For example, given two planes:(adsbygoogle = window.adsbygoogle || []).push({});

P1: 3x + y -2z = 4

P2: x + 2y + z = 1

There is a line of intersection between them. The direction vector can be solved by doing the cross product of the two normal vectors for each plane, but then a point must be included to find the exact equation for the line.

I realize row reduction can solve this easily, but without using matrices, I try:

Set 3x + y - 2z - 4 = 0 equal to 0 = x + 2y + z -1

Thus you have 2x - y - 3z = 3 [1] (which is the equation for a plane)

*Please let me know if my arithmetic is incorrect

So one solution to the subtraction of the two planes is (0,0,-1), but that is not a solution to the independent planes above. I also realize that [1] is a plane and although it includes points on the line, it is an entire plane itself with more than just that line as solutions. I am just trying to understand: why? If you have two planes, and equate them, shouldn't you get the similar solutions for both planes? If not, why not? I realize [1] is a plane and the result doesn't make sense geometrically, but in R2, if you have x = y and 3y = 2x, setting them equal to each other gives the solution(s) for the two lines. Does this property only hold for lines (objects that can be represented in R2)? Could anyone care to explain? Thank you!

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# Why equating two planes does not provide solutions?

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