Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $x_2^p$?

share|cite|improve this question

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?

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.