I can (grudgingly) accept that if the plane passes through origin, a vector in the plane could be written as $x,y,z$. If we assume there's a normal vector with coordinates $A,B,C$, then the dot product of the normal vector our first vector(assuming the vectors to be orthogonal and the dot product to be zero) could be written as $Ax+By+Cz=0$.
I'm struggling with interpreting this geometrically, but I've managed to convince myself that this expression would indeed describe a plane of some sorts, since fixing ,say, $z$ in place would yield a line determined by the other variables, and since an infinite number of lines for an infinite number of values for z would yield a plane.
Now I'm trying to rationalize why on earth the plane described by the dot product would necessarily run through the vector described by $y,x,z$.
The way I see it, this would hold true if and only if $A=B=C=1$?
Best Answer
Given a vector connecting the origin to any point on the plane $(x,y,z)$ it is by definition orthogonal to the/a normal vector to the plane $(A,B,C)$, therefore:
$$(x,y,z)\cdot (A,B,C)=0 \iff Ax+By+Cz=0$$