# For 3 straights $l_1,l_2,l_3$ in a 2-dimensional projective space $P(V)$ so that they don’t intersect. How can I prove the following statement:

Given 3 straight lines $$l_1,l_2,l_3$$ in a 2-dimensional projective space $$P(V)$$ so that for two on two lines: $$l_i \cap l_j=\emptyset$$ And $$\{V_i\vert i=1,2,3\}\subset V$$ and $$l_i= P(V_i)$$

Can somebody help me with my proof of the following statement:

1. prove that $$\forall x\in l_1:\exists! y\in l_2 \wedge \exists! z\in l_3: x,y,z$$ are collineair

what I have tried so far: I have chosen a base $$\{v_1,v_2,v_3\}$$ for V. Then defined the parameter equations: $$l_i \leftrightarrow V_i \leftrightarrow x=\lambda_{0,i}v_0+\lambda_{1,i}v_1+\lambda_{2,i}v_2$$ Then I thought to prove that statement in $$\mathbb{E}^3$$? Such as: take $$x\in V_1$$. Because $$l_1,l_2, l_3$$ are hyperplanes in the projective plane $$\mathbb{R}P^2$$, they have to be hyperplanes and therefore planes in $$\mathbb{E}^3$$, and because $$V_1 \cap V_2 \cap V_3 = \emptyset \implies V_1^{\perp}=V_2^{\perp}=V_2^{\perp}$$and therefore we find that all planes have the same orthogonal vector $$(a_0,a_1,a_2)$$ and thus also the same cartesian-equation in the projective space: $$P(V_i)=l_i\leftrightarrow 0=a_0x_0+a_1x_1+a_2x_2$$ now I wanted to define a plane $$V_\perp \leftrightarrow x + vct\{(a_0,a_1,a_2),b\}$$with (b the vector between x and the origin). Since the intersection between each plane $$V_i$$ and that perpendicular plane $$V_\perp$$ would be 3 parallel lines, and then each straight would be represented as a point in the projective space which would lay then on the same projective line ($$l_\perp=P(V_\perp))$$

I ended up stuck in this problem. I would appreciate very much some help to make me understand the nature of this problem and how I can prove that statement on this manner, and also if there was an alternative way through the use of duality?

• This probably is about lines in a $3$ dimensional projective space, such that any $2$ do not intersect. So you work in $\mathbb{R}P^3$,( or $P^3(k)$ , $k$ a field ) Commented Jan 14, 2023 at 8:04