Explicit basis for orthogonal subspace in polynomial form Let $x_1,x_2,x_3$ be three real numbers, not all zero, and
$x=(x_1,x_2,x_3)$. We know that the orthogonal space
$H_x=\lbrace (h_1,h_2,h_3) \in {\mathbb R}^3 | \ h_1x_1+h_2x_2+h_3x_3=0 \rbrace$
has dimension exactly $2$.
My question : Can we find an explicit
basis in polynomial form, i.e. can we find six polynomials
$P_1,P_2,\ldots,P_6$ in ${\mathbb R}[x_1,x_2,x_3]$ such that,
for any $(x_1,x_2,x_3)\neq (0,0,0)$, the two vectors $v_x=(P_1(x_1,x_2,x_3),P_2(x_1,x_2,x_3),P_3(x_1,x_2,x_3))$ and $w_x=(P_4(x_1,x_2,x_3),P_5(x_1,x_2,x_3),P_6(x_1,x_2,x_3))$ form a basis of
$H_x$ ?
My thoughts : 1) Since we can take $w_x = x \wedge v_x$ (wedge product), it suffices to find $w_x$ (the constraints are that $w_x$ must always be nonzero and always in $H_x$, as long as $x\neq (0,0,0)$).


*It is trivial to find a $w_x$ that works "most of the time" ; for example $w_x=(x_2+x_3,x_3-x_1,-x_1-x_2)$ works except on the line directed by $(1,-1,1)$. It is unclear however, at least for me, if a global solution exists.


*The analogous problem in two dimensions has an easy solution :
$(-x_2,x_1)$ is a basis vector of the orthogonal space to $(x_1,x_2)$.
 A: This is not possible. Consider $A=\mathbb{R}[x_1,x_2,x_3]$, the polynomial ring ($\mathbb{R}$ can be replaced by any field). Let $A^3\to A$ be the $A$-module map given by the vector $(x_1,x_2,x_3)$ and let $N$ be the kernel. If $v_x,w_x$ (I think of them as polynomial vectors) existed, clearly they belong to $N$. If your condition is satisfied, one sees that these generate $N$ modulo any maximal ideal of a non-zero point. Thus, $N/\left<v_x,w_x\right>$ is supported only at the origin. Since $N$ has rank 2, this says the submodule $\left<v_x,w_x\right>$ is free of rank 2 and by depth considerations, this implies $\left<v_x,w_x\right>=N$. This is impossible, since the ideal $(x_1,x_2,x_3)$ has depth one and thus homological dimension 2, not one.
EDIT:  The above argument is incomplete. My usual bias made me deal with it as though reals is algebraically closed. The argument only says that support of $N/\left<v_x,w_x\right>$ is a variety which contains no real points except possibly the origin. I will leave it here and not delete it just in case some one sees a way out.
