# Distance between affine space and point

Let $A,A'$ two affine subspaces of a finite Euclidean Vectorspace $V$. Let $p,p'$ two points, such that $d(A,p)=d(A',p')$. $\dim(A)=\dim(A')$

I would like to show that there exists a movement $\alpha:V->V$ such that $\alpha(A)=A'$ and $\alpha(p)=p'$

I only know, if $\alpha$ is a movement, then the following properties hold:

(1) $d(\alpha(v),\alpha(w))=d(v,w)$

(2) $\alpha$ is affine

(3) $\alpha$ affine and for a basis $a_1,..,a_n$ it holds that $d(\alpha(a_i),\alpha(a_j))=d(a_i,a_j)$

How can I use this to prove the statement above ?

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What is $d$? Euclidean distance? – user1551 Dec 2 '12 at 21:33
yes, d is the euclidean – Voyage Dec 2 '12 at 21:33

Specifically, take an orthonormal basis $\{u_1,\ldots,u_m\}\ (m=\dim A)$ of $A$ and an orthonormal basis $\{v_1,\ldots,v_m\}$ of $A'$. Complete them to two orthonormal bases $\{u_1,\ldots,u_n\}\ (n=\dim V)$ and $\{v_1,\ldots,v_n\}$ of $V$. For any $x\in V$, define $\alpha(x) = p' + \sum_i \lambda_iv_i$ whenever $x-p=\sum_i \lambda_iu_i$. Now you are done.