Finding the Shortest Distance between Skew Lines

[nameVoid]

when trying to find the distance between 2 skew lines i see how we can take the a vector between points on each and the cross product of the direction vectors to find the distance but is this looks only to be the distance from one line to a particular point on the other

 

[nameVoid]

my question is if this is the shortest distance between the lines it does not appear to me that it is when using arbritrary points on each line

 

[verty]

I see what you mean, you are correct, it won't be the shortest distance. The vector must be perpendicular to both lines to give the shortest distance. Can you think of a way to find a vector perpendicular to both lines?

 

[Solkar]

when trying to find the distance between 2 skew lines i see how we can take the a vector between points on each and the cross product of the direction vectors to find the distance but is this looks only to be the distance from one line to a particular point on the other

It's not right ahead the minimal distance d between the lines g_1, g_2 you find by selecting those arbitrary points p_1, p_2 on each, you do this for getting representations of the parallel planes to be constructed:

A plane P can be described in many ways, one of them is
- a normal n
- and a fixed value λ, for which

<n,p> = λ (¹)

holds for all p ∈ P:

Given one point p in the plane for which <n,p> yields a certain value λ(p)
Now consider a second point q ∈ P:

The vector
r := q – p
is parallel to the plane, thus it's orthogonal to n
<n, r> = 0

With
λ(q) := <n,q> = <n, p + r> = <n, p> + <n, r>,
the last "=" due to bi-linearity of <.,.>
one gets
λ(q) = <n,q> = <n, p> + 0 = <n,p> = λ(p).

So the, seemingly arbitrary, λ(p), we found is actually the plane-specific λ we've been looking for.

---

Because finding the minimal distance d between the lines g_1 and g_2 is equivalent to finding the distance of the constructed parallel planes p_1, p_2 , one gets for d
d = abs(λ_1 - λ_2),¹For shortness sake, I do not distinguish points from vectors here

 

[HallsofIvy]

The "shortest distance between two lines" (also know as the distance between the two lines. When we have regions that have more than one line between them, by "the distance" between them we typically mean the shortest distance between them) is measured along the line perpendicular to both lines. And that means that the line must line in the unique plane perpendicular to both.

As an example, suppose the lines are given by the equations x= t+ 3, y= 2t- 1, z= 3t and x= 2s+1, y= s- 2, z= s+ 4. Any plane perpendicular to the first line is of the form x+ 2y+ 3z= A, for some number, A, and any plane perpendicular to the second line is of the form 2x+ y+ z= B, for some number B. Find A and B such that those are the same plane and find the distance between the points where the two lines intersect each other.

 

[nameVoid]

so the cross product between 2 vectors is always perpendicular to both vectors but what about the length of this vector

 

[HallsofIvy]

No, the shortest distance between two points is a number. It is measured along a straight line. And the topic here is the shortest distance between to skew lines, not two points.

 

[nameVoid]

it would be obviously the shortest distance between two parrallel linesbut considering two skew lines where the distance varies

 

[haruspex]

so the cross product between 2 vectors is always perpendicular to both vectors but what about the length of this vector

The length of the cross-product is not necessarily what you want.
You need to start by specifying the form in which the two lines are defined. Something like r₀+λ₀s₀, r₁+λ₁s₁?

 

[HallsofIvy]

so the cross product between 2 vectors is always perpendicular to both vectors but what about the length of this vector

The length of the cross product is not relevant. If you are given the two lines as <x, y, z>= <a, b, c>+ <d, e, f> t and <x, y, z>= <p, q, r>+ <u, v, w> s, then <a, b, c> and <u, v, w> are vectors in the direction of the two lines. Their cross product is perpendicular to both and so perpendicular to the plane you want to find.

 

[LCKurtz]

Let's say you divide the cross product of the two direction vectors by its length to make at a unit vector $\hat n$ perpendicular to both lines. Then if $P_0,\, P_1$ are points, one on each line, and you let $\vec V=\overline{P_0P_1}$ be the vector displacement between the points, then the distance $d$ between the lines is$$
d =|\vec V \times \hat n|$$The advantage of this formula is that you are normally given two points and two direction vectors, and you don't need anything else.

[Edit] Typo alert, that should be $d =|\vec V \cdot \hat n|$

 

[haruspex]

If you are given the two lines as <x, y, z>= <a, b, c>+ <d, e, f> t and <x, y, z>= <p, q, r>+ <u, v, w> s, then <a, b, c> and <u, v, w> are vectors in the direction of the two lines.

<d, e, f> and <u, v, w>

 

[nameVoid]

if v=p0p1 then d=|v|cost=v.(axb)/|axb| now this is obviously the length of v is not going to be constant with arbtrary points

 

[LCKurtz]

if v=p0p1 then d=|v|cost=v.(axb)/|axb| now this is obviously the length of v is not going to be constant with arbtrary points

Please quote the post to which you are responding. It it's mine there was a typo. Still, what is your point?

 

[haruspex]

if v=p0p1 then d=|v|cost=v.(axb)/|axb| now this is obviously the length of v is not going to be constant with arbtrary points

As far as I can see, you have still not given an explicit statement of the vector form in which the two lines are specified. Is it a pair of points on each line? One point plus a direction vector for each line? Something else?

 

[nameVoid]

well we have p0 and p1 as points on each line and vectors a and b as directions

 

[haruspex]

well we have p0 and p1 as points on each line and vectors a and b as directions

OK, so you can use the formulae LCKurtz gave you. But LCK was using P₀, P₁ as labels, not vectors, hence gave the vector form of the line joining them as $\overline{ P_0P_1 }$. If you have p₀, p₁ as vectors, what will you write instead?

now this is obviously the length of v is not going to be constant with arbitrary points

Why not? See what happens if you replace p₀ by another point on that line, p₀+at.

 

[nameVoid]

length of v is not constant where v is the vector between points d=lvlcost

 

[nameVoid]

the example in the book takes parameter 0 for both lines but again i do not understand how this distance is the shortest distance considering that d is proportional to v

 

[micromass]

nameVoid, could you please quote the post you're replying to? It's a lot easier for us. Just hit "quote" on the right-hand side.

 

[nameVoid]

lvlcost

 

[LCKurtz]

the example in the book takes parameter 0 for both lines but again i do not understand how this distance is the shortest distance considering that d is proportional to v

It is not proportional to the length of $v$ because there is a $\cos t$ in the formula.

Look at the picture. It doesn't matter which vector across between the lines you use. They all have the same projection on the normal direction (dashed) which gives the distance between the lines.

 

[nameVoid]

v.(axb)=lvllaxblcost
d=lvlcost=v.(axb)/laxbl

 

[LCKurtz]

v.(axb)=lvllaxblcost
d=lvlcost=v.(axb)/laxbl

I give up. You won't quote the post to which you are replying and you just spout formulas as if that is some kind of conversation. Not with me. I'm gone.

 

[nameVoid]

ok you are telling me that the scalar projection of v onto axb is not equal to lvlcost

 

[haruspex]

v.(axb)=lvllaxblcost
d=lvlcost=v.(axb)/laxbl

Using cos t isn't going to tell you anything because you don't know what t is. Stick to the vectors.
1. Write v as a function of p₀ and p₁
2. Consider an alternative definition of the first line using p₀'=p₀+λa instead of p₀. Write the corresponding v'.
3. Expand v'.(axb) and compare it with v.(axb)

 

[nameVoid]

d=lvlcost=v.(axb)/laxbl
if you are trying to tell me that the distance is not affected by the length of v then you are telling me that v along axb is not equal to lvlcost

 

[nameVoid]

ld=lvlcost

 

[nameVoid]

ldl=lvlcost

 

[haruspex]

d=lvlcost=v.(axb)/laxbl
if you are trying to tell me that the distance is not affected by the length of v then you are telling me that v along axb is not equal to lvlcost

No, I'm telling you that if you replace, say, p₀ by a different point on the line p₀+aλ then the vector v defined by p₀ and p₁ will change in both magnitude and direction. It will change in such a way that d does not change. But don't take my word for it, follow the steps I've laid out for you and see for yourself.
I shall post no more on this thread until you have posted an attempt to do so.

 

[nameVoid]

ok so it seems to be consistant so there is only one lunit vectorl which is perpendicular to both lines and therefore the shortest distance must be in this direction i believe that this can be understood by taking all vectors to the origin and then raising one direction vector away from the other you will see that the shortest distance is along the cross product vector in a way its raising the point of intersection and then meaduring the distance