$\Lambda, \Theta$ subspaces in $\mathbb{E^n}$

Question

Let $\Lambda, \Theta$ be two subspaces in $\mathbb{E^n}$ such that $\Lambda, \Theta$ are skew and $dim(\Lambda )+dim(\Theta)=n-1$.

If we call $\Psi$ the hyperplane such that $\Lambda\subseteq \Psi$ and $\Theta || \Psi$ with $P \in \Theta $ a generic point, then $d(\Lambda, \Theta)=d(P,\Psi)$.

How could I prove this? Thank you in advance.

Answer 1

I will give you the general idea. I don't myself know how to fill all the details.

By definition of subspace, $\Lambda,\Theta$ are cosets of some linear subpaces $V,U$ respectively.

$d(\Lambda,\Theta)=d(\lambda,\theta)$ for some $\lambda\in\Lambda, \theta\in\Theta$

Hence

$\Lambda=\lambda+V,\Theta=\theta+U$

Moreover

$(\lambda-\theta) \perp U, (\lambda-\theta) \perp V$

$\dim (V+U)=\dim V+\dim U -\dim (U\cap V)=\dim V+\dim U=n-1 $

where $\dim (U\cap V)=0$ because $\Lambda, \Theta$ are skew .

$\Lambda=\lambda+V\subseteq\lambda+(U+V)$

$\Theta=\theta+U\parallel \lambda+ (U+V)$

By uniqueness of a hyperplane $\Psi$ such that $\Lambda\subseteq \Psi$ and $\Theta\parallel \Psi$, we get $\Psi=\lambda+(U+V)$.

Since $\lambda-\theta \perp \lambda + (U+V)=\Psi$ and $\lambda\in\Psi$,

$d(\theta,\Psi)=d(\theta,\lambda)$

Given an arbitrary point $P\in\Theta$,

$d(P,\Psi)=d(\theta,\Psi)=d(\theta,\lambda)=d(\Theta,\Lambda)$

As I said I don't know how to prove many facts used on the way. I can list them if you wish.