I am trying to get into projective geometry, and here is one task which I find hard and don't really know how to approach.
Let be $$L_1:= \alpha_1 x_0 + \beta_1 x_1 + \gamma_1 x_2 $$ $$L_2:= \alpha_2 x_0 + \beta_2 x_1 + \gamma_2 x_2 $$ two lines in the projective plane $ \mathbb{P}^2 $.
How can I find their intersection point? Thanks a lot in advance!
Lines in $\Bbb P^2$ correspond to planes (containing the origin) in $\Bbb R^3$ and points correspond to lines.
Setting $n_i:=(\alpha_i,\beta_i,\gamma_i)$, a point $p=(x_0,x_1,x_2)$ is on line $L_i$ iff $n_i\cdot p=0$ for the usual dot product of $\Bbb R^3$, i.e. iff $n_i\perp p$.
(Note that if $p$ satisfies this, all scalar multiples of $p$ will also do, so indeed it holds for the whole line of $p$ in $\Bbb R^3$, which is a single point in $\Bbb P^2$.)
Now, the usual cross product $n_1\times n_2$ produces a vector which is orthogonal to both $n_i$. That (and its any nonzero scalar multiple) will represent the intersection of the two lines $L_i$.
Note that the dual problem has basically the same solution:
If $p_1,p_2$ points are given in $\Bbb P^2$, then $p_1\times p_2$ produces the coefficients of the equation of the line $p_1p_2$.