Determine projective planes in $\mathbb{RP}^4$

2018-05-08 15:39:56

In $\mathbb{RP}^4$ let $\pi_1$ and $\pi_2$ be planes and $l$ a line, given by

\begin{eqnarray*}

π_1:&\quad x+3z-s=0,&\quad 2x+3y+t=0,\\

π_2:&\quad -x+z+2t=0,&\quad 3x+y=0,\\

l:&\quad -13x+3z=0,&\quad 7y+3t=0,\quad -38y+3s=0.

\end{eqnarray*}

I want to determine the planes $π$ that contain $l$ and intersect with each $π_1$ and $π_2$ in a line.

I already found that $π_1 \cap π_2 = (-1:3:13:-7:38)$ and that $l$ does not intersect with either one of the two planes.

Unfortunately i am stuck with the further procedure and thus appreciate any help very much!

If a plane $\pi$ intersects the plane $\pi_i$ in a line $m$ and contains the line $l$, then $\pi$ contains both $m$ and $l$ and hence the two lines intersect. In particular $l$ intersects $\pi_i$, which you have shown not to be the case, hence no such plane $\pi$ exists.

  • If a plane $\pi$ intersects the plane $\pi_i$ in a line $m$ and contains the line $l$, then $\pi$ contains both $m$ and $l$ and hence the two lines intersect. In particular $l$ intersects $\pi_i$, which you have shown not to be the case, hence no such plane $\pi$ exists.

    2018-05-08 17:37:19