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Math Help - Minimal dimension where plane and line can be non-paralel without having intersection

  1. #1
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    Minimal dimension where plane and line can be non-paralel without having intersection

    In which dimension, is it possible to construct plane and line, which are not parallel, but don't have any common point either (analogy of skew lines for line and plan)?
    I suspect it is 4D but i can't prove it. The only approach I could come with, was proving it is not possible in 3D and finding example for 4D.
    However veryfing example requires, i believe, finding normal vector of the plane (and direction vector of the line, i figured this one out) which i was not able to do in 4D.
    In book of exercises it is not considered difficult one, so i suspect it has easier solution too.
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    Re: Minimal dimension where plane and line can be non-paralel without having intersec

    with a single point (the 0-vector), we have no lines. thus a trivial vector space won't work.

    in one dimension, we have no planes, so a vector space of dimension 1 won't work.

    in two dimensions, we have just one plane (the entire space), so every line in the space is a subset of the (only) plane, and so intersects it at every point of the line.

    in three dimensions, we now have many planes and lines to choose from. so here, the challenge is to show that if a line is not parallel to a plane, it must intersect the plane. this is a bit tricky, as you have to think, what does it MEAN for a line to be "parallel to a plane"? it might be useful to define a parallel line as one lying in a parallel plane to our given plane. can you think of a way to create a non-parallel plane out of your given line, created from lines parallel to the given line, such that EVERY ONE of the parallel lines intersects the given plane?

    in four dimensions (say x,y,z,w) consider the xy-plane and the w-axis. these intersect at the origin (0,0,0,0). clearly the w-axis is perpendicular to the xy-plane, so it is NOT parallel. now shift the xy-plane 1 unit up along the z-axis. prove the plane:

    P = {(x,y,1,0): x,y in R} and the line L = {(0,0,0,w): w in R} have no intersection

    (if you like you can write P = s(1,0,0,0) + t(0,1,0,0) + (0,0,1,0), and L = u(0,0,0,1) + (0,0,0,0)).
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  3. #3
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    Re: Minimal dimension where plane and line can be non-paralel without having intersec

    Thank you, for the clue part with four dimensions it sure is done.

    For 3D, can't I say this: In 3D plane is given by one cartesien equation of type P:{(x,y,z);a1*x+b1*y+c1*z=d1.}
    line is given by two cartesien equations (or by intersection of two cartesien equations) L:{(x,y,z);a2*x+b2*y+c2*z=d2 , a3*x+b3*y+c3*z=d3}

    Point of intersection of line L and plane P is a soultion of system of three equations with three variables.This in turn means that P and L either
    a)have one solution,(they intersect in singular point ) -that's in case matrix A formed by P's and L's left sides of equations is regular,
    either
    b)matrix A is singular, than there is either no solution-that's in case if rank(A|d)>rank(A) or infinity of solution that's in case rank(A|d)=rank(A)
    infinity solution case have necessarily dimension 1, because ker(A)+rank(A)=3 and rank(A)=2.Therefore it is a line. However, if rank(A|d)>rank(A), it is always possible to choose such vector of right sides d, that rank(A|d)=rank(A).This choice of right sides doesn't affect normal vector of the plane, nor direction vector of line, because these depends only on left side's coefficients.Hence, it translates only as moving of line, or plane conserving their relative position.
    =>Therefor, every line which has no intersection with plane can be "moved" without changing its direction vector, so it would be entirely on this plane (sorry, not native speaker).This means that every line which has no intersection with plane has constant distance from the plane, so it is parallel. (So it is not possible to have both non-parallelity and not having intersection.)

    ?
    I feel quite sure with individual steps but in in 4D the approach gave wrong results, so i'm not sure if it is because of method or because of some mistake i made in 4D and maybe not in 3D.
    Last edited by athelred; February 25th 2013 at 12:03 AM.
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