# Show there exists a unique plane $\pi_0$ which intersects with 3 other planes $\pi_0 \cap \pi_i$ in a line in projective space $\mathbb{R}P^4$

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.

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.
• 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! – Belgium_Physics Jun 24 '19 at 9:59