Find the Shortest Distance Between Two Lines: L1 and L2

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To find the shortest distance between the lines L1 and L2, the approach involves constructing a function for the squared distance between the two lines, defined as the norm squared of the distance vector function d(s,t) = r(t) - u(s). This method simplifies the process since minimizing the squared distance is easier than minimizing the distance itself. The function is dependent on two variables, t and s, and extrema can be found using standard methods without needing a second derivative test, as the squared distance has only one minimum. This technique provides a clear pathway to determine the shortest distance effectively. Understanding this approach is crucial for solving similar problems in vector calculus.
brad sue
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Hi,
Can someone help me with this:

What is the shortest distance between the two lines:

L1: r= (1,0,0) +t( 2,3,4)
L2 u= (2,1,0) +s(1,2,0)

Thank you very much for your help

B
 
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There is probably a shorter way in terms of vectors and stuff, but here's a way:

Build the function "distance squared btw the lines", defined as the norm squared of the distance vector function

\vec{d}(s,t) = \vec{r}(t) - \vec{u}(s)

The reason we work with the function distance squared instead of the function distance itself is that the distance is minimum when the distance squared is minimum, and as you'll see, it is much easier to find when d² is minimum then when d is minimum.

This function is a function of the two variables t and s. Use the usual method to find the extrema of the function. (no need to use the second derivative test to determine the nature of each extrema; yuo know the function distance quared as no maxima and only one minimum :wink:.
 
quasar987 said:
There is probably a shorter way in terms of vectors and stuff, but here's a way:

Build the function "distance squared btw the lines", defined as the norm squared of the distance vector function

\vec{d}(s,t) = \vec{r}(t) - \vec{u}(s)

The reason we work with the function distance squared instead of the function distance itself is that the distance is minimum when the distance squared is minimum, and as you'll see, it is much easier to find when d² is minimum then when d is minimum.

This function is a function of the two variables t and s. Use the usual method to find the extrema of the function. (no need to use the second derivative test to determine the nature of each extrema; yuo know the function distance quared as no maxima and only one minimum :wink:.

OK thank you very much
 
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