Does the sign of $a(x-x_0)+b(y-y_0)+c(z-z_0)$ represent which side of the plane $M$ is?

Question

For $P(x_0, y_0, z_0)$ on a plane of normal vector $N(a, b, c)$ and a random point $M(x, y, z)$

does the sign of $a(x-x_0)+b(y-y_0)+c(z-z_0)$ represent which side of the plane $M$ is on (and is null if $M$ is on the plane)?

Answer 1

Answer is yes, as your product is $ \vec N \cdot \vec{PM} $

To see it, set your plane as $ \mathcal P : z = 0 $ ; your normal vector would be $\vec N (a=0,b=0,c=1)$, your point $P$ would have coordinates $(x_0,y_0,0)$ and your product would be $a(x-x_0)+b(y-y_0)+c(z-z_0) = z$, which obviously is positive or negative depending on which side of the plane you're on.

That's an illustration more than a demonstration, but given that scalar product is invariant by isometric transformation we're not that far...

PS: please use Mathjax on M.SE : Should MathJaX be pointed out to new users?