Subspace intersecting many other subspaces V is a vector space of dimension 7. There are 5 subspaces of dimension four. I want to find a two dimensional subspace such that it intersects at least once with all the 5 subspaces.
Edit: All the 5 given subspaces are chosen randomly (with a very high probability, the intersection is a line).
If i take any two of the 5 subspaces and find the intersection it results in a line. Similarly, we can take another two planes and find another line. From these two lines we can form a 2 dimensional subspace which intersect 4 of the 5 subspaces. But can some one tell me how we can find a two dimensional subspace which intersect all the 5 subspace.
It would be very useful if you can tell what kind of concepts in mathematics can i look for to solve problems like this?
Thanks in advance.
Edit: the second paragraph is one way in which i tried the problem. But taking the intersection of the subspace puts more constraint on the problem and the solution becomes infeasible.

Assuming your vector space is over $\mathbb R$, it looks to me like "generically" there should be a finite number of solutions, but I can't prove that this finite number is positive, nor do I have a counterexample.  We can suppose your two-dimensional subspace $S$ has an orthonormal basis $\{ u, v \}$ where  $u \cdot e_1 = 0$ (where $e_1$ is a fixed nonzero vector).   There are 10 degrees of freedom for choosing $u$ and $v$.  The five subspaces are
the kernels of five linear operators $F_j$ of rank 3; for $S$ to have nonzero intersection with ${\rm ker} F_j$ you need scalars $a_j$ and $b_j$ with $a_j^2 + b_j^2  = 1$ and $F_j (a_j u + b_j v) = 0$.  This gives 5 more degrees of freedom for choosing points $(a_j, b_j)$ on the unit circle, minus 15 for the equations $F_j (a_j u + b_j v) = 0$, for a net of 0 degrees of freedom, and thus a discrete set of solutions (finite because the equations are polynomial).
For actually finding solutions in particular cases, I found Maple's numerical solver fsolve worked pretty well - the system seems too complicated for the symbolic solvers.
