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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?

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    $\begingroup$ $(A,B,C) = k(a, b, c)$ for any $k \neq 0$ $\endgroup$
    – abel
    Jan 15, 2015 at 4:06

2 Answers 2

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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.

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  • $\begingroup$ Oh, okay, thanks. So from the last step I was at, how can I directly prove that $(a, b, c)$ is of the form $(\lambda A, \lambda B, \lambda C)$? $\endgroup$ Jan 15, 2015 at 4:11
  • $\begingroup$ Your equations say that both $(a, b, c)$ and $(A, B, C)$ are orthogonal to ${\bf v} := (x_2 - x_1, y_2 - y_1, z_2 - z_1)$, but of course there's a plane's worth of such vectors, so these equations aren't enough to force $(a, b, c)$ and $(A, B, C)$ to be parallel. This suggests that you need to pick another vector $\bf w$ parallel to the plane and not a multiple of ${\bf v}$, and consider simultaneously the conditions normality to $\bf v$ and $\bf w$ impose on a vector $(a, b, c)$. $\endgroup$ Jan 15, 2015 at 4:22
  • $\begingroup$ Now I'm confused -- if $(a, b, c)$ and $(A, B, C)$ are orthogonal to the same plane, wouldn't they have to be parallel? I can see how $(A, B, C)$ could be a multiple of $(a, b, c)$, but wouldn't that imply parallelism? For that matter, I don't quite see how I've shown that (A, B, C) itself is orthogonal to the plane. That's my end goal, but where I got stuck I had still only established that both $Ax_2 + By_2 + Cz_2 = Ax_1 + By_1 + Cz_1$ and $ax_2 + by_2 + cz_2 = ax_1 + by_1 + cz_1$, where (a, b, c) is assumed to be orthogonal. $\endgroup$ Jan 15, 2015 at 4:32
  • $\begingroup$ The point is that the last pair of equations in the post encode exactly the fact that $(a, b, c)$ and $(A, B, C)$ are both orthogonal to the particular vector $(x_2 - x_1, y_2 - y_1, z_2 - z_1)$, but do not encode the fact that those vectors are both orthogonal to the plane. $\endgroup$ Jan 15, 2015 at 4:34
  • $\begingroup$ Oh, I see. Thank you! $\endgroup$ Jan 15, 2015 at 4:37
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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 $(l,m,n) = \lambda (A,B,C)$.

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