Intuitively, why does the dot product of any point in the plane with the normal vector always give the same answer?

Question

A few days ago I asked: Help with Proposition $2.3.3$ from Elem. Differential Geometry by Pressley

But now I have a similar question:

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If $\alpha$ and $\beta$ are points on the plane, why does $\alpha \cdot \mathbf N= \beta \cdot \mathbf N$. I know one answer is because $$(\alpha - \beta) \cdot \mathbf N=0$$ $$\alpha \cdot N = \beta \cdot N$$

But looking at it this way: $$\| \alpha\|\|\mathbf N\|\cos\theta_1 = \| \beta\|\|\mathbf N\|\cos\theta_2$$ $$\| \alpha\|\cos\theta_1=\| \beta\|\cos\theta_2$$

Why does the dot product of any point in the plane (not vector ON the plane) with the normal vector always give the same answer?

Answer 1

If the vector is IN the plane then it will always be orthogonal with the normal vector. That is how it is defined. So we can pick whatever vectors sitting in the plane we like and the normal vector will always pass through them perpindicularly so the angle between any vectors and the normal vector will be equal.

Answer 2

Here is a picture.

$A$ and $B$ are points in the plane $P$, while $N$ is the subspace normal to $P$. Cleverly, I've called the origin $O$.

Recall that the dot product $A \cdot N$ is merely the length of the projection of $A$ onto $N$.

The green lines from $A$ and $B$ meet at the projection $\operatorname{proj}_N(A)$ of $A$ onto $N$ which just so happens to be the projection $\operatorname{proj}_N(B)$ of $B$ onto $N$.

The projections are the same, and hence have equal lengths. In other words, the dot products are equal.

Answer 3

Drop a perpendicular $d$ from the origin to the plane in question (it will be parallel to the normal). Then $$\| \alpha\|\cos\theta_1=\|d\|$$ for any $\alpha.$