Dgdrxrt

I need to show that there is a unique line (in $mathbbP^4$ I assume, or could they also mean in $mathbbR^5$?) 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 $dim(langle L, Mrangle cap N)=0$, so a single point in $mathbbP^4$.

I think I have to use this dimension, but I'm not sure what it means. It says $langle L, Mrangle$ is the union of all lines through $L$ and $M$. So is $langle L, Mrangle cap N$ 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..

The key here is that $langle L,Mrangle$ here is a hyperplane, three-dimensional in $mathbbP^4$. 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 $langle L,Mrangle$. Similarly, the union of lines through $x$ intersecting $M$ is another 2-d plane in $langle L,Mrangle$. 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: $L=ax_1+bx_2mid a,bin mathbbR$, $M=cy_1+dy_2mid c,din mathbbR$, $N=ez_1+fz_2mid e,fin mathbbR$ where $x_1,x_2,y_1,y_2,z_1,z_2$ are pairs of points on each line, with coordinates in $mathbbR^5setminus 0$. All of these use the description of $mathbbP^4$ as the quotient of $mathbbR^5$ minus the origin by $mathbbR^*$.

Now, the six 5-tuples $x_1,x_2,y_1,y_2,z_1,z_2$ in $mathbbR^5$ 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 $6times 6$ identity.

Since the lines don't lie in a hyperplane, our six points must span $mathbbR^5$, 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 $$alpha(Ax_1+Bx_2)+beta(Cy_1+Dy_2)+gamma(Ez_1+Fz_2)mid (alpha,beta,gamma)in mathbbR^3$$
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 $mathbbR^5$, 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 $alpha+beta+gamma=0$.