What is the difference between these two planes?

Question

As part of an exercise three equations defining planes are given, of which two were very similar:

$$ \left\{ \begin{array}{c} x + 5y - 2z = 5 \\ -x + 5y + 2z = -4 \end{array} \right. $$

I saw that adding the left hand sides eliminated all variables except $y$:

$$ ( x + 5y - 2z ) + ( -x + 5y + 2z ) = 0x + 10y + 0z = 10y$$ $$ 5 - 4 = 1$$ So: $$10y = 1 \implies y = 0.1$$ Which gives: $$ x + 0.5 -2z = 5 \implies x -2z = 4.5$$ $$ -x + 0.5 + 2z = -4 \implies -x + 2z = -4.5 \implies x - 2z = 4.5$$

I concluded the two equations defined the same plane, but the solution to the exercise given by the book implies three different planes. Where did I make a mistake?

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The full exercise:

Do these three planes intersect in (1) a point, (2) a line, (3) two lines (if two planes are parallel), (4) in three parallel lines:

$$ \left\{ \begin{array}{c} x + 5y - 2z = 5 \\ 2x - 4z = 1 \\ -x + 5y + 2z = -4 \end{array} \right. $$

The book gives (4) as the solution, but if the first and third equation simplify to the same $x−2z=4.5$, there would be no intersection, since that is a parallel plane to $2x - 4z = 1$.

Answer 1

You just found parametric equations of the intersection line of the two planes: $$\begin{bmatrix} x\\y\\z \end{bmatrix}=\begin{bmatrix}\frac92\\\frac1{10}\\0 \end{bmatrix}+t\begin{bmatrix}2\\0\\1 \end{bmatrix}.$$ The two equations cannot represent the same plane, since they have non-collinear normal vectors: $$\begin{bmatrix}1\\5\\-2 \end{bmatrix}\enspace\text{and}\enspace\begin{bmatrix}-1\\5\\2 \end{bmatrix}.$$

Answer 2

I'm not sure what you are doing, probably looking for intersection? Two equations describe the same plane iff they are proportional. This condition should be checked. It is not satisfied.

You may also observe, that the point $(0,1,0)$ lies on the first plane, but not on the second one. This also suffices to prove that the equations describe two different planes. Of course two different planes may intersect or not, this is another question.

Answer 3

You can simply check if both planes are equal by looking at both equations simultaneously. So we have $$ x+5y−2z-5=−x+5y+2z+4 $$ Reducing that equation we get

$$ 2x-4z-9=0 $$

Since not all variables cancel out those two planes are definitely not the same. They do intersect therefore.

Answer 4

You can interpret a plane equation $$ a_x x + a_y y + a_z z = b $$ as scalar product $$ (a_x,a_y,a_z) \cdot (x,y,z) = b $$ where $(a_x,a_y,a_z)$ is a normal vector of the plane and $b$ relates to the signed distance $d$ of the plane to the origin via: $$ d = \frac{b}{\lVert (a_x,a_y,a_z) \rVert} = \frac{b}{\sqrt{a_x^2 + a_y^2 + a_z^2}} $$ This gives $$ (1,5,-2) \cdot (x,y,z) = 5 \\ $$ and $$ (-1, 5, 2) \cdot (x,y,z) = -4 \iff \\ (1, -5, -2) \cdot (x,y,z) = 4 \\ $$ which tells you that these are quite different planes.

Here is an image of the scene, it suggests that the last answer might apply.

(Large version)