0
$\begingroup$

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?

$\endgroup$
3
  • $\begingroup$ 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"? $\endgroup$ Aug 27, 2019 at 15:37
  • $\begingroup$ I will update my question to explain this. $\endgroup$ Aug 27, 2019 at 15:42
  • $\begingroup$ Have posted a new question following on from this one here - math.stackexchange.com/questions/3336283/… $\endgroup$ Aug 27, 2019 at 19:26

1 Answer 1

2
$\begingroup$

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 $\frac{|D|}{||(A,B,C)||}$.

$\endgroup$
4
  • $\begingroup$ 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? $\endgroup$ Aug 27, 2019 at 15:59
  • 1
    $\begingroup$ @RyanWalter Not quite. $(A,B,C)$ is always normal to the plane regardless of the value of $D$. $\endgroup$
    – amd
    Aug 27, 2019 at 16:04
  • $\begingroup$ Okay, in that case my misunderstanding goes slightly deeper than I first thought. I will re-post an updated question later. Thanks for your help @amd and kccu $\endgroup$ Aug 27, 2019 at 16:13
  • $\begingroup$ Have posted a new question following on from this one here - math.stackexchange.com/questions/3336283/… $\endgroup$ Aug 27, 2019 at 19:25

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .