How is $r$ used in vector equation for plane?

Question

How is $r$ used in vector equation for plane?

Since the equation is:

$$r \cdot n = a \cdot n$$

And the $r$ is a variable that remains in the final equation. But what is it used for?

Answer 1

Lets rename $a$ to $r_0$, just to use the same letters as most literature, and also the linked pictures.

If you write your equation like $$(r-r_0)\cdot n=0$$ the equation is much more intuitive.

That equation is the so called normal form of a plane. All vectors $r$ that fulfill that equation define your plane.

Here is a picture that illustrates everything that follows. Look at it wihle reading through the rest of this answer.

(Source)

The vector defined by $(r-r_0)$ is the vector "from $r_0$ to $r$", as you can see in the picture. In the picture the vector $r$ is on the plane, so the red vector "lies exactly in the plane".

If the vector $r$ would be located below or above the plane, the red vector $(r-r_0)$ would "point out of the plane".

That property can be translated to "the red vector $(r-r_0)$ should be orthogonal to $n$".

And that is exactly what the scalar product $$(r-r_0)\cdot n\overset{!}{=} 0$$ does.

The reason is simple if we look at the (geometrical) definition $$x\cdot y = |x||y|\cos(θ),$$ with $|x|$/$|y|$ being the length of vector $x$/$y$ and $θ$ being the angle between $x$ and $y$. If these two vectors are orthogonal it is $θ=90°$, and $x\cdot y=0$. Otherwise the scalar product is not $0$.