Given two planes, P1, P2. Find parametric equations for the line of intersection between the two planes.
P1 = 2x -3y + 4z = 3
P2 = x + 4y - 2z = 7
The Attempt at a Solution
Let N1 be the normal vector to P1, and N2 be the normal vector to P2. Then,
N1 = (2, -3, 4) ; N2 = (1, 4, -2)
Then, the direction numbers for our desired line will come from the components of the cross product between the normal vectors, because it will be parallel to our desired line.
Then, N1xN2 = (-10, 8, 11).
Then, Eq. of line: x = x? -10t ; y = y? +8t ; z = z? + 11t
I currently need to find a point that would be on our desired line, so that I can complete the equation for this line. The book we're using unsatisfactorily just says in it's example version of this type of problem to "let z = 0" in both the given equations of the two planes in it's example, and then solve the system that emerges from that.
My problem is, why? How would I know ahead of time to let one of these three unknowns be 0?(In this case, the back of the book suggests to let y = 0 for my problem here, but I want to know what line of reasoning would let me come to that conclusion).