In the 3-dimensional real projective space $\mathbb{P}^{3}_{\mathbb{R}}$ we are given the point $P[0,1,0,1]$, and the lines $r_1$,$r_2$, of equations:

$$r_1: \bigg\{ \begin{array}{l} x_0-x_2+2x_3=0 \\ 2x_0+x_1=0 \\ \end{array} \qquad r_2: \bigg\{ \begin{array}{l} 2x_1-3x_2+x_3=0 \\ x_0+x_3=0 \\ \end{array} \quad $$ Prove that there exist exactly one line $\bar r$ which passes through P and intersects $r_1,r_2$, and find its cartesian equations.

How to procede? I'm a rookie with projective geometry.

  • 2
    $\begingroup$ Have you done any work on this? Hint: You have a point and a line $r_{1}$, they determine a plane. Find where the second line $r_{2}$ intersects that plane. $\endgroup$ – Morgan Rodgers Feb 18 '18 at 6:39
  • $\begingroup$ Does this argument work in projective spaces as smooth as in affine spaces? $\mathbb{P}^{3}_{\mathbb{R}}$ is in corrispondence with the 4-dimensional complex space $\mathbb{C}^4$, is it possible that in this environment can occur something "pathological"? $\endgroup$ – M.M. Feb 18 '18 at 6:49
  • $\begingroup$ I don't think $\mathbb{P}^{3}_{\mathbb{R}}$ is in correspondence with $\mathbb{C}^{4}$, or at least I don't see where you are deriving that correspondence. It corresponds to $\mathbb{R}^{4}$. And I'm not sure what you mean by something pathological. Your point $P$ is a one-dimensional subspace (of $\mathbb{R}^{4}$), $r_{1}$ and $r_{2}$ are two-dimensional. $\endgroup$ – Morgan Rodgers Feb 18 '18 at 6:55
  • $\begingroup$ Sorry about $\mathbb{C}^4$, I meant $\mathbb{R}^4$, I was doing other things in the meanwhile and I got destracted. Anyways, thanks. $\endgroup$ – M.M. Feb 18 '18 at 7:07
  • $\begingroup$ I’d say that, to the contrary, the “pathological” cases occur in affine spaces, where you have to deal specially with parallelism. $\endgroup$ – amd Feb 18 '18 at 8:10

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