In $\mathbb{P}_\mathbb{R}^3$ the following quadric are given:
$Q_1: x_0^2+x_2^2+x_3^2=0$, $Q_2: x_0^2+x_1^2-x_2^2-2x_0x_3+2x_2x_3=0.$
Say if there is a homography $\omega: \mathbb{P}_\mathbb{R}^3\to \mathbb{P}_\mathbb{R}^3$ such that $\omega(Q_1) =Q_2$.
