Find the vertices of the “triangle” in $P^{2}(R)$ whose sides are the projective lines $P(U_{1})$, $P(U_{2})$, $P(U_{3})$.

Question

So $U_{1}$, $U_{2}$ and $U_{3}$ are the 2-dimensional vector subspaces of $R^{3}$ defined by $x_{0}$=0, $x_{0}$+$x_{1}$+$x_{2}$=0, $3x_{0}$-$4x_{1}$+$5x_{2}$=0 respectively. What is a way to find the vertices of triangle in $P^2$(R)? Thanks.

Answer 1

Recall that a plane in $\mathbb R^3$ is a line in $\mathbb P^2(\mathbb R)$ and that a line in $\mathbb R^3$ is a point in $\mathbb P^2(\mathbb R)$.

The plane $x_0 = 0$ in $\mathbb R^2$ corresponds to the projective line parametrized by $(0:s:t)$ for $s,t \in \mathbb R$. Similarly the plane $x_0 + x_1 + x_2 = 0$ in $\mathbb R^2$ corresponds to the projective line $(u:v:-u-v)$. So the intersection of these lines is the point $(0:1:-1)$ and we did found one vertex of your triangle.

Similarly, the two other vertices can be obtained in a same way : $(9 : -2 : - 7)$ and $(0 : 5 : 4)$.