P is a variable point in space. Q is a fixed point on the z-axis. The plane normal to PQ through P cuts the x-axis at R and the y-axis at S. Find the locus of P such that PR and PS are at right angles.
Solution
Answer: sphere centre Q, radius QO, where O is the origin, excluding the two circles formed by the intersection of the sphere with the y-z plane and the x-z plane.
Let Q be (0, 0, r). Let P be (a, b, c). If R is (x, 0, 0), then PR and PQ are perpendicular, so their dot product is 0, so a(a - x) + b2 + c(c - r) = 0, hence ax = (a2 + b2 + c2 - cr). Similarly, if S is (0, y, 0), then by = (a2 + b2 + c2 - cr). We require PR and PS perpendicular so a(a - x) + (b - y)b + c2 = 0, hence ax + by = a2 + b2 + c2. So a2 + b2 + c2 - 2cr = 0 and hence a2 + b2 + (c - r)2 = r2, which shows that P lies on the sphere centre Q radius QO.
Conversely, suppose P lies on the sphere. Then SP and SO are tangents to the sphere and hence equal. Similarly, RP = RO, so PRS and ORS are similar. Hence ∠RPS = ∠ROS = 90o. However, if P lies in the z-x plane, then S is not a finite point and if P lies in the y-z plane, then R is not a finite point. So we must exclude points lying on these two circles of the sphere.
© John Scholes
jscholes@kalva.demon.co.uk
5 Mar 2002