1
$\begingroup$

I am aware there was already a topic on this: Why are the coefficients of the equation of a plane the normal vector of a plane?

However, it did not answer the question, and instead explained it the exact same way as the Paul website, AND used the normal vector that the original author was asking about.

So why does: $~ax+by+cz=d~$ produce a normal vector $~\vec n=⟨a,b,c⟩~$?

$\endgroup$
1
  • $\begingroup$ It might help to consider the simpler case of a line. in standard form $ax + by = d$ the vector $(a,b)$ is perpendicular to the line. $\endgroup$
    – Doug M
    Sep 13, 2019 at 5:27

3 Answers 3

5
$\begingroup$

There are two ways to understand this.

  1. The usual way of deriving the equation of a plane is done by considering, a priori, some vector normal to this plane. We usually do this by picking three points in the plane. These determine two vectors, say $u$ and $v,$ parallel to this plane. It follows that their product $u×v$ is a normal vector to the plane. Now for any vector from one of the known points with coordinates $(p,q,r)$ to an arbitrary point $(x,y,z)$ in the plane, we have that the vector $(x-p,y-q,z-r)$ is perpendicular to $u×v,$ whose coordinates we may take as $(l,m,n).$ Then it follows that the plane equation is $(x-p,y-q,z-r)\cdot (l,m,n)=0.$ If this is multiplied out and put in the form you know, then it takes the form $$lx+my+nz=C(p,q,r,l,m,n),$$ where $C$ is a constant -- I've only emphasized what its made of. But the important point to note is that the coefficients of the variables in the equation are exactly those of the normal $u×v.$ So this always gives us a normal vector the plane.

  2. The other way comes from interpreting the equation $ax+by+cz=d$ quite differently, as follows. The equation says that the plane is the set of all points the projection of whose position vector $(x,y,z)$ onto a fixed position vector $(a,b,c)$ is a fixed quantity, $d.$ It follows from the fact that the vector $(a,b,c)$ can only be the normal position vector to the plane, otherwise the projection of an arbitrary vector $(x,y,z)$ onto it couldn't be constant. You may draw a picture to convince yourself of this.

$\endgroup$
2
$\begingroup$

First, we will use the fact that when two vectors have dot product zero, they are perpendicular.

Take any point $(x_0,y_0,z_0)$ on the plane. So $ax_0+by_0+cz_0=d$.

Then for any other point $(x,y,z)$ on the plane, we get $((x,y,z)-(x_0,y_0,z_0))\cdot (a,b,c)=d-d=0$.

But clearly, $(x,y,z)-(x_0,y_0,z_0)$ is a vector parallel to the plane.

$\endgroup$
1
$\begingroup$

If you consider the plane $ax+by+cz=0$ you can view this equation as the scalar product $$(a,b,c)\cdot(x,y,z)=0.$$ This means that the vector $(a,b,c)$ is orthogonal to the considered plane. Since your plane is parallel to the plane $ax+by+cz=d$ you have that the vector is orthogonal to your plane.

enter image description here

$\endgroup$
4
  • $\begingroup$ but what's preventing a,b,c from being in line with the plane since it just needs to be perpendicular to x,y,z $\endgroup$ Sep 13, 2019 at 4:49
  • $\begingroup$ @centurysword Not just one $(x,y,z)$ -- it has to be perpendicular to every $(x,y,z)$ in the plane. And since this particular plane goes through the origin, all of those vectors are parallel to the plane. $\endgroup$
    – David K
    Sep 13, 2019 at 4:53
  • $\begingroup$ just to make sure I understand this right, what determines if it goes point up or down (i know it's based on perspective) $\endgroup$ Sep 13, 2019 at 4:59
  • $\begingroup$ Of course $(x,y,z)$ is any point not a particular one. $\endgroup$ Sep 13, 2019 at 14:16

You must log in to answer this question.

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