Linear Algebra: Changing General form into vector form I encountered this little rule in one of my answer books:
The direction vector of the general equation ax + by = c is d = [b, -a].
I have never seen this rule formally described so I'm wondering if someone could help me find a theorem that describes it fully.
What I also wonder if what happens if I want to convert say a three dimensional plane ax + by + cz = d into vector form? Does the same rule follow where d = [c, -b, -a]?
The context for this was in this chegg textbook answer
 A: The plane $ax+by+cz=d $ is perpendicular to the vector $(a, b, c) $ ...
A: If the direction vector of a line is $\mathbf d$, then all points on the line are of the form $\mathbf p_0+t\mathbf d$, where $\mathbf p_0=(x_0,y_0)$ is some known point on the line and $t\in\mathbb R$. Set $\mathbf d=(b,-a)$ and plug this into the equation of the line: $$a(bt+x_0)+b(-at+y_0) = ax_0+by_0=d.$$ The dot product $(a,b)\cdot(b,-a)=ab-ba=0$, so the vector $(a,b)$ is perpendicular (a.k.a. normal) to the line.  
In a similar fashion, the vector $(a,b,c)$ is perpendicular to the plane $ax+by+cz=d$. A plane is two-dimensional, though, so to convert this into parametric form you need two direction vectors instead of one. There’s an analogous rule to the one you have for a line: Since $a$, $b$ and $c$ are not all zero, at least two of $(a,b,c)\times(1,0,0)=(0,c,-b)$, $(a,b,c)\times(0,1,0)=(-c,0,a)$ and $(a,b,c)\times(0,0,1)=(b,-a,0)$ are non-zero. These vectors are all orthogonal to $(a,b,c)$. Choose any two of these vectors and call them $\mathbf u$ and $\mathbf v$. Then all of the points on the plane are of the form $\mathbf p_0+s\mathbf u+t\mathbf v$. I’ll leave verifying this to you.
