# Unclear if a vector is perpendicular to a plane

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.

• You need the cross product instead of the dot product. Apr 1, 2022 at 9:53
• 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. Apr 1, 2022 at 9:54
• @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. Apr 1, 2022 at 10:07
• 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? Apr 1, 2022 at 10:09
• @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. Apr 1, 2022 at 10:13

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 $$n and use $$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.