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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

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