Shortest distance between two lines (vector algebra)

[FelixISF]

1. The problem statement, all variables and given/known data
line l1 : x=2 y= -1 + p z= 2p
line l2 : x=-1 + t y=1-3t z=1-2t

Find the shortest (exact) distance between l1 and l2.

2. Relevant equations
That's what I am looking for!


3. The attempt at a solution

Thanks!

[CFDFEAGURU]

You have to show your attempt at a solution before we can help you.

Thanks
Matt

[FelixISF]

I don't see your point. I have tried to solve the problem but didn't manage to do so, cause I can't find the right formulas.

All I am asking for is the information on what formulas might be helpful.

Do you expect me to write some attempt here, even if the wrong equations were used? whats the point?

[CFDFEAGURU]

Yes, you are expected to post your work. If it is wrong then we can help to steer you onto the right path. So yes, write your attempt and indicate where you think you went wrong and then we can help.

That is my point.

Thanks
Matt

[FelixISF]

I tried to use the following equation:

d = sqrt((x1-x2)^2+(y1-y2)^2+(z1-z2)^2), but it doesnt give me the right answer.
whats wrong then ?

thanks :)

[CFDFEAGURU]

The distance formula is,

d = sqrt((x2-x1)^2+(y2-y1)^2+(z2-z1)^2)

Well, the first thing you should do is substitute in you values for the quantity (x2-x1), square that quantity, and then simplify it if you can. Do the same for the other two quantities. Add them together and then take thier square root.

Thanks
Matt

[LCKurtz]

1. The problem statement, all variables and given/known data
line l1 : x=2 y= -1 + p z= 2p
line l2 : x=-1 + t y=1-3t z=1-2t

Find the shortest (exact) distance between l1 and l2.

2. Relevant equations
That's what I am looking for!


3. The attempt at a solution

Thanks!

I will give you a few hints. The first is that you don't want to use distance formulas. From your two lines you can get direction vectors D1 and D2, right? And the cross product N = D1 X D2 would be normal to both lines. From your lines you can also get points P1 and P2 on line1 and line2 and use them to get a vector V across from one line to the other. Can you see that the distance between the lines would be the absolute value of the component of V on N? Try that.

[fawk3s]

d = sqrt((x2-x1)^2+(y2-y1)^2+(z2-z1)^2)
d = sqrt((x1-x2)^2+(y1-y2)^2+(z1-z2)^2)

there's no difference between these 2..

[toomajj]

Can you please explain this a little more? I see that N=D1(cross)D2 is normal to both vectors, hence normal to the plane of the two vectors D1 and D2. Also the vector V we formed from one line to another has to be in the plane of the vectors D1 and D2. Therefore N is normal also to V; now how can V have any component on another vectos normal to it; the component of V on N is zero. Am I right?
Thanks
TJ

[Mark44]

1. The problem statement, all variables and given/known data
line l1 : x=2 y= -1 + p z= 2p
line l2 : x=-1 + t y=1-3t z=1-2t

Find the shortest (exact) distance between l1 and l2.

2. Relevant equations
That's what I am looking for!
The solution doesn't count as a relevant equation.

3. The attempt at a solution

Thanks!

[The Chaz]

d = sqrt((x2-x1)^2+(y2-y1)^2+(z2-z1)^2)
d = sqrt((x1-x2)^2+(y1-y2)^2+(z1-z2)^2)

there's no difference between these 2..

Shhh! It's rude to point out that a post is completely useless. Oh wait...