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Intersection of Lines (Vectors)

  1. Aug 26, 2013 #1

    FeDeX_LaTeX

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    1. The problem statement, all variables and given/known data
    Show that if the two lines

    ##\frac{x - c_1}{d_1} = \frac{y - c_2}{d_2} = \frac{z - c_3}{d_3}## and

    ##\frac{x - d_1}{c_1} = \frac{x - d_2}{c_2} = \frac{x - d_3}{c_3}##

    intersect, they lie in the plane ##r.(c \times d) = 0##

    where c = c1i + c2j + c3k and d = d1i + d2j + d3k

    3. The attempt at a solution

    I've shown that they intersect at ##x = c_1 + d_1, y = c_2 + d_2, z = c_3 + d_3##. It seems intuitively obvious that the lines lie in the plane r.(c x d) = 0, but I just can't seem to properly show this. Any advice?
     
  2. jcsd
  3. Aug 26, 2013 #2

    verty

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    It is possible for the lines to intersect at infinitely many points, for example let ##c_i = d_i = i##. In this case, ##r . (c × d) = 0## isn't a plane. So I think one must assume we are talking about lines that intersect at exactly one point, if at all.

    Given this, and assuming your proof is correct that they intersect at c+d if they intersect at all, showing that the points (c, d, c+d) are in that plane will suffice because each line goes through two of those points.
     
  4. Aug 26, 2013 #3

    FeDeX_LaTeX

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    Thanks -- I assume that the lines are non-parallel. All three points lie in the plane.
     
  5. Aug 26, 2013 #4

    haruspex

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    The question is a little odd. Those two lines necessarily intersect (at x = c+d).
    Given an equation like ##\frac{x - c_1}{d_1} = \frac{y - c_2}{d_2} = \frac{z - c_3}{d_3}##, you can set each term to equal some scalar parameter t, then express the line as a vector equation. It should then be obvious that the line lies in the subspace (be it a line or a plane) generated by c and d.
     
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