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horses
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Homework Statement
Find two lines in R3 in parametric form which satisfy the following conditions. Also, find the points on the lines which achieve the closest distance.
Conditions:
1. They are not parallel to any of the coordinate planes
2. They do not intersect and are not parallel
3. They are a distance of 2 apart at the points which have parameter values t = 0 and u = 0
4. The lines are NOT closest when the parameter values are t = 0 and u = 0.
Homework Equations
The Attempt at a Solution
I know how to find the shortest distance between two skew lines, but how do I find where that distance occurs in terms of points on the lines if all I have are parameters? I'm thinking that I need to use projections from the lines onto a plane then find the intersection of the projections, but when I try to do this I either get an unsolvable equation or infinite number of answers, since I'm dealing with parameters rather than exact coordinates.
This problem would be easier if the closest distance was 2 and occurred when t = 0 and u = 0, since I could start with two lines distance 2 apart parallel to the xz plane, then rotate them around the z axis 45 degrees using a rotation matrix. If the closest distance was 2 and occurred when t = 0 and u = 0, my lines could be (-t*sqrt(2)/2, -t*sqrt(2)/2, t) and (u*sqrt(2)/2 - sqrt(2), u*sqrt(2)/2 + sqrt(2), u) but unfortunately that is not the case here.
In terms of the distances, for 2 lines
L1 = (a,b,c) + t[v1,v2,v3]
L2 = (d,e,f) + u[w1,w2,w3]
distance between them must be 2:
2 = sqrt((a-d)^2 + (b-e)2 + (c-f)^2)
so 4 = (a-d)^2 + (b-e)2 + (c-f)^2
2 sets of lines I tried:
(0,0,2) + t[3,1,-1]
(0,0,0) + u[1,2,3]
for these the shortest distance seems to be 2/sqrt(6), which is indeed less than 2, but where does this distance occur? and
(0,0,3) + t[1,2,-1]
(0,0,1) + u[1,-1,1]
while for these its 6/sqrt(14), but I still don't see how I can find where this distance occurs.