Non-mathematician here, so maybe it's a no-brainer for all of you...

I have a construct with three N-dimensional vectors. Vectors $\mathbf{u}$ and $\mathbf{v}$ together define a 2D plane, question is if the third vector $\mathbf{w}$ is perpendicular to that plane. Problem is that the vector dot-product can not be evaluated, the only information I have is the following: when choosing vectors in that plane with the same length as $\mathbf{w}$, then for each dimension $n$ I can find a vector which has a value in that dimension larger than that of $\mathbf{w}$. When trying to visualize this in 3D I have the idea that $\mathbf{w}$ cannot be perpendicular to the plane, but I have no idea if this is correct, if it's also the case for N dimensions and how a proof or at least a logical argument would look like.

  • $\begingroup$ You need the cross product instead of the dot product. $\endgroup$
    – QBrute
    Apr 1, 2022 at 9:53
  • $\begingroup$ I don't think I fully understand what you are asking. But maybe you mean this:- Given any two vectors lying in a plane , we can find a third vector $\mathbf{u\times v}$ which is perpendicular to the plane . Also the term dimension has a very specific meaning in Vector spaces and linear algebra. However I think you are more interested in the $2D$ and the $3D$ space. $\endgroup$ Apr 1, 2022 at 9:54
  • $\begingroup$ @QBrute: with the dot-product I could check perpendicularity of $\mathbf{w}$ via $(a\mathbf{u}+b\mathbf{v},\mathbf{w})=0$. But the dot product cannot be evaluated analytically / closed form in this case. $\endgroup$ Apr 1, 2022 at 10:07
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    $\begingroup$ I'm not sure how you intend to derive the properties of a vector without knowing what the vector is. How are you constructing it? $\endgroup$ Apr 1, 2022 at 10:09
  • $\begingroup$ @Mr.Gandalf Sauron: No, I do not want to find a third vector which is perpendicular to the plane, I already have the third vector $\mathbf{w}$ and I want to know if it is perpendicular to the plane given the characteristics of the components in each dimension like I described (choose vectors on the plane, make them the same length as $\mathbf{w}$, and then I see that for every dimension $n$ I can find a vector on the plane with a value in that dimension larger than that of $\mathbf{w}$). Besides that: the crossproduct seems to be ill-defined I think in more than 3 dimensions. $\endgroup$ Apr 1, 2022 at 10:13

1 Answer 1


I think I see the error in my 'conjecture'. It seems to work in 3D, but not in higher dimensions.

A rough sketch for a counter-example:

If you have a vector $\mathbf{w}=(w_1,w_2,\cdots,w_{N-1},w_N)$ and assume all $w_n$ are positive for simplicity, you can make a vector $\mathbf{u}=(u_1,u_2,\cdots,u_{N-1},u_N)$ where you take $u_n=w_n+\delta$ ($\delta$ very small) for all $n<N-1$ and use $u_{N-1}$ and $u_N$ to tune for $|\mathbf{u}|=|\mathbf{w}|$ and $\mathbf{u}$ perpendicular to $\mathbf{w}$. Then follow an equivalent recipe for vector $\mathbf{v}$, where you at least choose $v_{N-1}=w_{N-1}+\delta$ and $v_N=w_N+\delta$, and have the other vector components tuned to $|\mathbf{v}|=|\mathbf{w}|$ and $\mathbf{v}$ perpendicular to $\mathbf{w}$.

So in general, already in 4D, you have enough freedom to construct vectors $\mathbf{u}$ and $\mathbf{v}$ which form a plane/disc, have characteristics described in my original question with respect to $\mathbf{w}$, but nevertheless $\mathbf{w}$ is perpendicular to the plane/disc.


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