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?
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 $2x – 4z = 1$.
Best Answer
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: $$\begin{bmatrix}1\\5\\-2 \end{bmatrix}\enspace\text{and}\enspace\begin{bmatrix}-1\\5\\2 \end{bmatrix}.$$