Line through a point and two non-intersecting lines in a projective space I have just read an introduction chapter to projective geometry and am trying to solve a few of the problems that are listed in my book. Quickly, I stumbled upon something that I cannot solve:
Suppose $l$ and $m$ are two lines in the projective space $\mathbb{P}^3$ of dimension 3, such that $l \cap m = \emptyset$. Prove that for any point $p \in \mathbb{P}^3 \setminus (l \cup m)$ there is a unique line that passes through $l$, $m$ and $p$.
I find it quite easy to solve this problem in the context of affine spaces. But since a projective space is not a vector space, the methods that I used for affine spaces do not work here.
I started thinking I might have to suppose that $l$ and $m$ are the projections of two 2-dimensional subspaces of $\mathbb{R}^4$, say $L$ and $M$ respectively, with $L \cap M = \{0\}$. And also that $p$ is a projection of $P$, which is a 1-dimensional subspace of $\mathbb{R}^4$, but not a subspace of $L$ or $M$. Then the task would become to find a 2-dimensional subspace of $\mathbb{R}^4$ that contains $P$ and intersects with $L$ and $M$. The projection of this subspace into $\mathbb{P}^3$ would be the required line.
But at this point I'm getting lost in my own thoughts. Any help would be greatly appreciated.
Edit:
I managed to write down something resembling clear reasoning and wanted to share it.
Let $\pi(W)$ denote the projective space associated with a vector space $W$. Denote $p = \pi(P)$, $l = \pi(L)$ and $m = \pi(M)$. Since $l$ and $m$ do not intersect, $L$ and $M$ must intersect trivially. Since $p \notin l$ and $p \notin m$, $P$ must not be a subspace of $L$ or $M$. This implies that $P = \text{span}\{x\}$ where $x$ is a linear combination of two vectors, one from $L$ and one from $M$.
Hence, we may suppose that there are four linearly independent vectors $t,u,v,w \in \mathbb{R}^4$ such that $L = \text{span}(t,u)$, $M = \text{span}(v,w)$ and $x = t + v$ (so that $P = \text{span}\{t + v\}$). We now construct the 2-dimensional subspace $V$ of $\mathbb{R}^4$ with the following requirements:

*

*$P$ must be a subspace of $V$ or, equivalently, $t + v \in V$.

*$V \cap L \neq \{0\}$ or, equivalently, $\lambda_1 t + \lambda_2 u \in V$ for some $\lambda_1, \lambda_2 \in \mathbb{R}$ (not both $0$).

*$V \cap M \neq \{0\}$ or, equivalently, $\mu_1 v + \mu_2 w \in V$ for some $\mu_1, \mu_2 \in \mathbb{R}$ (not both $0$).

Consider the possible values for $\lambda_1$, $\lambda_2$, $\mu_1$ and $\mu_2$:

*

*If $\mu_2 = 0$ and the other three were non-zero, we would have that $t+v, \lambda_1 t + \lambda_2 u, v \in V$. These three vectors are linearly independent, which would imply that $\text{dim} V \geq 3$.

*Similarly, in the cases that only $\lambda_1 = 0$, $\lambda_2 = 0$ or $\mu_1 = 0$ it would follow that $\dim V \geq 3$.

*If all scalars were non-zero, we would have $\dim V = 4$.

*If we choose $\lambda_1 = \mu_1 = 0$ and $\lambda_2 \neq 0 \neq \mu_2$, we would have that $t+v, u, w \in V$, and once again it would follow that $\dim V \geq 3$.

Since the above choices of the scalars leave $\dim V \geq 3$, we must choose $\lambda_2 = \mu_2 = 0$ and $\lambda_1 \neq 0 \neq \mu_1$. This makes $V = \text{span}\{t, v\}$ the only possible choice of $V$.
Finally, we can verify that $\pi(V)$ is a line that passes through $p$, $l$ and $m$:

*

*Since $P$ is a subspace of $V$, $p = \pi(P)$ must be contained in $\pi(V)$.

*Since $V \cap L \neq \{0\}$, we must have $\pi(V) \cap \pi(L) = \pi(V) \cap l \neq \emptyset$.

*Since $V \cap M \neq \{0\}$, we must have $\pi(V) \cap \pi(M) = \pi(V) \cap m \neq \emptyset$.

*Since $\dim V = 2$, it follows that $\dim \pi(V) = 1$, therefore, $\pi(V)$ is a line.

Lastly, we can claim uniqueness of $\pi(V)$ from the fact that the above construction led us to only one solution.
 A: Let $\mathbf p$  be a non-zero vector from $\mathbb R^4$ which represents $p\in \mathbb P^3$.
Since $\mathbf L \cap \mathbf M = 0$ and $\dim(\mathbf L) + \dim(\mathbf M) = 4$ we can claim that $\exists \mathbf u\in \mathbf L, \mathbf v\in \mathbf M: \mathbf p = \alpha\mathbf u + \beta\mathbf v$ and $\mathbf u, \mathbf v$ are unique up to a scale.
The plane $\mathbf N$ spanning $\mathbf u,\mathbf v$ has non-trivial intersections with both $\mathbf L$ and $\mathbf M$ and it contains $\mathbf p$, which means that its projection to $\mathbb P^3$ satisfies the requirement. This proves the existence.
To prove uniqueness of $\mathbf N$ we note that if we pick $\mathbf u \in \mathbf L\cap \mathbf N$, $\mathbf v \in \mathbf M\cap \mathbf N$ (where both $\mathbf u$ and $\mathbf v$ are unique up to a scale) then $\mathbf p$ can be represented as a linear combination of $\mathbf u$ and $\mathbf v$. But as has been mentioned above, such vectors are unique up to a scale, which means that the plane thus defined is unique too.
