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This is Theorem of Three Perpendiculars, How to prove variation of it for N dimensional case (or to prove that it is false)?

Let $\mathbb{R}^N$ is N dimensional Euclidean space and $h_1, h_2$ are two intersecting hyperplanes (each has dimension N-1), $\mathbb{S}$ is intersection of hyperplanes $h_1$ and $h_2$. Let $x_1$ is a point at $h_1$, $x_2$ is an orthogonal projection of $x_1$ to the second hyperplane $h_2$, $x_1^p$ is an orthogonal projection of point $x_1$ to $\mathbb{S}$, $x_2^p$ is orthogonal projection of $x_2$ to $\mathbb{S}$.

How to prove that $x_1^p$ is equal $x_2^p$?

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Okay, I write solution I have just found.

Hyperplane $h_1$ is $\{x| a_1^Tx +b_1 = 0\}$, for some vector $a_1$ and number $b_1$. Similar $h_2 = \{x| a_1^Tx + b_2 = 0\}$. Intersection set is defined as $\mathbb{S} = \{x|a_1^Tx +b_1 = 0, and\ a_1^Tx + b_2 = 0\}$.

Projection point $x_1^p$ is a solution of next optimization problem: $x_1^p = \arg\max ||x-x_1||^2,\ with\ respect\ to\ x\in\mathbb{S}$.

Similarly, $x_2^p = \arg\max ||x-x_2||^2,\ with\ respect\ to\ x\in\mathbb{S}$.

Because $x_2$ is a projection of $x_1$ in $h_2$, we have $x_2-x_1 = \alpha a_2$, where $\alpha$ is some scalar.

Next step is $||x - x_2||^2 = ||x - x_1 - \alpha a_2|| ^2 = ||x - x_1|| - 2((x - x_1)\cdot \alpha a_2) + ||\alpha a_2||^2 = ||x-x_1|| ^2 + const$, where we used that $x^Ta_2 = b_2$ because of the constraint set.

So, essentially optimization problems are the same, which means that points $x_1^p$ and $x_2^p$ coincide.

Do you like the proof?

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