Solved by Henry K. Shin

Question:

Lines and Locus

Let L1 and L2 be two skew (i.e. nonintersecting) lines with perpendicular direction vectors. Find, with proof, a precise general description of the locus of points equidistant from L1 and L2.

Answer:

Without a loss of generality, by rotating and contracting or expanding,
we can assume the direction vector of L_1 is (0,0,1) and passes through
(0,0,0) and hence the equation L_1(t) = t(0,0,1).  Similarly, we can
assume, without a loss of generality, that the line L_2 has a directional
vector of (0,1,0) (note (0,1,0) dot (0,0,1) = 0) and passes through
(1,0,0).  Since the perpendicular distance from P(x,y,z) to a line is (A
cross P)/(|A| * |P|) where A is directional vector of a line, we have
sqrt(x^2 + y^2) = sqrt(z^2 + (x-1)^2) or x^2 + y^2 = z^2 + (x-1)^2 which
gives x = (1/2)(z^2 - y^2).  This has a form x = (z^2)/(a^2) - (y^2)/(b^2)
with a = b = sqrt(2), a hyperbolic paraboloid

-- Henry K. Shin