Prove that if 
and 
are two planes which are not parallel then they intersect in a line.
Proof. Let the Cartesian equations of and be given by
![\[ ax + by + cz + d = 0, \qquad \text{and} \qquad a'x + b'y + c'z + d' = 0 \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-1e37aa4eca40ad96fd178ba3b89ea73e_l3.png "Rendered by QuickLaTeX.com")
respectively. Then, the intersection is given by the common solutions 
of these two equations. Since and are not parallel, we know they do not have the same normal vector so that 
for all 
. Further, since the normals are nonzero, we know each equation has at least one nonzero coefficient. Without loss of generality, let 
. Then,
![\[ x = \frac{-by - cz - d}{a}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-4e12c6596e952e886e21ad5dd0f5ef97_l3.png "Rendered by QuickLaTeX.com")
Substituting into the Cartesian equation for we have
![\[ (b'a - ba') y + (c'a - ca') z + (d'a - da') = 0 \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-73e4a0ac95fae5b0b0195da3205df9cb_l3.png "Rendered by QuickLaTeX.com")
is the set of solutions for the points on 
. But, we know at least one of 
or 
is nonzero, otherwise 
. Hence, we have the equation for a line. Therefore, is a line
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. :)
“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).
There is nothing wrong with that part. Your counter example is wrong. ac’ would be 3*6. ca’ would be 1*6. If ac’=ca’ and ab’=ba’ then
=(a/a’) provided a and a’are nonzero.
Your alternative proof sounds fine, however.
I do take issue in the original proof asserting that something of the form a”y +b”z +c” =0 defines a line when a” and b” are nonzero. It could, for example, define a plane parallel to the x axis.
There was some accepting error in my post. Obviously I didn’t read the fine print. What I meant to say was:
There is nothing wrong with that part. Your counter example is wrong. ac’ would be 3*6. ca’ would be 1*6. If ac’=ca’ and ab’=ba’ then
[a,b,c]=(a/a’)[a’,b’,c’] provided a and a’are nonzero.
Your alternative proof sounds fine, however.
I do take issue in the original proof asserting that something of the form a”y +b”z +c” =0 defines a line when a” and b” are nonzero. It could, for example, define a plane parallel to the x axis.
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.