Plane Equation Where $D \neq0$

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?

• What exactly do you mean by "But seeing as though the equation of a plane is really just the normal vector dotted with a vector which lies on the plane"? Aug 27, 2019 at 15:37
• I will update my question to explain this. Aug 27, 2019 at 15:42
• Have posted a new question following on from this one here - math.stackexchange.com/questions/3336283/… Aug 27, 2019 at 19:26

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 $$\frac{|D|}{||(A,B,C)||}$$.
• Okay I think I understand what you mean. The normal vector ($A$, $B$, $C$) and the vector ($x$, $y$, $z$), go from the origin to the position specified by their components, which if the plane intersects the origin, means that the vector ($x$, $y$, $z$) will lie on the plane, whilst ($A$, $B$, $C$) will be perpendicular to the plane. If the plane does not pass through the origin however, then ($x$, $y$, $z$) no longer lie on the plane since its tail is at the origin. For the same reason, ($A$, $B$, $C$) will not be perpendicular to the plane. Is this correct? Aug 27, 2019 at 15:59
• @RyanWalter Not quite. $(A,B,C)$ is always normal to the plane regardless of the value of $D$.