Line in an affine space of dim3 I'm trying to understand the theory of the affine space $A^3(R)$ and its affine subspace  of the line.
A line is defined by the intersection of two planes:
$r : \begin{cases}AX+BY+CZ+D=0\\
A'X+B'Y+C'Z+D'=0\end{cases}$
The direction of r is the one-dimensional subspace of V defined by:
\begin{cases}AX+BY+CZ=0\\
A'X+B'Y+C'Z=0\end{cases}
The directional vector if r is (and I don't understand why):
$l=\begin{vmatrix} B & C \\ B' & C' \end{vmatrix}$, 
$m=-\begin{vmatrix} A & C \\ A' & C' \end{vmatrix}$,
$n=\begin{vmatrix} A & B \\ A' & B' \end{vmatrix}$
because $(l,m,n)$ is the solution of the homogenous system.
Then I've tried with an example.
I have the line
r : \begin{cases}x-y+2z+1=0\\
-x+y-z+2=0\end{cases}
the subspace is defined by
 \begin{cases}x-y+2z=0\\
-x+y-z=0\end{cases}
 with the coefficient matrix
$$A=\begin{bmatrix}1 & -1 & 2\\-1 & 1 & -1\end{bmatrix}$$
that canbe reduced to
$$\begin{bmatrix}1 & -1 & 0\\0 & 0 & 1\end{bmatrix}$$
and the solution is $(t,t,0)$ with $t \in R$
PS: i've checked the exercise


The directional vector if r is (and I don't understand why):
$l=\begin{vmatrix} B & C \\ B' & C' \end{vmatrix}$, 
  $m=-\begin{vmatrix} A & C \\ A' & C' \end{vmatrix}$,
  $n=\begin{vmatrix} A & B \\ A' & B' \end{vmatrix}$
because $(l,m,n)$ is the solution of the homogenous system.

You can look at this in different ways:


*

*The line is given as the solution set to a system of two planes. This system has an infinite number of solutions (all the points of the line, of course!) and the solution set is 1-dimensional. In parametric form, the solution set is simply a parametric representation of the line. You can solve the system and verify that the general solution is given by the formulas above, so this gives you the directional vector.

*Since the line is given as the intersection of two planes and those planes are given in cartesian form, you can simply read the normal vectors of both planes: $(A,B,C)$ and $(A',B',C')$ respectively. The direction of the intersection (and thus of the line) is perpendicular to both normal vectors so you can find the directional vector by taking the cross product of the normal vectors of the two planes: $(A,B,C) \times (A',B',C')$; you'll find the formulas given above.


Using this formula on your example will give you the directional vector $(-1,-1,0)$. Note that any non-zero multiple of a directional vector represents the same direction so this agrees with the solution set $(t,t,0)$ you found (take $t=-1$) by manually solving the system.
