I’m trying to prove the following result:
Given $3$ pairwise disjoint lines $L_1$,$L_2$ and $L_3$ in $\mathbb{P}^3(k)$, we can find a change of coordinates such that $L_1=V(Z,T)$, $L_2=V(X,Y)$ and $L_3=V(X-Z,Y-T)$.
I’ve started by writing $$L_i=\{U(x_i:y_i:z_i:t_i)+V(x_i’:y_i’:z_i’:t_i’)\mid U,V\in k\}$$ Then since $L_1$ and $L_2$ are disjoint we can’t find any $A,B,C,D\in k$ (not all $0$) such that $$\begin{align*}&A(x_1,y_1,z_1,t_1)+B(x_1’,y_1’,z_1’,t_1’)\\&+C(x_2,y_2,z_2,t_2)+D(x_2’,y_2’,z_2’,t_2’)=0\end{align*}$$ and so these vectors are linearly independent. Then we can invert the matrix $$\begin{pmatrix}x_1 & x_1’ & x_2 & x_2’\\ y_1 & y_1’ & y_2 & y_2’\\ z_1 & z_1’ & z_2 & z_2’\\ t_1 & t_1’ & t_2 & t_2’\\\end{pmatrix}$$ to give a change of coordinates mapping $L_1$ to $V(Z,T)$ and $L_2$ to $V(X,Y)$.
But I can’t see why in general we should be able to map $L_3$ to $V(X-Z,Y-T)$, since a linear map is determined by its action on a basis. Any help would be much appreciated.