Dimension in $\mathbb{P}^4$ of $\langle L,M \rangle \cap N$ with $L,M,N$ pairwise non-intersecting and not in one hyperplane

Question

Given three lines, $L, M, N \in\mathbb{P}^4$, not in one hyperplane and not pairwise intersecting, I need to calculate

$$\dim(\langle L,M\rangle\cap N).$$

By the dimension of intersection theorem for projective spaces we have

$$\dim(\langle L,M \rangle \cap N) = \dim\langle L,M \rangle + \dim N - \dim\langle \langle L,M \rangle N \rangle.$$

But I do not know how to interpret the angle bracket notation for two lines. I do know:

What I know: A projective line through two points $P=(p_0:...:p_n), Q=(q_0:...:q_n)$ is defined by first moving these points to $\mathbb{R}^{n+1}$, and then we have $$ PQ=\langle P,Q\rangle=\{\lambda p_0 + \mu q_0 : ... : \lambda x_n + \mu q_n \mid (\lambda,\mu)\neq (0,0)\}. $$

I also think that $\dim N$ must be equal to 2, since a projective line is a plane in Euclidian space, so given two points of $N$, we have that $N$ is the span of two lines passing through these two points respectively.

Furthermore I know that $\dim \mathbb{P}^4 = 4$.

Answer 1

$\langle L,M\rangle$ is itself a hyperplane if $L$ and $M$ don't intersect.
Since it doesn't contain $N$, it will intersect $N$ in a single point.