Generating a basis of the orthocomplement of a vector in $\mathbb{E}^{n}$ There is a vector $v\in\mathbb{E}^{4}$. I can define an orthogonal complement with basis vectors $e_{1},e_{2},e_{3}$ by the rule:
$$e_{i}=P_{i}\cdot v,$$
where 
$$P_{1}=\begin{pmatrix}0 & 0 & -1 & 0\\
0 & 0 & 0 & 1\\
1 & 0 & 0 & 0\\
0 & -1 & 0 & 0
\end{pmatrix}
 , P_{2}=\begin{pmatrix}0 & 0 & 0 & -1\\
0 & 0 & -1 & 0\\
0 & 1 & 0 & 0\\
1 & 0 & 0 & 0
\end{pmatrix}
 , P_{3}=\begin{pmatrix}0 & 1 & 0 & 0\\
-1 & 0 & 0 & 0\\
0 & 0 & 0 & 1\\
0 & 0 & -1 & 0
\end{pmatrix} .$$
If components of $v$ are continuous function on a curve, then components of $e_{i}$ will be the continuous functions too. Thus this rule allows me to define a local chart in some neighborhood of the curve. 
I know that the same trick works in $\mathbb{E}^{2}$ and $\mathbb{E}^{8}$ sapces. 
Question: Is there a general way to uniquely define a basis for orthogonal complement of the given vector in any dimansional euclidean space?
 A: No, and in fact the examples $\Bbb E^2, \Bbb E^4, \Bbb E^8$ mentioned in the question are the only (nontrivial) Euclidean spaces in which one can do this. (Of course, one can do this for $\Bbb E^1$, for which the orthocomplement of a nonzero vector is trivial.)
More precisely:
Suppose one has matrices $Q_1, \ldots, Q_{n - 1} \in M(n, \Bbb E)$ such that for any vector $v \in \Bbb E^n$ the ordered set $(P_a \cdot v)$ is a basis of $\langle v\rangle^{\perp}$. On the other hand, for all $v$ in the $(n - 1)$-sphere $\Bbb S^{n - 1} \subset \Bbb E^n$, we may identify the tangent space $T_v \Bbb S^{n - 1}$ to $\Bbb S^{n - 1}$ at $v$ with $\langle v \rangle^{\perp}$, and so the vector fields $v \mapsto Q_a \cdot v$, $a = 1, \ldots, n - 1$, comprise a global frame on $\Bbb S^{n - 1}$, that is, $\Bbb S^{n - 1}$ is parallelizable. It was proven (both by Kervaire and Bott-Milnor, IIRC) that the only parallelizable spheres are $\Bbb S^0, \Bbb S^1, \Bbb S^3, \Bbb S^7$, so we must have $n = 1, 2, 4, 8$.
One can construct the sets $(Q_a)$ uniformly the four cases: For each $n$, consider the unique real division algebra ($\Bbb R$, $\Bbb C$, $\Bbb H$, or $\Bbb O$) of dimension $n$ and pick a (real) orthogonal basis $(1, f_1, \ldots f_{n - 1})$ of it. By construction, $(f_a v)$ is a basis of $\langle v \rangle^{\perp}$, so we can take $Q_a$ to be the matrix representation w.r.t. the above basis of the left multiplication map $x \mapsto f_a x$. For example, taking the usual basis $(1, i, j, k)$ of $\Bbb H$, we get the matrices
$$
   Q_1 = -P_3, \quad Q_2 = P_1, \quad Q_3 = P_2 ,
$$
where the matrices $P_a$ are as in the question.
Here are some notes on the subject.
