
#1
Jun111, 02:12 PM

P: 656

1. The problem statement, all variables and given/known data
x1 = (0,0,4) + s(2,0,1) x2 = (4,2,2) + t(5,1,1) 2. Relevant equations 3. The attempt at a solution First of all I find the common perpendicular (vector cross product) Make it unit Find a arbitrary line joining the two 2 lines (set s = t= 0) Then scalar produce the unit normal and the line between the vectors to get the minimum distance. I get 2/sqrt(14), while the answers say just sqrt(14). How so? Also I am just doing the method, but what does it mean geometrically. Why find a line perpendicular, or use a unit vector? Thanks Thomas 



#2
Jun111, 02:32 PM

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The shortest distance from a point to a line is always along the line perpendicular to the given line. That is because the hypotenuse of a right triangle is longer than either leg.
As a result of that, the shortest distance between two skew lines is along the line perpendicular to both given lines. 



#3
Jun111, 02:32 PM

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RGV 



#4
Jun111, 02:39 PM

P: 656

Shortest Distance between 2 skew lines (vectors)How does finding this vector help. For that matter, how does finding an arbitray line between the 2 vectors help Thanks Thomas 



#5
Jun111, 02:45 PM

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#6
Jun111, 03:00 PM

P: 558

Your answer is correct.
Geometrically meaning: imagine the 2 parallel planes that each contains one of the lines. Easier: imagine 2 parallel planes and on each plane draw one line. Connect the 2 lines by a 3rd line AB. In A draw the perpendicular of the 2 planes. The length of the perpendicular is the projection of AB onto the perpendicular of the planes in A (or B, the same). 



#7
Jun111, 06:58 PM

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#8
Jun111, 07:24 PM

P: 656

why would you want to minimize d(s,t)^2 and not d(s,t). am I missing something stupidly ovbious?




#9
Jun111, 11:24 PM

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