Proving two planes are parallel (question about the equation)

Question

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 $-5$ ?

Answer 1

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.

Answer 2

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).