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 $A=B=C=1$?
I will shamelessly repost Pablo Caneiro Elias answer on quora, which helped me work through this problem. The original post can be found here, so please give it an upvote.
The best way of understanding this is to look at the dot product operator. To simplify let’s focus on R3 , although the dot product can be applied in Rn . The dot product is a function $R3×R3→R:⟨u,v⟩→(uxvx+uyvy+uzvz)$ and it is also defined as the operator $⟨u,v⟩=|u||v|cosθ$ where $θ$ is the angle between two vectors u and v and symbol |w| denotes the Euclidean length of the vector w. As for vocabulary that will be used, a normalised vector w is a vector for which |w|=1 .
For two non zero vectos u and v , the operator ⟨u,v⟩ returns zero if and only if the two vectors are perpendicular to each other (since $cos(π2)=0$ ). So, you can tell whether vectors are perpendicular or not using the dot product. For arbitrary vectors u and v , $⟨u,v⟩=|u||v|cosθ=uxvx+uyvy+uzvz$ .
Lets consider, at first, only planes containing the origin $(0,0,0)∈R3$: if the plane contains the origin, all vectors in the plane can be seen as points in the plane, because the coordinates of the vectors are the same as of the points in the plane in that case (and any linear combination of vectors in the plane remains in the plane).
The most convenient (and intuitive) way of visualizing a plane that contains the origin is as a restriction in space of all vectors that are perpendicular to the so called plane's normal vector n . We can use our previously introduced dot product operator to write that restriction mathematically as ⟨n,w⟩=0,w∈R3.
Then, to check whether the point w belongs to the plane, just plug it in the dot product above. If the result is zero, then yes, point w lies in the plane. Otherwise it doest not lie in the plane.
Now, take $n=(A,B,C)$ as the plane's normal, and $w=(x,y,z)$ as an arbitrary point in R3 . Expanding the dot product you have $⟨n,w⟩=|n||w|cosθ=Ax+By+Cz=0$ as the mathematical restriction of all points that belong to the plane. It is the traditional plane equation. It comes from the dot product operator.
But what if the plane does not contain the origin? Then we have a problem: we are checking for orthogonality of vectors using plane’s normal vector. But since the plane now does not contain the origin, a point in the plane is a vector in R3 that is not contained in the plane. This means that if we use our previous mathematical restriction model, points in the plane are going to fail the restriction.
So let’s now consider an extra plane parameter p which is the position of the plane. For now we are going to assume the plane's normal vector to be normalised ( $|n|=1$ - we will check the general case later), which is a perfectly reasonable assumption since you can always normalise a vector n by scaling it by $1|n|$.
Now to check whether an arbitrary point w in R3 lies in the plane positioned at p , we create a vector from p to the given arbitrary point w . Then we check whether this new vector is in the plane. So we write:
$⟨n,(w−p)⟩=0 $
So, if the vector (w−p) is perpendicular to plane’s normal vector, then the point w lies in the plane. What we did here? We reduced the general case of planes that don’t contain the origin to the case we’ve already seen of planes that contain the origin, by translating the arbitrary point w by −p .
Since the dot product can be distributed as regular product, we have:
$⟨n,(w−p)⟩=⟨n,w⟩−⟨n,p⟩=0⟹⟨n,w⟩=⟨n,p⟩$
Now, ⟨n,p⟩ depends only on intrinsic plane parameters and is typically called D .
So we have: $Ax+By+Cz=D$ as a general equation of a plane that does not contain the origin (if p is not the zero vector).
It happens also, that D is the plane’s (signed) distance to origin. Why? Well, just notice that, ⟨n,p⟩ is the length of the orthogonal projection of the plane’s position vector into plane’s normal vector (which is perpendicular to the plane by definition). To check that, notice that $⟨n,p⟩=|n||p|cosθ$. Since $|n|=1$ , we have $⟨n,p⟩=|p|cosθ$. So it is the (signed) distance between the plane and the origin (the sign depends on the direction of the plane’s normal vector).
Now, what if the plane normal vector is not normalised, e.g, $|n|≠1$ ? Well then you can still check whether a point is in the plane or not, since non zero orthogonal vectors have zero dot product despite their lengths, but in this case the value D will be scaled by the plane’s normal vector length. Then |D| is not the plane’s distance from origin (unless its zero, then the plane contains the origin). Its easy to verify that by using the dot product bilinear property: $⟨λu,v⟩=λ⟨u,v⟩=⟨u,λv⟩$ .
Working through this problem required a few epiphanies:
(1) In the case of a plane passing through origo, the dot product position vector of an arbitrary point w on the plane can be written as $Ax+By+Cz=0$. This also defines the relationship between $x,y$ and $z$ in point $w$, since the coordinates have the same nominal values as the components of the vector pointing to $w$. Since the relationship between the variables can be defined in the very same way for every single point in the plane, it would be fair to say that the equation $Ax+By+Cz=0$ really defines the plane.
(2) In the cae of a plane that doesn't pass through origo, we can write the dot product as $Ax+By+Cz-Ax_p-By_p-Cz_p$ and assuming we compare every point $w$ with the same point $p$, $-Ax_p-By_p-Cz_p$ can be written as a constant, say $D$.
(3) This still feels quite unintuitive, since it's hard to grasp why the dot product between the normal vector and the distance between $w$ and $p$ would have anything to do with the coordinates of point $w$. The key insight is that the coordinates of $w$ are represented as the $x,y$ and $z$ in $Ax+By+Cz+D=0$.
By the same logic as in (1), the relationship $x,y,z$ for every point in the plane can be described by the equation $Ax+By+Cz+D=0$, since the values for $x,y$ and $z$ will be nominally the same as the ones described in the equation (Assuming that is that the normal vector and the vector $pw$ is orthogonal).
These were actually (at least for me) huge conceptual leap and I have spent the better part of a week working through the details.