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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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  • $\begingroup$ I see that the origin comes into play... $\endgroup$ – Winston Jan 16 '17 at 23:07
  • $\begingroup$ Have you ever heard of "hyperplanes"? $\endgroup$ – WalterJ Jan 16 '17 at 23:08
  • $\begingroup$ 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. $\endgroup$ – mathreadler Jan 16 '17 at 23:15
  • $\begingroup$ @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. $\endgroup$ – pjs36 Jan 16 '17 at 23:17
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    $\begingroup$ 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. $\endgroup$ – amd Jan 17 '17 at 6:41
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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 $s_k \neq 0$.

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