vectors perpendicular to a vector-geometrically

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I will ask my question using $\mathbb R^3$ as a reference.

Vectors of the form $[x,y,z]$ perpendicular to $[a,b,c]$ are those where $[a,b,c]\cdot [x,y,z]=0$, that is where $ax+by+cz=0$.The $x,y,z$ components of these vectors must satisfy the equation of the above plane.

At first, I was like that makes sense asthe vector $[a,b,c]$ is perpendicular to the plane $ax+by+cz=0$. But, it kind of doesn't because all those other parallel planes $ax+by+cz=d$ where d is non-zero contain the very same vectors (whose components satisfy $ax+by+cz=0$.) So, what is the meaning geometrically of the plane $ax+by+cz=0$ and the fact that the vector [a,b,c] is perpendicular to it? How would you represent the collection of vectors perpendicular to a given vector or explain it?

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One interpretation is that one of the vectors can not help in constructing the other vector by linear combinations. The closest fit will always be for length 0 of that vector in any linear combination. So if we geometrically view a vector ${\bf v} = [x,y,z]$ as constructed by summing a set of other vectors ${\bf e}_k$:

$${\bf v} = \sum_{\forall i} s_i {\bf e}_i$$

For any ${\bf e}_k = [a,b,c]$ perpendicular to $\bf v$ there would be no point in having $s_k \neq 0$.