[Linear Algebra] - Find the shortest distance d between two lines

[CoolMan2017]

Let L1 be the line passing through the point P1=(−2,−11,9) with direction vector d2=[0,2,−2]T, and let L2 be the line passing through the point P2=(−2,−1,11) with direction vector d2=[−1,0,−1]T Find the shortest distance d between these two lines, and find a point Q1 on L1 and a point Q2 on L2 so that d(Q1,Q2)=d. Use the square root symbol to get the exact value

 

[MarkFL]

Hello and welcome to MHB! :D

We ask that our users show their progress (work thus far or thoughts on how to begin) when posting questions. This way our helpers can see where you are stuck or may be going astray and will be able to post the best help possible without potentially making a suggestion which you have already tried, which would waste your time and that of the helper.

Can you post what you have done so far?

 

[HallsofIvy]

Looks pretty straight forward to me- though, frankly, I don't see what it has to do with "Linear Algebra"- just basic Calculus.

Line L1 can be written in a vector equation as (x, y, z)= (-2, -11, 9)+ (0, 2, -2)t= (-2, -11+ 2t, 9- 2t) and L2 as (x, y, z)= (-2, -1, 11)+ (-1, 0, -1)s= (-2- s, -1, 11- s). The distance between any two points, one on the first line, the other on the second is $D= \sqrt{(-2- (-2-s))^2+ (-11+ 2t- (-1))^2+ (9- 2t- (11- s)^2}= \sqrt{s^2+ (-10+ 2t)^2+ (s- 2t- 2)^2}$.

Minimizing that distance is the same as minimizing its square:
$D^2= s^2+ (-10+ 2t)^2+ (s- 2t- 2)^2$.

Set the derivatives with respect to s and t to 0 and solve the two equations for s and t.