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For 2D vectors, computing the vector orthogonal to a given $v$ is straightforwardly done using Cross, as for example shown here.

However, Cross does not seem to work for more than two dimensions, and I couldn't find any function giving a complete set of normal vectors to a given $v$.

For example, I'm looking for a function that, given v={1,0,0}, will give me back {{0,1,0},{0,0,1}}, or some other equivalent set of vectors (that is, I want a basis for the orthogonal space $v^\perp$).

An easy way to do this in 3D is with something like:

normalVecs[v_] := Cross[v, #] & /@ IdentityMatrix@3 // Orthogonalize // Select[Norm@# > 0 &];
normalVecs @ {1, 0, 0}

but using Cross does not generalize to more than 3 dimensions.

What is the best way to do this?

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    $\begingroup$ Isn't it just NullSpace[{v}] that you ask? $\endgroup$
    – MeMyselfI
    Nov 25, 2018 at 14:49
  • $\begingroup$ @MeMyselfI Ah! I knew there had to be some function doing this, but didn't think of looking with that name. I think that would be the natural answer to this question $\endgroup$
    – glS
    Nov 25, 2018 at 14:51

2 Answers 2

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general solution

normalVecs[vec_] := Module[{l = Length@vec}, 
Cross[vec,##&@@#]&/@ Subsets[IdentityMatrix@l,{l-2}] //DeleteCases@Table[0,l]]  


normalVecs[{1, 0, 0, 0, 0, 0}]    

{{0,0,0,0,0,1},{0,0,0,0,-1,0},{0,0,0,1,0,0},{0,0,-1,0,0,0},{0,1,0,0,0,0}}

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  • $\begingroup$ Note that, unlike Nullspace[{v}], this does not produce a minimal basis of the orthogonal vector space. See for example normalVecs[{1, 1, 0}] which outputs two vectors that are the same up to an overal constant. It is clear that the built in NullSpace is preferable. $\endgroup$
    – Kvothe
    Apr 19, 2023 at 23:04
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One can find a set of orthogonal vectors, orthogonal to each other and to the given vector v, using NullSpace[{v}]. (Credit goes to MeMyselfI, see their comment on the question.)

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