determine the smallest distance between the lines: used critical points.

eckimz

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Jun 22, 2018
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Here is my work, I think its done, although, my professor sent us a list without answers, as usual haha.

It gives 2 lines: π‘Ÿβƒ—1 = 〈3 βˆ’ 𝑑; 4 + 4𝑑; 1 + 2𝑑βŒͺ and π‘Ÿβƒ—2 = βŒ©π‘‘; 3; 2𝑑βŒͺ.

First I determined that the second line I'm using the parameter (s) instead of t.

After that, I wrote my distance as d=sqr (3-t-s)Β²+(4+4t-3)Β²+(2+2t-2s)Β²
I changed to dΒ², cause it will still be the smaller distance.

Ok, 2n I did the partials for d/dt=0 and d/ds=0

3rd n 4th are the answers for dt, dtt, dts, ds, dss, dst.

5th I have done the system, used gauss. I've checked it all on symbolab.

Got t and s. t=-1/24 and s=11/8.

Applied s and t on dΒ².

dΒ²= [(3-t-s)Β²+(4+4t-3)Β²+(2+2t-2s)Β²]
dΒ²= [(3-(-1/24)-(11/8))Β²+(4+4(-1/24)-3)Β²+(2+2(-1/24)-2(11/8))Β²

Got, 109/6 u.d.

Tested by D=| dss dst | and gave me 420-36= 384. So I have that D>0 and Dtt >0, that should be exactly the point of the smaller distance at relative minimum.
| dts dtt |

Does anyone see any flaws? Thanks.


(some notes in my language, nothing important, just saying what I'm doing so I can study the steps, just like I did here in english)
 

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eckimz

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Jun 22, 2018
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Ive found this aswell d(r1,r2)=|(v1,v2, A1A2)/|V1XV2| on youtube, but I have no idea how to do by analytical geometry (I should haha, but I don't), I can't recall, so Id rather try this way, maybe its wrong tho, idk, im pretty much confused after learning this formula haha
 

pka

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Jan 29, 2005
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Here is my work, I think its done, although, my professor sent us a list without answers, as usual haha.
It gives 2 lines: π‘Ÿβƒ—1 = 〈3 βˆ’ 𝑑; 4 + 4𝑑; 1 + 2𝑑βŒͺ and π‘Ÿβƒ—2 = βŒ©π‘‘; 3; 2𝑑βŒͺ.
Comment: I find your images so absolutely messy that I don't even pretend to read them.
I assume that you can prove that these are skew lines; do not intersect and are not parallel.
Given two skew lines: \(\displaystyle \ell_1: P+t\vec{D}~\&~\ell_2: Q+s\vec{E}\)
The distance between the lines is \(\displaystyle d({\ell _1},{\ell _2}) = \frac{{\left| {\overrightarrow {PQ} \cdot (\vec{D} \times \vec{E})} \right|}}{{\left\| {\vec{D} \times \vec{E}} \right\|}}\)
 
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