L_{1} is the line { (t + 1, 2t - 4, -3t + 5) **:** t real } and L_{2} is the line { (4t - 12, -t + 8, t + 17) **:** t real }. Find the smallest sphere touching L_{1} and L_{2}.

**Solution**

The sphere has diameter PQ, where P(s + 1, 2s - 4, -3s + 5) is on L_{1}, and Q(4t - 12, -t + 8, t + 17) is on L_{2}. PQ is as short as possible. It can also be characterised as perpendicular to L_{1} and L_{2}.

PQ^{2}/2 = 7s^{2} + st + 9t^{2} + 25s - 52t + 457/2. At the minimum the two partial derivatives must be zero, so 14s + t + 25 = 0, s + 18t - 52 = 0. Solving s = -2, t = 3. This gives P (-3, -12, 11), Q(0, 5, 20), the radius (= PQ/2) as 1/2 √251 and the centre as (-3/2, -7/2, 31/2).

*As originally set, there was a typo, so that the y-coordinate in L _{1} appeared as 2t + 4 instead of 2t - 4. That did not change the principle, but made the arithmetic horrendous!*

© John Scholes

jscholes@kalva.demon.co.uk

15 Feb 2002