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Intersection of Lines in R4

  1. Sep 22, 2010 #1
    To find the intersection of two lines in R3, you set the lines equal, right?
    [a,b,c] + d[e,f,g] = [h,i,j] + k[l,m,n]
    Then split these into three equations,
    1. a + d(e) = h + k(l)
    2. b + d(f) = i + k(m)
    3. c + d(g) = j + k(n)

    And solve for k and d, correct?

    If k and d are consistent, then these values are used to find the point of intersection.

    My question is: Is this same method used in R4, with two lines of 4 dimensions?
  2. jcsd
  3. Sep 23, 2010 #2
    Probably you could get some help after explaining your notation....
  4. Sep 25, 2010 #3


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    Yes, it is exactly the same. You would have one line given by, say,
    [tex]\begin{pmatrix}x_1 \\ x_2 \\ x_3 \\ x_4\end{pmatrix}= \begin{pmatrix}a_1t+ b_1 \\ a_2t+ b_2 \\ a_3t+ b_3 \\ a_4t+ b_4\end{pmatrix}[/tex]
    and the other by
    [tex]\begin{pmatrix}x_1 \\ x_2 \\ x_3 \\ x_4\end{pmatrix}= \begin{pmatrix}c_1s+ d_1 \\ c_2s+ d_2 \\ c_3s+ d_3 \\ c_4s+ d_4\end{pmatrix}[/tex]
    then you would set them equal getting 4 equations in the two unknown values s and t:
    [itex]a_1t+ b_1= c_1s+ d_1[/itex]
    [itex]a_2t+ b_2= c_2s+ d_2[/itex]
    [itex]a_3t+ b_3= c_3s+ d_3[/itex]
    [itex]a_4t+ b_4= c_4s+ d_4[/itex]

    Of course, it would be unlikely for 2 numbers to satisfy all 4 equations!

    In the plane, the "typical" behavior is for lines to intersect- the only non-intersecting lines are parallel lines, a very unusual situation. And, in two dimensions, we are solving two equations in two unknowns- there would not be a unique solution only if the determinant of the coefficient matrix were 0.

    In three dimensions, the "typical" behavior is for lines to be skew- not intersecting. In order to be able to solve three equations for two unknown values, the equations must NOT be independent.

    And the situation is worse in four dimensions. Two lines intersecting is a very special situation indeed.
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