I need to show that there is a unique line (in $\mathbb{P}^4$ I assume, or could they also mean in $\mathbb{R}^5$?) that intersects three lines $L,M,N$ which are pairwise non-intersecting and not in the same hyperplane.
In a previous exercise I worked out that $\dim(\langle L, M\rangle \cap N)=0$, so a single point in $\mathbb{P}^4$.
I think I have to use this dimension, but I'm not sure what it means. It says $\langle L, M\rangle$ is the union of all lines through $L$ and $M$. So is $\langle L, M\rangle \cap N$ then the set of all lines through all three lines?
Other things I tried:
I know that $\langle L, M, N \rangle = 4$, since these lines are not in the same hyperplane.
Let $P$ be a point on $L$. Let $\Pi$ be the plane containing $M$ and $P$. In the same way we can construct a plane $\Pi'$ through $N$ and $P$. But I don't know if I'm going in the good direction, I simply don't know how to visualize this stuff in higher dimensions..
Best Answer
The key here is that $\langle L,M\rangle$ here is a hyperplane, three-dimensional in $\mathbb{P}^4$. There is, as you have noted, a single point $x$ where $N$ meets this hyperplane.
Now, the union of all lines through $x$ that intersect $L$ is a 2-d plane in $\langle L,M\rangle$. Similarly, the union of lines through $x$ intersecting $M$ is another 2-d plane in $\langle L,M\rangle$. The intersection of these planes will be a line; it can't be a plane because then $L$ and $M$ would intersect, and a dimension argument says it's at least a line. That line passes through $x$, so it's one of the lines that defines each of the planes, and thus it's the line we seek.
Now, seeing how this fits together, an explicit construction:
Define the lines $L$, $M$, $N$ as follows: $L=\{ax_1+bx_2\mid a,b\in \mathbb{R}\}$, $M=\{cy_1+dy_2\mid c,d\in \mathbb{R}\}$, $N=\{ez_1+fz_2\mid e,f\in \mathbb{R}\}$ where $x_1,x_2,y_1,y_2,z_1,z_2$ are pairs of points on each line, with coordinates in $\mathbb{R}^5\setminus \{0\}$. All of these use the description of $\mathbb{P}^4$ as the quotient of $\mathbb{R}^5$ minus the origin by $\mathbb{R}^*$.
Now, the six 5-tuples $x_1,x_2,y_1,y_2,z_1,z_2$ in $\mathbb{R}^5$ must have a nontrivial linear dependence relation; there are constants $A,B,C,D,E,F$ not all zero so that $Ax_1+Bx_2+Cy_1+Dy_2+Ez_1+Fz_2 = 0$. These constants can be found by Gaussian elimination on the matrix with rows $x_1,x_2,y_1,y_2,z_1,z_2$ augmented with a $6\times 6$ identity.
Since the lines don't lie in a hyperplane, our six points must span $\mathbb{R}^5$, and the constants $A,B,C,D,E,F$ are unique up to constant multiples. Also, since $L$ and $M$ don't intersect, there's no nontrivial linear combination of $x_1,x_2,y_1,y_2$ that's zero - which means that at least one of $E$ and $F$ is nonzero. Similarly, at least one of $A$ and $B$ is nonzero, and at least one of $C$ and $D$ is nonzero.
The line we seek is the set of linear combinations $$\{\alpha(Ax_1+Bx_2)+\beta(Cy_1+Dy_2)+\gamma(Ez_1+Fz_2)\mid (\alpha,\beta,\gamma)\in \mathbb{R}^3\}$$ Since the three vectors $Ax_1+Bx_2$, $Cy_1+Dy_2$, $Ez_1+Fz_2$ are linearly dependent, that's a $2$-dimensional subset of $\mathbb{R}^5$, which projects to a line. It intersects $L$ at $Ax_1+Bx_2$ for $\alpha=1,\beta=0,\gamma=0$ and similarly for the other two lines. We could also eliminate the redundancy by restricting to $\alpha+\beta+\gamma=0$.