Are all the points on any given plane perpendicular to the normal vector to that plane?

Question

The normal to a plane given by $2x - y + 3z = 5$ is $(2, -1, 3)$. Does this mean all the points on that plane are perpendicular to $(2, -1, 3)$? I'd like to think so, but then a point $(1, 1, \frac 43)$ on the plane is not perpendicular to $(2, -1, 3)$.

Thanks.

Answer 1

Not all the points of the plane are perpendicular to the normal vector (unless we consider a point as a vector and the plane pass by the point $(0,0,0)$) but if $A$ and $B$ are two points of this plane then the vector $\vec{AB}$ is perpendicular to the normal vector.

Answer 2

If you take any two points $(x, y, z)$ that satisfy $2x - y + 3z = 5$, call them $A$ and $B$, and then form the vector from $A$ to $B$ by subtraction, $A-B$, then you will find that $A-B$ is perpendicular to the normal to the plane.

Try it with $A = (1, 1, {4\over 3})$ and $B = (0,-2,1)$.

In general, if $ax_1 + by_1 + cz_1 = d$ and $ax_2 + by_2 + cz_2 = d$, then you should be able to prove that $(a,b,c)\cdot\left[(x_1,y_1,z_1)-(x_2,y_2,z_2)\right] = 0$.