# 3-Variable Equations

I’m taking Algebra II this year, and we just finished the chapter on 3 variable equations. This also covered how they can be used to describe planes, and how the solution to them is the intersection of the three planes described in the three equations.

Anyway, I started wondering about how they came up with problems for us to solve - do they just start out with values for x,y and z and just guess and check until they find three equations that work? Or is there a more efficient way of doing it?

One problem in particular seemed especially difficult - it was three planes that intersect in one line, so combining any of the three equations together would always end up with the same equation…and of course, all four equations had to work with the same values for x. What process could they use to do that?.

Re-reading my post, I realized it was bit unclear - I’m wondering how they came up with the equations for us to solve, not how to solve them.

My guess would be that they work backwards, taking points on a graph, then working backwards to get a question that would yield those numbers.

I wouldn’t be surprised if there are pretty standard ways to do it. For instance, if you want to find three planes that intersect in a line, you just calculate the direction vector, rotate it to get two lines in the first plane, and then rotate that plane a couple times.

It should all be pretty standard vector geometry. In fact, I wouldn’t be surprised if it can be done by computer.

Let’s pick a point: (x,y,z)=(3,7,-3)

Take an equation ax+by+cz=d and pick random numbers for a,b,c:

a=8, b=-5, c=-1 and plug in with (x,y,z) to get:
83-57-1*-3=-8 so equation 1 is:

(1) 8x-5y-z=-8

Repeat 2 more times:

a=-4, b=1, c=2 gives d=-11
a=5, b=-2, c=-2 gives d=7
In each case I picked 3 numbers and calculate d.

And our set of equations is:
(1) 8x-5y-z=-8
(2) -4x+y+2z=-11
(3) 5x-2y-2c=7

Note that since I randomly choose integers I will get an integer solution which is nice for beginners. You don’t have to choose integers. (Hmm, I wonder if my equations are independent? Exercise for the reader.)

To get a plane that passes thru a given line, take two points on the line, say (-1,5,2) and (6,3,-2). Randomly choose a and b but solve for c and d. E.g,
a=2, b=1 gives
-2+5+2c=d
12+3-2c=d
which solves to c=3 and d=9. So 2x+y+3c=9 passes thru the two points.

Lather, rinse, repeat. Check for suitable independence.

If you don’t care which line the planes pass thru you can get equations by doing linear combinations of a base set. E.g., randomly pick:
3x-5y-7z=10
-2x-6y+9z=1

And multiply each by a constant and add (I’ll choose 1 for the constants):
x-11y+2z=11
Which has a solution of (who cares at this point). As long as the second equation is not a constant multiple of the first I should be okay. But it’s harder to set up simple answers this way.