If I have two planes:
$$5x + y - z = 7$$
$$-25x -5y + 5z = 9$$
I can see that from the first plane I get the vector $\langle5,1,-1\rangle$ from the coefficients and then from the second plane I get coefficients of $\langle -25,-5,5\rangle$.
Straight away I can see that vector $2$ is just vector $1$ multiplied by $-5$, but and I would have thought this was enough to suggest parallel, but I doubt myself because of the answers at the en do the equations.
If I multiplied all of plane $1$ by $-5$ I'd get $-25x -5y + 5z = -35$. It is the $-35$ and the $9$ that confuse me, if the planes are equal should these two numbers also be multiples of $-5$ ?
$\endgroup$ 12 Answers
$\begingroup$Recall that planes are parallel in $3$ dimensions if they do not share any points. Since every point $P=(x,y,z)$ on the first plane satisfies $$5x+y-z=7$$
we know (as you pointed out) that every point on the first plane also satisfies
$$-25x-5y+5z=-35$$
However, since $-35\ne 9$ no point $P$ on the first plane can satisfy the equation of the second plane, so the two planes do not share any points and are therefore parallel.
However if the second equation had $-35$ instead of $9$ on the right-hand side, then the two planes would indeed be the same (since one equation is satisfied exactly when the other is).
$\endgroup$ $\begingroup$That is sufficient to show that the planes are parallel. I'm not sure how much you know about vectors but the way to do it with vectors is:
Let $\mathbf n = (5,1,-1),\;\mathbf m = (-25,5,5) = -5\mathbf n$ then the equations of your planes are:
\begin{align}
\Pi_1:&&\mathbf r\cdot\mathbf n &=7\\
\Pi_2:&&\mathbf r\cdot\mathbf m &=9\\
\end{align}
which is equivalent to
\begin{align}
\Pi_1:&&\mathbf r\cdot\mathbf n &=7\\
\Pi_2:&&\mathbf r\cdot\mathbf n &=-\frac 95\\
\end{align}
because $m=-5n.$ Clearly at most one of those equations is satisfied for any given $\mathbf r$ so the planes must be parallel as they cannot intersect.
