Two sides of a plane

Question

Consider a plane $ax+by+cz-d=0$ . Consider two points on the opposite sides the plane $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$. Why do $ax_1+by_1+cz_1-d$ and $ax_2+by_2+cz_2-d$ have opposite signs? Can someone explain the theory behind this?

Answer 1

You can write the equation of the plane as ${\bf n}\cdot {\bf r} + d = 0$, recall that if two vectors are parallel, the inner product is positive, and negative if the vectors are anti-parallel. So the vector ${\bf n}$ is perpendicular to the plane, if ${\bf r}_1$ is on the side of the plane to which ${\bf n}$ points, the inner product is positive, whereas if it is on the other side, the inner product is negative

Answer 2

I think your question is just one about outward of inward facing normal vector. For these points, you can orient the plane with $N$ or $-N$ where $N = (a,b,c)$ is the normal for the plane. So for $(x_1,y_1,z_1)$ the plane is defined by $((x,y,z) - (x_1,y_1,z_1)) \cdot N = 0$, and for the other we have that $((x,y,z) - (x_2,y_2,z_2)) \cdot -N = 0$, hence the difference in sign.