Find the shortest distance between two lines:

[Simkate]

Given the following two skew lines:
L1: (0, 4, -3) + s(-1, 1, 3)
L2: (1, 2, 5) + t(-3, 2, 5)

Find the shortest distance.



MY WORK::
Cross-product of the lines (-1, 1, 3) X (-3, 2, 5) = (-1, -4, 1) with length 3*sqrt(2)
Vector between the points (0, 4, -3) - (1, 2, 5) = (-1, 2, -8)

Dot product of those results (-1, 2, -8) . (-1, -4, 1) = -15

Remove sign and divide by cross-product length 15 / (3 sqrt(2)) = 5/sqrt(2)




I was wondering if the VECTOR between the points is right or is the reverse
(1,2,5)-(0,4,-3)= (1, -2, 13) ?

 

[Mark44]

Given the following two skew lines:
L1: (0, 4, -3) + s(-1, 1, 3)
L2: (1, 2, 5) + t(-3, 2, 5)

Find the shortest distance.



MY WORK::
Cross-product of the lines (-1, 1, 3) X (-3, 2, 5) = (-1, -4, 1) with length 3*sqrt(2)
Vector between the points (0, 4, -3) - (1, 2, 5) = (-1, 2, -8)

Dot product of those results (-1, 2, -8) . (-1, -4, 1) = -15

Remove sign and divide by cross-product length 15 / (3 sqrt(2)) = 5/sqrt(2)




I was wondering if the VECTOR between the points is right or is the reverse
(1,2,5)-(0,4,-3)= (1, -2, 13) ?

The two lines can be thought of as lying in two parallel planes. The cross product of the directions of the two lines gives you a vector that is perpendicular to these planes.

Your first vector between the two points on the lines, <-1, 2, -8> is correct, but the opposite, <1, -2, 8> would also work. Was <1, -2, 13> a typo?

What you want is the projection of the vector <1, -2, 8> in the direction of the vector <-1, 4, 1> (I didn't check your cross-product work). That will give you the shortest distance between the two lines.