Defining a plane using only 2 points (and an axis)

Question

Suppose you have a $3$-dimensional space in which there are 2 points ($A$ and $B$) defined (non identical). Now, you can define a line that goes through them but you cannot define a unique plane, because there are infinitely many planes that are rotating along that line.

Now, I've been having trouble defining the following in strict mathematical terms. That's also why I'm asking here, so you'll have to forgive me for being vague or imprecise.

Suppose that you also chose one axis towards which you want the plane to lay 'flat'. Imagine you model this using table and a piece of paper. The axis along the table are $X$ and $Y$ and the axis up is $Z$. You can rotate the paper along two imaginary points, but at one point it would lay flattest towards your $Z$ axis. There would be a clean slope from one point to the other.

How would you figure out the equation of this plane which would be defined using only two points and an 'flatness axis' (for my lack of better words). My thoughts led me towards using a third point ($C$) which would form a vector ($AC$) perpendicular to the vector ($AB$) where the $Z$ coordinate of point $C$ would be the same as point's $A$. Maybe.

And how would you generalize it to $n$-dimensional cases? (2 points, one flatness axis)

EDIT:

I have managed to reduce my problem to the following question: I have a vector $\vec n$ and $\vec m$. I want to find vector $\vec o$, which is perpendicular to the vector $\vec n$ and lies on the plane defined using vectors $\vec n$ and $\vec m$.

Answer 1

What you (probably) need (your question is not so clear) is the equation of a plane passing through two given points $A$ and $B$ and parallel to a given line $r$.

If you know a direction vector $\vec r$ of the line, then the vector $$ \vec n= (B-A)\times \vec r $$ is perpendicular to the requested plane, whose equation is then $$ n_x x+n_y y+n_z z=n_x x_A+n_y y_A+n_z z_A. $$

EDIT.

A vector $o⃗$, which is perpendicular to vector $n⃗$ and lies on the plane defined by vectors $n⃗$ and $m⃗$ can be computed as follows: $$ \vec o = \vec n\times(\vec n\times\vec m)= \vec n\,(\vec n\cdot\vec m)-\vec m\,(\vec n\cdot\vec n). $$

Answer 2

Given $P$ and $Q$ it suffices consider the infinitely many vectors $\vec n=(a,b,c)$ orthogonal to $P-Q$ then the plane equation is

$$ax+by+cz+d=0$$

and $d$ can be found by the condition that $P$ (or $Q$) belongs to the plane.

Answer 3

This 'flatness axis' is usually called 'normal vector'. It is perpendicular to the plane. Let's call it $\vec{n} = [a, b, c]$. Then we need only one point $P_0 = [x_0, y_0, z_0]$ together with the normal vector to define the plane. For an arbitrary point P on the plane, the vector $\vec{P} - \vec{P_0}$ must be perpendicular to $\vec{n}$. It means that the dot product must be zero $(\vec{P} - \vec{P_0})\cdot\vec{n} = 0$. And here comes the equation:$$(x - x_0)\cdot a + (y - y_0)\cdot b + (z - z_0)\cdot c = 0$$