Why equating two planes does not provide solutions?

1. Oct 19, 2014

MathewsMD

For example, given two planes:

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!

2. Oct 19, 2014

Staff: Mentor

For the intersection, both equations at the same time have to be satisfied. Your approach finds points in some third plane that is not relevant for this problem. You could call this third plane "the difference of the two planes", but I think there is no standard name for it because it does not have an interesting meaning. It is not even uniquely well-defined - you can use the same original planes to find a different "difference plane".

3. Oct 19, 2014

mathman

To find a point on both planes, set one coordinate = 0 and use the equations for the planes to get values for the other two coordinates.
This will work almost all the time, unless the line of intersection is parallel to the axis of that coordinate. In that case simply set another coordinate to 0.

4. Oct 19, 2014

FactChecker

If you follow the same process with these, you will make the same mistake. Suppose you say x - y = 0 = 3y - 2x => 0 = 4y - 3x. This gives you too many solutions, like (x,y) = (1, 3/4), that do not work in the original two equations. The problem is that combining two equations into one, without eliminating a variable, and not including another equation, looses too much information. Obviously, the only true solution is x = y = 0. You might want to review the correct process for solving simultaneous equations.

Last edited: Oct 19, 2014
5. Oct 20, 2014

Staff: Mentor

Minor quibble. You can't set one equation equal to another. You can set two numbers or expressions equal, but not two equations.

6. Oct 22, 2014

MathewsMD

Hmm...didn't I find both equations are expressions of 0? Thus I just replaced 0 with an equivalent expression? Do you mind expanding on why this is not allowed?

7. Oct 22, 2014

Staff: Mentor

An equation is not an expression. An equation is a statement about the equality of two expressions. The equation can be either true or false. It's harder to define expression, other than to say that it represents a number, and therefore has a value.
Nothing wrong with that.
Here's kind of a ridiculous example, just to make a point.
Eqn. 1: 2x + 3 = 5
Eqn. 2: x - 7 = 9

If I "set these two equations equal" I get (2x + 3 = 5) = (x - 7 = 9), which is pretty much nonsense, any way you look at it. My point is that you don't "set two equations equal."

8. Oct 22, 2014

FactChecker

It is valid, but the resulting single equation is weaker than the original two. Consider the two equations: x-2=0 and 3-x=0. Obviously there is no solution to both because the first one says x=2 and the second one says x=3. But if you set the two expressions of 0 equal, you get: x-2 = 3-x, which has a solution x = 0.5.

One way to solve simultaneous equations, to use one equation to solve for a variable and replace that variable in all the other equations. Keep doing this till you end up with an equation with a small number of variables. Solve that and back-substitute to calculate the variables that were replaced.
My trivial example: x-2=0 and 3-x=0; Use the first equation, x=2 to replace x in the second equation: 3-2=0 => 1=0 => not possible. That is the correct answer.
Your two line example: x=y and 3y=2x: Use the first equation to eliminate x from the second equation by substituting x with y: 3y=2y => y=0. Then x=y=0; That is the correct solution
Your two plane example: 3x + y -2z = 4 and x + 2y + z = 1; Use the second equation to solve for x (x = 1 - 2y -z) and replace x in the first equation: 3*(1-2y-z) + y -2z =4 => 3-6y-3z+y-2z=4 => -5y -5z = 1 => y = -z-1/5. So the modified set of equations is x = 1 - 2y -z and y = -z-1/5. To test this with one example, pick any value of z, calculate y from that, then calculate x and see if those values satisfy the original two equations. Let z=0. Then y= -0-1/5 = -1/5. Finally x = 1 - 2y -z = 1 - 2 (-1/5) -0 = 7/5. Check if (x,y,z)=(7/5, -1/5, 0) satisfy the original two equations. It does.

9. Oct 22, 2014

FactChecker

CORRECTION:
Should have been solution x=2.5. Sloppy arithmetic. Sorry.