and line 2 are not parallel.

    then you've satisfactorally solved your problem, haven't you?
     
  6. Feb 16, 2008 #5
    Your thinking looks to suffice for this case. As I view your problem, it is actually difficult to find 2 lines that intersect in r^3 even. If such a pair intersects at one point though (meaning they are coplanar), it is easy to make that pair not intersect by shifting (translating) one line out of said plane.
     
  7. Feb 16, 2008 #6

    Okay, your point is taken sir, and I accept your criticism. However, I do feel
    your being a little cute in saying all I have to show is that they do not
    intersect and are not parallel. That is merely repeating what the problem
    implies. I may be learning, but that doesn't mean I am stupid.

    I am given rather modest weapons in my text. I dare say I can use the
    cross product to show that two vectors (points in R^3) are not orthogonal. That is,
    if I can show perhaps P1 and B2 (points in R^3) are not orthogonal. Done. So, how
    to show they do not intersect? Perhaps I can determine they are not parallel by
    using the identity cos T = < x,y> /(||x|| ||y||).

    That is, cosine is a trigonometric function with angle T in [0,pi] and x and y can be
    any vector (point) in R^3 and <x,y> is inner product. And ||x|| is the norm.

    So, I am back to the crappy notion of how to show non-intersection?


    HF08
     
  8. Feb 16, 2008 #7
    Why are you so interested in orthogonality? I thought you were just trying to show that two lines (and their corresponding vector directions) are not parallel, which is easily accomplished by observing if one line's directional vector is a multiple of the other's.
     
  9. Feb 16, 2008 #8
    I shall refer to another text than the one I am currently using. I feel as your comment as suggested, there is a more simple method than what I am doing, which is being needlessly complicated and stupid.

    HF08

    PS - It is because the question in the section of my text is in love with cross products
    and orthogonality. Don't shoot the messenger. :-)
     
  10. Feb 17, 2008 #9

    Hurkyl

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    No offense is intended; even the brightest people make "simple mistakes". And many people have a mental block on this point -- they are so used to the idea that they have to "solve" something to get the answer that they have difficulty accepting that, to answer a question like this, they merely have to write down an answer (and verify that it really is an answer).

    Thus, I wanted to make sure we were okay with that before moving on.



    Are any of your weapons a test for two lines being parallel? What about the definition of parallel? Maybe it is possible to apply that directly, rather than using a fancier tool.


    I don't understand the aim; first off, points cannot be orthogonal. I assume you mean the two vectors whose tails are the origin and whose heads are at P1 and B2? If these vectors are orthogonal, what does that tell you about the lines?


    Well, if they intersect, then there has to exist a point Q having both of these properties:
    (1) Q lies on line 1
    (2) Q lies on line 2

    So, you merely want to show that there is no point Q that has both of these properties. Can you translate these geometric conditions into algebra? If so, then you can apply all you know about algebra!
     
  11. Feb 17, 2008 #10
    Yeah to be honest I am getting pretty confused the more you try and explain your method! I mean yes, if you have the 2 directional vectors of the lines, let's call them [tex]\vec{l_{1}}[/tex] and [tex]\vec{l_{2}}[/tex], then [tex]\vec{l_{1}}\times \vec{l_{2}}[/tex] cannot be 0 because that implies they are parallel, but the difference between your approach and one that will answer the question is that you are currently trying to find an algorithm that "checks" the conditions you want given two lines, but you haven't found a way to actually FIND one, which is what you really want by what you're asking.
     
    Last edited: Feb 17, 2008
  12. Feb 17, 2008 #11
    Gentlemen,

    I apologise if I seemed too sharp. Thank you for your posts. I agree with your points. First, I need to develope better communication skills. I would appreciate suggestions on being more clear.

    Second, I feel we are making progress. I do have a result about lines (actually it is worded as vectors). I provide this result as follows:

    Assume x,y are two nonzero vectors and the Cauchy-Schwartz Inequality, then x and y are parallel. <x,y>[tex]\leq[/tex](||x|| ||y|| )

    I am assuming we are all familiar with norm and inner product definitions. Now, I am somewhat torn, as Mr.Hurkyl has detected. Do I expend further efforts on the theorem above or do I go into the realm of algebra?

    I am still working on the problem and I will post again. I am going to retire for a few hours and post in the afternoon.

    Thank You,
    HF08
    are
     
  13. Feb 17, 2008 #12
    Let me ask this question in a new way

    I have obtained the following two lines. The brackets with numbers are vectors.
    Unfortunately, I was not able to put them in vertical notation. That is, pretend that the they are given in the correct notation.

    l[tex]_{1}[/tex]=[4,0,4]+t[-4,0,4]
    l[tex]_{2}[/tex]=[0,3,0]+t[0,-3,3]

    I have been asking alot of questions. I was trying to make this problem really hard.
    So, I will try to make it more algebraic and more simple. Let me ask this in a new way.
    How can I show that these two lines are not parallel and do not intersect?

    Thank You,
    HF08

    PS - I agree with Mr. Hurkyl. It was pathetic how long it took me to switch gears and
    write my line in this more agreeable fashion.
     
  14. Feb 17, 2008 #13

    Hurkyl

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    If they intersected at some point Q, then there would be a and b for which [itex]\ell_1(a) = Q[/itex] and [itex]\ell_2(b) = Q[/itex]....
     
  15. Feb 17, 2008 #14
    In other words, could I claim that there is some t such that l[tex]_{1}[/tex]=l[tex]_{2}[/tex]? This seems more direct. Sigh...I keep editing this. The latex should be saying l1=l2, but it's not working out right. I can't explain myself right if I can get this latex quick paste not to work right.

    I am going to get this. You know, I just got done typing up a proof for a problem that I consider much more involved. I don't understand why this is kicking my butt.

    HF08
     
    Last edited: Feb 17, 2008
  16. Feb 17, 2008 #15

    Hurkyl

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    If Q lies on both lines, there's no reason to think that [itex]\ell_1(t) = Q[/itex] and [itex]\ell_2(t) = Q[/itex] both have the same solution for t....

    It works better if you put entire formulas inside the [ itex ] ... [ /itex ] tags... if you just put individual symbols inside the tags, the formatting isn't cohesive. (And itex works better than tex, when mixed with normal text)
     
  17. Feb 17, 2008 #16
    To Mr. Hurkyl,

    I agree sir. It was more of a whim and wishful thinking. I shall consider my Calculus 3 text for this problem. Perhaps it has more tools for consideration than is provided in my current text. I will keep your comment in mind. Perhaps I should consult a linear algebra text. Perhaps I could instead write this as a system of linear equations?

    Regards,
    HF08
     
  18. Feb 17, 2008 #17

    Hurkyl

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    Probably. Doing geometry with lines is a fairly common source of linear equations.

    In fact, [itex]\ell_1(a) = \ell_2(b)[/itex] is a linear equation....
     
    Last edited: Feb 17, 2008
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