I was working on an assignment and I ran into this weird curve-ball situation where part of it is to find the mirrored point of $P=(3,10,9)$ in the plane $x-z=0$. I can see the plane is basically the x-axis stretching the z-axis.
So I start by doing the usual $r(t) = P + t(n)$ figuring it will give me $t$($n$ being the normal in the plane, so $n = (1,0,-1)$). But this is what happens:
$r(t) = (3,10,9) + t(1,0,-1)$
$<=> (3,10,9) + (t, 0, -t) $
$<=> x=3+t, y=10+0, z=9-t$
Then inserted into the plane:
$x-z=0 $
$<=> (3+t)-(9-t)=0 $
$<=> 3+t-9-t=0 $
$<=> -6=0$
In a gaussian elimination this would mean there is no solution, but what does it mean here? And how do I find the mirrored point now that I didn't find t?
Surely it can't just be that the mirror point is (-3,10,9)?
You made a mistake in $(3+t)-(9-t)=0$ $$3+t-9+t=0$$$$2t=6$$$$t=3$$ $$x=3+t\implies3+3=6$$$$y=10$$$$z=9-t\implies9-3=6$$ $$(x,y,z)=(6,10,6)$$