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Proving that planes are parallel

  1. Oct 1, 2012 #1
    In [itex]ℝ^{3}[/itex], how would I go about proving that two planes are parallel, given their equations? I know what the "word" parallel means, in the sense that two planes are always equidistant from one another, so that they must either never intersect, or that they must intersect at every point on their graphs.

    But how does this translate to an algebraic, or vector definition of parallel planes, given the equations for both planes?

    Say the equation for plane 1 is [itex] a_{1}x + b_{1}y + c_{1}z = d_{1} [/itex] and the equation for plane 2 is [itex] a_{2}x + b_{2}y + c_{2}z = d_{2} [/itex].

    Under what conditions would they be paralle, given the geometric definitions of parallel I have just given?

    Or are my definitions just incorrect?

  2. jcsd
  3. Oct 1, 2012 #2
    Construct the planes' normal vectors [itex]a_1 \hat x + b_1 \hat y + c_1 \hat z[/itex] for the first plane and similarly for the second. If the vectors are parallel, the cross product must zero.
  4. Oct 1, 2012 #3
    I see, but how does this follow from the geometric definition I posted?

  5. Oct 1, 2012 #4
    If the planes are parallel, the vectors normal to them are parallel. The cross product can then be used as a test of whether the vectors are parallel.
  6. Oct 1, 2012 #5
    Yes ok, but how does that follow from the fact that they must either never intersect or they must intersect at every point on their graphs if they are parallel?

    The two definitions are quite distinct, and I am looking for a proof that they are definitions of the same thing, that is that satisfying one of the definitions automatically satisfies the other.

  7. Oct 1, 2012 #6
    Let me resort to some projective geometry and geometric algebra then.

    Consider a projective 2d space. This has a third "projection" dimension with unit vector [itex]e_0[/itex] (to go with [itex]e_1, e_2[/itex]). Vectors in this space can be taken to represent points, and typically they're of the form [itex]e_0 + x^1 e_1 + x^2 e_2[/itex]. Any scalar multiple of this vector is taken to represent the same 2d point. (This is why it's called "homogeneous" coordinates.)

    To imagine the above, consider a 3d space with a plane offset 1 unit above the origin. Where vectors intersect this plane sets the true 2d coordinates of a given point, and clearly multiplying any such vector by a scalar will not change this point of intersection.

    Now, consider two vectors [itex]p[/itex] and [itex]q[/itex] and the 2d subspace that they span. This subspace should intersect the projective plane in a line, and we get the familiar result from geometry that two points are all that's needed to describe a line. Let me denote this line with the following: [itex]p \wedge q[/itex].

    Now let's consider two other vectors [itex]r, s[/itex], which form the line [itex]r \wedge s[/itex].

    Remember that even though both these objects represent lines in real space, in the projective space they are 2d subspaces through the origin. If these lines are parallel, then the 2d subspaces they define should intersect only at the origin or they intersect everywhere (if the lines are coincident). If not, then the 2d subspaces share a common 1d subspace (a common vector). Since vectors = points, then if we can find the common vector, then we can find the point of intersection. The common elements between two subspaces is called the meet of the subspaces, and this is where I think we can connect to your geometric definition:

    Theorem: Two non-identical subspaces are parallel if and only if the meet of those subspaces is only the origin. Two identical subspaces are always parallel.

    I've introduces this notion based solely on lines, but I think it should be straightforward to extend to planes (if more difficult to visualize in projective space). Now, what is the meet between two projected lines? This is where the "geometric algebra" part really comes in. That formalism gives us the power to use a "commutator product" between subspaces that acts like an exclusive or. Between two projected lines, if they have a common vector, the commutator product of the lines will give us a 2d subspace that is orthogonal to that common line, which can then be found by duality. The process for planes in a real 3d space is similar.

    I know that this may seem like a lot more overhead than you were looking for--projective geometry is a bit strange. But if you're interested, I can go into more detail and prove things more rigorously, where here I've only tried to be persuasive.
  8. Oct 2, 2012 #7


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    Hey BiPolarity.

    Remember if two planes are parallel, the linear equations that describe them are linearly dependent.

    If you have two normal vectors that are scalar multiples of each other, then can you use the basic formula of linear independence to show that there are non-zero constants for this to hold?

    (Recall: a set of vectors are linearly independent if a linear combination equalling zero implies all coeffecients are also zero).
  9. Oct 2, 2012 #8


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    Suppose we have two planes with equations

    [itex] a_{1}x + b_{1}y + c_{1}z = d_{1} [/itex] and [itex] a_{2}x + b_{2}y + c_{2}z = d_{2} [/itex].

    (We assume here that [itex] a_1, b_1,c_1[/itex] are not all 0, for then the solution set of the first equation alone is all the space, if [itex]d_1=0[/itex], and empty otherwise. Similarly for the second equation.)

    Now, if [itex]a_1=a_2[/itex], [itex]b_1=b_2[/itex], [itex]c_1=c_2[/itex], or if we can obtain this by multiplying one of the equations with a nonzero constant (which does not change the solution set), then the equations can have no common solution, and hence the planes do not intersect, so they are parallell, unless also [itex]d_1=d_2[/itex], in which case the equations are identical and the planes coincide.

    In every other case, solving the equation system gives a one-parameter solution, where one of the variables can be choosen freely. This means that the planes intersect along a line.
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