Proving the components of the vector perpendicular to the plane are the coefficients of the plane equation. I understand this intuitively, but I'm having trouble with the proof. Say I have a plane:
$$Ax+By+Cz = D,$$ and I choose points on the plane:
$$(x_1, y_1, z_1), \space (x_2, y_2, z_2).$$
I know that the vector $(x_2 - x_1, y_2 - y_1, z_2 - z_1)$ must be perpendicular to the vector I end with, which I'll call $(a, b, c)$.
$$(a, b, c) \cdot (x_2 - x_1, y_2 - y_1, z_2 - z_1) = 0,$$ therefore $$(ax_2 - ax_1 + by_2 - by_1 + cz_2 - cz_1) = 0.$$ Thus, $$ax_2 + by_2 + cz_2 = ax_1 + by_1 + cz_1.$$ Since they're both on the plane, we know that $$Ax_2 + By_2 + Cz_2 = Ax_1 + By_1 + Cz_1 = D,$$ leaving me with:
$$Ax_2 + By_2 + Cz_2 = Ax_1 + By_1 + Cz_1$$
$$ax_2 + by_2 + cz_2 = ax_1 + by_1 + cz_1$$
but that isn't enough to justify that $A = a, B = b,$ and $C = c$, is it?
 A: There are many vectors orthogonal to the plane
$$A x + B y + Cz = D$$
(where not all of $A, B, C$ are zero): We know from your standard argument that $(A, B, C)$ is, but so is are all of the vectors $(\lambda A, \lambda B, \lambda C)$, $\lambda \in \mathbb{R}$, in the line it spans.
A: The fundamental result here is that given some vector $(A,B,C)$ and any other vector $(x,y,z)$ there is a unique $\lambda \in \mathbb{R}$ and $(r,s,t)$ such that $(x,y,z) = \lambda(A,B,C)+ (r,s,t)$ and $(A,B,C) \cdot (r,s,t) = 0$.
(It is easy to compute the values, but that is not the point at the moment.)
We have a plane $P = \{ (x,y,z) \mid (A,B,C) \cdot (x,y,z) = D \}$.  
Suppose $(l,m,n)$ is perpendicular to the plane $P = \{ (x,y,z) \mid (A,B,C) \cdot (x,y,z) = D \}$. In particular, this means
$(l,m,n) \cdot (x_1-x_2,y_1-y_2,z_1-z_2) = 0$ for all $(x_k,y_k,z_k) \in P$, $k = 1,2$. It is not hard to see that this is equivalent to
$(l,m,n) \cdot (\delta_x,\delta_y,\delta_z) = 0$ for all 
$(\delta_x,\delta_y,\delta_z)$ such that $(A,B,C) \cdot (\delta_x,\delta_y,\delta_z) = 0$.
Now write $(l,m,n) = \lambda(A,B,C)+ (r,s,t)$ as above. Since $(A,B,C) \cdot (r,s,t) = 0$, we have
$(l,m,n) \cdot (r,s,t) = 0 = \lambda (A,B,C) \cdot (r,s,t) + \|(r,s,t)\|^2$, so we see $(r,s,t) = 0$ from which we get
$(l,m,n) = \lambda (A,B,C)$.
