Given 4 lines in Euclidean 3-space:
L1: x = 1, y = 0;
L2: y = 1, z = 0;
L3: x = 0, z = 1;
L4: x = y, y = -6z.
Find the equations of the two lines which both meet all of the Li.
Solution
A routine computation. Assume the line meets L1 at (1,0,a) and L2 at (b,1,0). Then it is (x - 1) = t(x - b), y = t(y - 1), (z - a) = tz. So it can only cut L3 if 1/b = 1 - a, and L4 if 6a = 6ab - 1. This gives a quadratic for a, which we can solve to get a = -1/2 or 1/3. Hence the possible lines are (1,0,-1/2) + t(-1/3,1,1/2) and (1,0,1/3) + t(1/2,1,-1/3).
© John Scholes
jscholes@kalva.demon.co.uk
4 Sep 1999