Given the plane equation:
$Ax + By + Cz = D$
It is said that if $D \neq 0$, then $-D$ represents the distance, in the direction of the normal vector, between the plane and the origin.
But seeing as though the equation of a plane is really just the normal vector dotted with a vector which lies on the plane (where $A$, $B$ and $C$ are the components of the normal vector, and $x$, $y$ and $z$ are the components of a vector which lies on the plane) how is it possible for $D \neq 0$? If the normal vector is perpendicular to the vector which lies on the plane, then the dot product ($D$) should also always be equal to $0$.
Why is this not the case?
Best Answer
You are mistaken. The expression $Ax+By+Cz$ is the dot product of the normal vector $(A,B,C)$ with the position vector $(x,y,z)$, i.e., the vector pointing from the origin to $(x,y,z)$. In general, the vector $(x,y,z)$ does not lie in the plane. In fact, it lies in the plane only in the case $D=0$. It might help you to draw a picture of this.
Another note is that $|D|$ represents the distance between the plane and the origin only if the normal vector is a unit vector. Otherwise the distance is $\frac{|D|}{||(A,B,C)||}$.