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For all unit vector $\nu \in \mathbb{R}^n$ consider an affine hyperplane $A_{\nu}$ orthogonal to the direction $\nu $.

Now consider n linearly independent unit vectors $\nu_ 1 , \nu_2, \dots, \nu_n \in \mathbb{R}^n$. I'm asking: is the intersection $A_{\nu_1} , A_{\nu_2}, \dots , A_{\nu_n}$ necessarily a point?

I think the answer is yes, but I'm not sure... In the case $n =2$ it is true (the intersection of two lines having different directions is a point). In the case $n=3$ is also true. But in general is it true?

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    $\begingroup$ When you say that $A_{\nu}$ is an affine hyperplane you must also specify a point on the plane. This does not change the answer to your question. $\endgroup$ – Eric Towers Dec 5 '18 at 14:24
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    $\begingroup$ Thank you for your contribution Eric. But I don't understand: am I wrong If I say "Consider an affine hyperplane $A_{\nu}$ orthogonal to $\nu$" even if I mean that we don't have information about the point but we only know that it is orthogonal to $\nu$? I hope my question is clear $\endgroup$ – Hermione Dec 5 '18 at 14:44
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    $\begingroup$ Any plane parallel to the plane you are thinking of is orthogonal to $\nu$. How about the perpendicular bisector of $\nu$? How about the linear subspace perpendicular to $\nu$? $\endgroup$ – Eric Towers Dec 5 '18 at 16:42
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Yes, it is just a point. It is a solution of a system$$\left\{\begin{array}{l}\langle\nu_1,x\rangle=\mu_1\\\langle\nu_2,x\rangle=\mu_2\\\vdots\\\langle\nu_n,x\rangle=\mu_n.\end{array}\right.$$This is a system of $n$ linear equations in $n$ unknowns and the fact that the $\nu_k$'s are linearly independent is equivalent to the assertion that the matrix of its coefficients is invertible. Therefore, the system has one and only one solution.

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  • $\begingroup$ Is it really $\mu_i$? or 0? $\endgroup$ – Stockfish Dec 5 '18 at 14:25
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    $\begingroup$ It would be zero if we were working with hyperplanes passing through the origin, but the OP wasn't assuming that. $\endgroup$ – José Carlos Santos Dec 5 '18 at 14:27
  • $\begingroup$ Ah correct, thank you for the reminder! $\endgroup$ – Stockfish Dec 5 '18 at 14:28

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