Exercise review: perpendicular-to-plane line Please, can you check the following execution is correct:
Problem text
I have a plane in affine space in  $\Bbb R^4$ described by two following equations:
\begin{Bmatrix}3x+y-z-q +1=0\\ -x-y+z+2q=0\end{Bmatrix}
I have the coords of a point $P$: (0,1,1,0)
Describe the locus of $Q$ points such that line $PQ$ is $\bot$ to the plane.
My solution
Now I look for the line (which $Q$ points belong to) perpendicular to the plane and passing trough point $P$.
In order for a plane described by equation $ax+by+cz+dq+e=0$ and a line whose coefficients are $l,m,n,t$ to be perpendicular, this must be true:
$\frac{a}{l}=\frac{b}{m}=\frac{c}{n}=\frac{d}{t}$
So, the equations of line passing by generic point $P(x_0,y_0,z_0,q_0)$ is:
$\frac{x-x_0}{a}=\frac{y-y_0}{b}=\frac{z-z_0}{c}=\frac{q-q_0}{d}$
Then, in my specific problem I get:
$\begin{Bmatrix}x=y-1 \\ y=2-z \\ z=\frac{q}{2}+1 \end{Bmatrix}$
 A: The two equations
$$3x+y-z-q=0\ ,\qquad -x-y+z+2q=0$$ (I have omitted the "$+1$", as it amounts only to a translation and does not affect directions) determine two hyperplanes (to be exact: $3$-dimensional subspaces) $U_1$, $U_2$ in ${\mathbb R}^4$. Their intersection is a two-dimensional subspace $U\subset{\mathbb R}^4$. The lines $P\vee Q$ should be orthogonal to this $U$. There is a fundamental principle in linear algebra which says: "The orthogonal complement of an intersection is the span of the orthogonal complements." The orthogonal complement of $U_1$ is the  $1$-dimensional subspace $\langle a\rangle\subset{\mathbb R}^4$ spanned by $a:=(3,1,-1,-1)$, and the orthogonal complement of $U_2$ is the  $1$-dimensional subspace $\langle b\rangle$ spanned by  $b:=(-1,-1,1,2)$. The quoted principle then implies that $U^\perp$ consists of all linear combinations $x=\lambda a+\mu b$. As $\vec {PQ}$ has to be orthogonal to $U$ it follows that $\vec {PQ}$ has to be such a linear combination. Therefore the locus of the admissible $Q$ has the following parametrization:
$$Q=(0,1,1,0)+\lambda(3,1,-1,-1)+\mu(-1,-1,1,2)\qquad\bigl((\lambda,\mu)\in{\mathbb R}^2\bigr)\ .$$
(There are two errors in your argument: (a) Two linear equations in ${\mathbb R}^4$ determine a two-dimensional subspace $U$. Its orthogonal complement $U^\perp$  is not a line, but again a two-dimensional plane, because $4-2=2$. (b) A vector which is orthogonal to the intersection $U_1\cap U_2$ does not have to be orthogonal to $U_1$ and $U_2$ separately.)
