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Finding intersection of three planes

  1. Sep 25, 2014 #1

    I'm having trouble with this question, any help would be much appreciated! :)

    Q1: Given the three vectors:

    n1 = (1, 2, 3)
    n2 = (3, 2, 1)
    n3 = (1, −2, −5)

    Find the intersection of the three planes ni*x = 0. What happens if n3 = (1, −2, −4)? Why is this different?
  2. jcsd
  3. Sep 25, 2014 #2
    Are you familiar with vector spaces and matrix rank?
  4. Sep 25, 2014 #3


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    Your original n3=(1,-2,-5) = n2-2n1 so the three planes don't intersect. The pairwise intersections (of the planes) are 3 parallel lines.

    The other n3 would work since it is not a linear combination of the others.
  5. Sep 26, 2014 #4
    Actually they do intersect, just not in a single point as can be shown by the Rouché-Capelli theorem.
  6. Sep 26, 2014 #5


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    You don't really need to know linear algebra- just the basics of systems of equations.
    The planes defined by the first three vectors are
    x+ 2y+ 3z= 0
    3x+ 2y+ z= 0
    x- 2y- 5z= 0.

    Find the general solution to that system (there is NOT a unique solution because the determinant of coefficients is 0). What does that define, geometrically. The second set of equations do NOT have 0 determinant so have a unique solution. What that solution is should be obvious.
  7. Sep 26, 2014 #6


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    I forgot that the three parallel lines could be coincident.
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