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Consider $3$ planes $\pi_1, \pi_2, \pi_3$ in the projective space $\mathbb{R}P^4$. They intersect two by two in a point. All the $3$ planes together spans the projective space and $\pi_1 \cap \pi_2 \cap \pi_3 = \emptyset$. Now I want to show there exists a unique plane $\pi_0$ which intersect the other planes in a line such that $\pi_0 \cap \pi_i$ is a line for $i = 1,2,3$.

You can split the proof in $2$ parts. First one needs to show there exists such a plane. Afterwards one needs to show that plane is unique. So to start with the existence, I thought of making a guess for the dimension of the plane and verifying with the dimension relation if it fulfills all the requirements. But I couldn't really find a guess. I also thougt about dualizing the statement, but this neither helped me. I can't check the uniqueness if I haven't find the plane yet.

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Let $\pi_0=\operatorname{span}\{\pi_1\cap\pi_2,\pi_2\cap\pi_3,\pi_3\cap\pi_1\}$. This gives existence.

For uniqueness, note that $\pi_0$ must contain $\operatorname{span}\{\pi_0\cap\pi_i\mid i=1,2,3\}$, the span of three lines. So the three lines are coplanar, so pairwise intersection is nonempty. Hence $\pi_0\cap\pi_1\cap\pi_2\neq\varnothing$, but $\pi_1\cap\pi_2$ is a point.

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  • $\begingroup$ That was the guess I was looking for! I took $\pi_0 = span\{\pi_1 \cap \pi_2, \pi_2 \cap \pi_3\}$, now I see my mistake, thanks! $\endgroup$ – Belgium_Physics Jun 24 '19 at 9:59

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