# Why are the coefficients of the equation of a plane the normal vector of a plane?

Why are the coefficients of the equation of a plane the normal vector of a plane?

I borrowed the below picture from Pauls Online Calculus 3 notes:

http://tutorial.math.lamar.edu/Classes/CalcIII/EqnsOfPlanes.aspx

And I think the explanation he provides is great, however, I don't understand how one of the concepts work.

If the equation of a plane is $ax+by+cz=d$ how is it that $\overrightarrow n = \langle a,b,c \rangle$? From the picture below I suppose I can see this since $\overrightarrow r_0$, if continued past the plane, would clearly be perpendicular, but what about $\overrightarrow r$? That one is clearly not perpendicular if extended past the plane?

Sorry if what I'm asking is confusing.

• $r = (x, y, z)$ and $r_0$ are points on the plane so that $r-r_0$ is parallel to (or in) the plane and must be perpendicular to n. Then $(r-r_0).n = 0$ and the coefficients of n multiply the coordinates x, y and z. – Paul Nov 11 '15 at 15:35

Given a point $O=(x_0,y_0,z_0)$ and a vector $\vec n=\langle a,b,c\rangle$, we can describe a plane as the set of points $P=(x,y,z)$ such that $\vec{OP}\cdot \vec n=0$. In other words, the set of vectors, perpedicular to $\vec n$.
$$a(x-x_0)+b(y-y_0)+c(z-z_0)=0\implies ax+by+cz=ax_0+by_0+cz_0.$$
Letting $d=ax_0+by_0+cz_0$ gives the usual equation of the plane.