Given non-parallel lines L,M in Euclidean 3-space, there exist unique planes S,U meeting in L, and T,V in M, with both S,T perpendicular to both U,V. An algebraic proof of this turned out to be surprisingly nontrivial --- is there a more intuitive synthetic demonstration? [The appropriate generalisation of this this apparently tedious little lemma provides one crucial link in the classification of isometries for a quadratic inner-product space. It may well be a (special case of) some well-known theorem concerning bases of vector subspaces, though I didn't recognise it as such.] Fred Lunnon