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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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    I see that the origin comes into play... – 2017-01-16
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    Have you ever heard of "hyperplanes"? – 2017-01-16
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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. – 2017-01-16
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    @Winston Yes, that darn origin is pretty much the issue here. Vectors $[x, y, z]$ satisfying $ax + by + cz = d$ are *not* perpendicular to $[a, b, c]$: we see their dot product with $[a, b, c]$ is not $0$, it's $d$ :) But, if we change our frame of reference from the origin to a point in the above plane (say, $[d, 0, 0]$) and think of vectors as "emanating" from our new reference point, things match back up. We're getting into "affine geometry" here, and distinctions are subtle. – 2017-01-16
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    @pjs36 I totally get that. The components of the vectors $[x,y,z]$ which satisfy $ax+by+cz=d$ are not perpendicular to the vector $[a,b,c]$. The plane $ax+by+cz=d$, however, is perpendicular to the vector $[a,b,c]$ in so much as every vector in the plane is perpendicular to that vector $[a,b,c]$. – 2017-01-16
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    Right! The issue is that we don't get vectors in the plane by simply picking their coordinates. Let's focus on the specific plane $x + y + z = 5$, which should be perpendicular to $[1, 1, 1]$ in some sense. To generate vectors in our plane, we need to consider *differences* in coordinates, not the coordinates themselves. So because $[5, 0, 0]$ and $[0, 0, 5]$ are in our plane, we get the vector $[5, 0, 0] - [0, 0, 5] = [5, 0, -5]$: And *that's* something perpendicular to $[1, 1, 1]$. – 2017-01-16
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    @pjs36 I agree completely. But, then what is the meaning of $ax+by+cz=0$ if we were to think about what's going on geometrically. Is it a collection of points that satisfy $ax+by+cz=0$ or a plane perpendicular to the vector $[a,b,c]$. Are these ultimately the same notion as each point in this plane can be associated with a vector as the origin exists within it? I'm trying to help students visualize haha. – 2017-01-16
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    @pjs36 It's like if I were a student, and you were to ask me draw the set of objects perpendicular to a vector, I'd likely draw a plane (with intuition of a height of a pyramid in mind or something). If I'm clever, I'd draw a bunch of planes. How does this computation help this visualization? – 2017-01-16
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    Ok, I see a little bit better the question (I don't want to try to answer in the comments, but I don't fully feel I understand the question, even now). Is there a specific interaction that motivated this, or a general lack of geometric intuition, from students, or inconsistent reasoning, etc? – 2017-01-16
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    Ok, let me see. Suppose I ask the question to students, find the collection of vectors perpendicular to some given vector [2,3,5]. They might say it's all the vectors [x,y,z] where 2x+3y+5z=0 - yes dot product =0 good. And I may say, could you represent perpendicularity to a vector with a picture? They may draw a plane or "infinite" parallel planes with a vector sticking out or whatever. And I may say good to the "infinite" planes drawing and ask what's the meaning of 2x+3y+5z=0 in so much as it's both one of those planes and the solution above? – 2017-01-16
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    The properties of $\mathbb R^n$ let you play fast and loose with vectors and points—you can identify points with their displacement vectors from the origin, you can treat vectors attached to different points interchangeably, and so on. Vectors attached to a point in the plane $ax+by+cz=d\ne0$ aren’t strictly speaking the same ones as those emanating from the origin. This makes computation in $\mathbb R^n$ relatively simple—you can ignore all of those subtle differences—but I suspect that it’s also contributing to some of your doubts. – 2017-01-17

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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 $s_k \neq 0$.