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Prove that the intersection of two planes which are not parallel is a line

Prove that if M and M' are two planes which are not parallel then they intersect in a line.

Proof. Let the Cartesian equations of M and M' be given by

    \[ ax + by  + cz + d = 0, \qquad \text{and} \qquad a'x + b'y + c'z + d' = 0 \]

respectively. Then, the intersection is given by the common solutions (x,y,z) of these two equations. Since M and M' are not parallel, we know they do not have the same normal vector so that (a,b,c) \neq t (a', b', c') for all t. Further, since the normals are nonzero, we know each equation has at least one nonzero coefficient. Without loss of generality, let a \neq 0. Then,

    \[ x = \frac{-by - cz - d}{a}. \]

Substituting into the Cartesian equation for M' we have

    \[ (b'a - ba') y + (c'a - ca') z + (d'a - da') = 0 \]

is the set of solutions for the points on M \cap M'. But, we know at least one of (b'a - ba') or (c'a - ca') is nonzero, otherwise (a,b,c) = t(a',b',c'). Hence, we have the equation for a line. Therefore, M \cap M' is a line. \qquad \blacksquare


  1. Sean W Perry says:

    How bout this one:

    Rather than take both planes in generality, wlog, (by an affine change of variables,) one plane can be taken to be z = 0. Let this be plane P0 and the other plane be P1 defined by ax+by+cz+d = 0. That P0 and P1 are not parallel means that a and b cannot both be 0. Their intersection is given by ax+by+d = 0. Confined to the plane z = 0, this defines a line. :)

  2. S says:

    “But, we know at least one of (b’a – ba’) or (c’a – ca’) is nonzero, otherwise (a,b,c) = t(a’,b’,c’)” is incorrect. Take (a, b, c) = (3, 2, 1) and (a’, b’, c’) = (6, 8, 6).

    To solve this exercise, you can use the end of section 13.13 (to show that for each point in the intersection, there is a line going through it and which is also in the intersection) and theorem 13.10 (to show that there is exactly one such a line).

  3. Anonymous says:

    Let X1 and X2 be two points in the plain M1:
    X1 = p + m1 S + n1 T
    X2 = p + m2 S + n2 T
    => X1 – X2 = kS + pT and since M1 not parallel to M2, X1-X2 must not parallel to M2. So let line L1 as X1-X2 which not parallel to M2 and according to exercise 18 L2 must have one intersect point P1 in M2.
    Same way can get P2 and P3 at which both intersect with M2.
    Suppose P1 P2 P2 not in the same line so can get P1, P2 and P3 must in a only plain, but P1, P2, P3 in both M1 and M2 at the same time which can get M1 = M2, contradict with the case that M1 not parallel to M2. So, P1, P2, P3 must in the same line. The intersecting points must be in the same line.

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