The problem is known as finding the shortest distance between two skew lines and the points on the two lines
There are several approaches to solve this problem.
It can be done with knowledge of Analytic Geometry up to methods using multivariable calculus.
APPROACHE 1 USING MULTIVARIATE CALCULUS.
- Easily can be shown that the two lines are skew
- The directing vectors of the line passing through A and B and line passing through D and C are respectfully u =< −3, 0, 2 > and v = <1, 1, 4 >
- The parametric equation for line one are x = 5 −3t, y=−1 , z=2 +2t Equation (1)
- The parametric equation for line two are x = k, y=1 +k , z=2 +4k Equation (2)
- The distance between to two points on the above lines is D=√[ (k+3t−5)2 +(k+2)2 +(4k-2t)2]
To minimize D suffices to minimize the quantity Ω(t,k)= (k+3t−5)2 +(k+2)2 +(4k-2t)2 ⇒
Ω(t,k)= 18k2+13t2-10kt-6k-30t+29
Then we need the second derivative test for functions of two variables of the function Ω.
Ωt= 26t-2k-30 and Ωk=36k-2t-6
Solving the system 26t-2k-30=0 and 36k-2t-6=0 we find t=285/ 209 and k=84/209.
Substituting these values back to Equation (1) and Equation (2) we get the points
(x1, y1,z1)= (10/11, -1, 52/11) and (x2, y2,z2) =(84/209, 293/209, 754/209)
APPROACH 2 USING ANALYTIC GEOMETRY
- Lines L1 and L2 belong to two parallel planes P1 and P2. Find the equations of those two planes .The normal vector for both planes is the cross product of the directing vectors of the two lines.
- Find the equation of the plane P3 containing L1 perpendicular to P2.
- Find the equation of the plane P4 containing L2 perpendicular to P1.
- Find parametric equations of line L3 which is the intersection of planes P3 and P4
- The points of interest are the intersections of L3 with lines L1 and L2