(1) Let P1 be the closest point to P0 in L1; then let P2 be the closest point to P1 in L2; then, P3 the closest point to P2 in L1, P4 the closest point to P3 in L2, and so to infinity.

(2) Let P0' be the closest point to P0 in L1 and P0" the closest point to P0 in L2, and let P1 be the midpoint of P0' and P0"; now take P1' as the closest point to P1 in L1, and P1" as the closest point to P1 in L2, and let P2 be the midpoint of P1' and P1", and so to infinity.

(3) Let P0' be symmetrical to P0 with regard to L1, and P0" symmetrical to P0 with regard to L2, and let P1 be the midpoint of P0' and P0"; now let P1' be symmetrical to P1 with regard to L1, and P1" symmetrical to P1 with regard to L2, and let P2 be the midpoint of P1' and P1", and so to infinity.

Can you prove that no matter which algorithm I pick, the sequence P0, P1, P2, ... will converge to the same limit? Could you generalize that for three intersecting planes? For four three-dimensional spaces?