Let O be a point not in the plane p and A a point in p. For each line in p through A, let H be the foot of the perpendicular from O to the line. Find the locus of H.
Answer
circle diameter AB, where OB is the normal to p
Solution
Let B be the foot of the perpendicular from O to p. We claim that the locus is the circle diameter AB. Any line in p through A meets this circle at one other point K (except for the tangent to the circle at A, but in that case A is obviously the foot of the perpendicular from O to the line). Now BK is perpendicular to AK, so OK is also perpendicular to AK, and hence K must be the foot of the perpendicular from O to the line.
© John Scholes
jscholes@kalva.demon.co.uk
7 March 2004
Last corrected/updated 7 Mar 04