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I doubt if I am asking this question correctly but for what it’s worth I have a set of orthogonal vectors for which I would like to pick from another set the closest orthogonal vector from it to my initial set, more like arranging vectors in order of how closely orthogonal they are to each another. here’s what I mean say we have the first set of vectors as

$A = \{ [1,0,0,0,0] , [0,1,0,0,0]\}$

and the set for which the closest orthogonal vector to the first set has to be picked as

$B = \{ [0,0,0,0,1], [0,0,1,0,0] , [0,0,0,1,0]\}$

What vector magic can one hope to perform on $B$ such that only the vector that’s closest to the ones in $A$ is isolated. Merely looking at $B$ one can tell that $[0,0,0,0,1]$ is 2 axes farther away from (the resultant of) $A$ and $[0,0,0,1,0]$ is 1 axis farther away from (the resultant of) $A$ unlike $[0,0,1,0,0]$ which is the closest. Is there a way one can compare orthogonal vectors this way such that “closeness” is a function of how many extra axes (orthogonal-vectors) that can exist between a vector from $B$ and all in $A$

N.B. each vector in $B$ is perpendicular to all in $A$, vectors in $A$ are also orthogonal to each other, I am just generally interested in comparing orthogonal vectors in this manner

UPDATE: This question is asking how to select $[0,0,1,0,0]$ from $B$ as the orthogonal vector closest to (the resultant of) $A$.

Or if for example:
$A = \{ [0,1,0,0,0] \}$ and
$B = \{ [0,0,0,0,1], [0,0,1,0,0], [1,0,0,0,0], [0,0,0,1,0]\}$

how to select $[1,0,0,0,0]$ and $[0,0,1,0,0]$ from $B$ as the orthogonal vectors closest to $A$

Or if for example:
$A = \{ [1,0,0,0,0] \}$ and
$B = \{ [0,0,0,0,1], [0,0,1,0,0], [0,1,0,0,0], [0,0,0,1,0]\}$

how to select $[0,1,0,0,0]$ from $B$ as the orthogonal vector closest to $A$

linker
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  • Simply find the angle between a vector in $B$ and those vectors in $A$. The angle can be found using the dot product. – Hosam Hajeer Mar 29 '22 at 23:11
  • @Calmdownandhavesometea angles between all vectors in $B$ is 90, any vector in $B$ would also make a 90 with any in $A$ – linker Mar 29 '22 at 23:14
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    The vectors in $A$ are not all orthogonal to each other. Also, the order in which we write the coordinates is just the arbitrary order in which we write them and has no geometric significance. Any pair of axes is equally as "close" as any other pair; there is no "2 axes farther away". – David K Mar 29 '22 at 23:15
  • So all vectors in $B$ are equally "close" to the vectors in $A.$ You have set up a completely symmetrical example, all answers equally good. – David K Mar 29 '22 at 23:16
  • sorry I made a mistake with A, all of them should be orthogonal to each other – linker Mar 29 '22 at 23:16
  • @DavidK question updated – linker Mar 29 '22 at 23:18
  • I think you are using vector vocabulary where it's unnecessary and confusing. I think you are given two sets A and B each of which contains sequences of 0s and 1s with just one 1 of the same length. You measure the distance between two such strings as the distance apart of the positions of the 1s. Then you want the element of B that's nearest to A in some sense. If I am right, please completely rewrite your question using this new terminology, and provide a few more examples. (If I'm wrong, ignore this comment.) – Ethan Bolker Mar 30 '22 at 00:00
  • Hi @EthanBolker, I am sure I didn’t ask the question correctly. The vectors in both sets $A$ and $B$ aren’t compulsorily unit vectors parallel to some arbitrary axis $[x,y,z,w,v]$, it’s just for the sake of simplicity I made them that way. Yes I want an element in $B$ closest to $A$ in some sense. I am making an update to the question with a few more examples – linker Mar 31 '22 at 14:00
  • You should show some examples where the vectors are not simply all 0s and 1s (with just one 1). Small examples that explain how you measure "closeness" will help. Construct several. How do you measure closeness when A has multiple vectors? – Ethan Bolker Mar 31 '22 at 15:03
  • Sorry I made a mistake asking this question, I tried implementing the suggestion @atreju made at https://math.stackexchange.com/q/4396103/585488 . I was trying to decide which of the normals to select from trying compute the btn to a 2-polytope from a 3-polytope existing within a 4-polytope, I had 3 possible normal solutions from my kernel, I ended up creating a larger matrix with the normals from the 3-polytope and 4-polytope included in order to get a kernel with a unique normal solution. – linker Apr 24 '22 at 18:33

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