Isometry from $\mathbb R^m$ to a subspace of $\mathbb R^n$

Question

Let $\{w_1,\dots,w_m\}$ be a basis of a subspace $U$ of $\mathbb R^n$ and $\{v_1,\dots,v_m\}$ be an orthogonal basis of $\mathbb R^m$. Is there a canonical way to obtain a linear transformation $T:\mathbb R^m\to U\subset \mathbb R^n$ such that $\{Tv_1,\dots,Tv_m\}$ is an orthogonal basis of $U$ without using an orthogonal basis for $U$ a priori?

For example is there a linear map from $\mathbb R^2$ to $2x+3y+4z=0$ which preserves an orthogonal basis?

Answer 1

Does the following come under the criteria "without using an orthogonal basis for $U$ a priori"? Consider any basis $B$ of $U$ and then using Gram Schmidt, get an orthogonal basis $B'$. Then, just map $w_i$s to distinct elements of $B'$.