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How do I show two planes are neither coincident, or parallel?

  1. Mar 24, 2009 #1
    1. The problem statement, all variables and given/known data

    Show that the two planes are neither coincident, parallel, nor distinct. Identify, geometrically, how the planes intersect and determine the angle between the two planes to the nearest degree.

    [tex]\pi[/tex]1: x + 2y - 4z + 7 = 0
    [tex]\pi[/tex]2: 2x - 2y - 5z + 10 = 0

    I found the normals to be

    n1 = (1, 2, -4)
    n2 = (2, -2, -5)

    3. The attempt at a solution

    Two planes, if they intersect, intersect in a line.
    An equation of a plane has the form
    nxx + nyy + nzz = d
    where \vec n = <nx,ny,nz> is the normal vector to the plane.
    It can be shown that the angle between two planes' normal vector equals the angle between the planes. You can use the dot product
    \vec n1 dot \vec n2 = |n1||n2|cos(\theta)

    But how do I like......SHOW that they aren't either of those without drawing a diagram? Because it looks like like the first part of the question wants me to SHOW they are neither parallel or coincident, and the second part is finding the angle. So how would I show that they are neither parallel or coincident without proving the angle? All I can really think of is the fact that n2 [tex]\neq[/tex]kn1[tex]\pi[/tex]
     
    Last edited: Mar 24, 2009
  2. jcsd
  3. Mar 24, 2009 #2
    Well, parallel planes don't intersect, so finding an intersection proves they're not parallel, right? Coincident planes lie on top of one another so showing that they intersect in a line proves that they aren't coincident. But, aren't these two planes distinct, or am I thinking of something else.
     
  4. Mar 24, 2009 #3
    I think that maybe my professor just made a typo, because that doesn't make sense, from everything I've found so far, suggest they are distinct. But how do I show that? Do I use the dot/cross product for the parallel and perpendicular things?
     
  5. Mar 25, 2009 #4

    Mark44

    Staff: Mentor

    I agree. The problem statement is incorrect. Two planes in three-dimensional space (as yours are) must be either parallel or coincident. If they are parallel, they are distinct. If they intersection in a line, I would say that they are distinct, since we can tell them apart, but maybe your prof means distinct in the sense that they share no points at all.

    The two planes you are given have different normals, so they must intersect in a line, so it's not possible to prove that they are not parallel, not coincident, and not distinct. It's a little like saying, n is a positive integer - prove that it is not even and not odd.
     
  6. Mar 25, 2009 #5
    an inhomogenous system of two equations in three unknowns of the form Ax=b. look at the solution set of this system. if there are no solutions, then the two planes are parallel. if the solution set is a plane, then the two planes are coincident. if the solution set is a line, then the two planes intersect in a line and cannot be distinct (no points of intersection, if i interpret this correctly). it is impossible for the solution set to be a point for a 2x3 system. hope this helps.
     
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