Reflection planes by Householder transform Let we have two intersecting 2-dimentional planes in $\mathbb{R}^n$. First one contains linearly independent vectors $a$ and $b$, and the second one contains linearly independent vectors $u$ and $v$. If it is important let denote vector that lies in the intersection of planes as $s$. I need to set the Householder transform that maps first plane to the second. It is not necessary to map $a$ to $u$ or $b$ to $v$, I only need to represent the rotation around the vector $s$ that maps first plane to the second.
My idea is to compute the bisector plane between given planes and than just reflect over this bisector plane. But I can not write explicit formulas for such transform. Thanks for the help!
 A: Begin by computing orthonormal bases for each of the planes using $s$.  I will assume that neither $a$ nor $u$ are a multiple of $s$.
$\{s,a\}$ is basis for the first plane.  Compute
$$
a^{\perp} = a - \frac{s^Ta}{s^Ts}s, \qquad w_1 = \frac{a^{\perp}}{\|a^{\perp}\|}\\
u^{\perp} = u - \frac{s^Tu}{s^Ts}s, \qquad w_2 = \frac{u^{\perp}}{\|u^{\perp}\|}
$$
Now, if $\hat s = \frac{s}{\|s\|}$, then $\{\hat s, w_1\}$ is an orthonormal basis for the first plane and $\{\hat s, w_2\}$ is an orthonormal basis for the second.
It suffices to find the Householder transformation which maps $w_1$ to $w_2$ (as in the answer to your previous question).
A: *

*Take for $c$ any vector orthogonal to $P_1$ and normalize it $c'=c/\|c\|.$ 

*Take for $w$ any vector orthogonal to $P_2$ and normalize it $w'=w/\|w\|.$ 
How is it possible ? 


*

*if $n=3$ take $c=a \times b$ and $w=u \times v$.

*if $n>3$, we do the same, but we use a recipe given in remark below.
(Thanks to @Omnomnomnom who has made the remark that cross product, as such, does not exist in $\mathbb{R}^n$, $n>3$). 
The next step is to compute $V=w'-u'$, normalized as $V'=V/\|V\|$, which is a (unit) vector orthogonal to the bissector plane. Now we are able to use the classical formula for symmetry matrix $S$ with respect to a plane having unit normal vector $V'$:
$S=I-2V'V'^T$ where $I$ is the identity matrix.
For example, if $n=2$, let $V'=\binom{a}{b}$ ; then $$S=\begin{pmatrix}1&0\\0&1\end{pmatrix}-2\begin{pmatrix}a\\b\end{pmatrix}\begin{pmatrix}a&b\end{pmatrix}=\begin{pmatrix}1-2a^2&-2ab\\-2ab&1-2b^2\end{pmatrix}.$$
Remark: Being given 2 vectors in $\mathbb{R}^n$, how can we obtain a vector that is orthogonal to both ? It suffices to take the first 3 coordinates of $a$ and $b$, compute their cross product in $\mathbb{R}^3$, then fill in the $(n-3)$ remaining coordinates with zeros.
