Following on from a previous question - Plane Equation Where D Not Equal to Zero:
I am trying to develop an intuitive understanding for the equation of a plane, specifically, plane equations where $D \neq 0$:
$Ax + By + Cz = D$
In the image below is a 3-D space. The orange parallelogram represents a plane which does not intersect the origin. Points $P_1$, $P_2$ and $P_3$ all lie on the plane.
These points can be represented with position vectors; where $\vec v_1$, $\vec v_2$ and $\vec v_3$ correspond to the points $P_1$, $P_2$ and $P_3$ respectively:
Based on how I currently understand it, the vectors pictured below: $(\vec v_1 - \vec v_2)$ and $(\vec v_3 - \vec v_2)$, are parallel to the plane, which is to say that if either of them were moved and placed directly onto the plane, they would lie completely flat across it. Taking the cross product of $(\vec v_1 - \vec v_2) \times (\vec v_3 - \vec v_2)$ should yield a vector ($\vec n$) which is orthogonal to both $(\vec v_1 - \vec v_2)$ and $(\vec v_3 - \vec v_2)$, and therefore the plane. I haven't attempted to draw the normal vector $\vec n$ because it seems to me that it would be around about on the y-axis - so it should be pretty easy to visualise:
Based on what I have just gone through, there are two facts that seem to be in contradiction of each other:
- The vector $\vec n$ is normal to both $(\vec v_1 - \vec v_2)$ and $(\vec v_3 - \vec v_2)$, $\vec n$ is therefore normal to the plane (as $(\vec v_1 - \vec v_2)$ and $(\vec v_3 - \vec v_2)$, are parallel to the plane).
- $Ax + By + Cz \neq 0$
The value of $D$ in the plane equation cannot be equal to $0$, since the plane does not pass through the origin. Conversely, either of the vectors $(\vec v_1 - \vec v_2)$ or $(\vec v_3 - \vec v_2)$ dotted with the normal vector $(Ax + By + Cz)$ should be equal to $0$ since both vectors are orthogonal to the normal vector (or at least it appears this way to me).