Determining if two lines in R3 Intersect

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rtareen

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How to prove that two lines in 3D are not parallel and do not intersect; also, how to find the distance between them?

Problem Given two lines: $$(l_1)=(x,y,z)=(3,-1,2)+t(1,1,0)$$ $$(l_2)=(x,y,z)=(0,5,2)+t(1,-2,1)$$ Explain why these two lines are not parallel and why they do not intersect each other. Also, find ...
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I am trying to teach myself Calc 3 and it’s hard because I have no teacher. So I look up stuff online and if I still don’t understand it I ask for help here.

Here is a link to a question asked by someone on stackexchange. The problem is to show the lines are not parallel, and then to show they do not intersect. Then it asks to find the minimum distance between the two lines.

Determining if two lines in R3 are parallel is easy. If they are parallel, then one of their slope vectors should be a scalar multiple of the other. In the linked problem, the two lines are not parallel.

What I don’t actually understand is determining if the lines intersect or not. What is the general method to determine if two lines intersect, and what is the intuition behind it?

I know you must start by setting the two lines equal to each other and solve the resulting system of equations. In this case we get:

(1) 3 + t = s
(2) -1 + t = 5 - 2s
(3) 0 = 2 + s

The top answerer to the question substitutes s=0 into the first two equations giving t = -3 and t = 6, and he says this is a contradiction. I don’t think this is how you solve a system of equations, but I could be wrong. How do we know it’s supposed to intersect at s = 0 and not s = 1 or s = 2.

And how do we know that if the system of equations has no solution then the lines do not intersect?



Finally, there is the problem of finding the minimum distance between the lines. First he gives some complicated distance formula ( I have no clue what the formula is, it’s not in my textbook), and then he gives another method by finding the projection of Q1Q2 onto n, which is perpendicular to both lines. How can a vector be perpendicular to a line, I think he means perpendicular to the slope vectors of the lines. Anyways, I don’t understand how to find the minimum distance at all.
 
First, I wouldn't call them "slope vectors"; I'd say "direction vectors". (And, yes, a vector is perpendicular to a line if it is perpendicular (orthogonal) to the direction vector for that line.)

But what this person did was to see that the third equation tells you that s=0; that isn't a random guess. This won't always be so easy; in general, you just try to solve the system of equations, eliminating one variable at a time, say, and see if there is a solution. It just happens that in this case, the third equation already had only one variable.

As for why this implies the lines do not intersect, that's because if they did intersect, there would be a solution to the system. The system of equations is written as a way to say "for some values of s and t, the values of x, y, and z are equal on both lines"; that's equivalent to saying, "the lines intersect at that point (x,y,z), which is at those values of s and t".

As for finding the distance between the lines, there are various ways to do it; what does your textbook say? Does it give any examples of this at all, or any sort of formula? It will be very helpful if you can show us what you are being taught. (Use the book's index if necessary.)
 
Dr. Peterson, I looked through the index and I found that there is nothing in my book about the minimum distance between two lines. It looks like this kind of problem was skipped. I cannot find it in either of the books I am
using.

I guess this means it must not be important, or do you think that it is?
 
Let [MATH]P[/MATH] and [MATH]Q[/MATH] be points on the two lines and [MATH]\vec V = \overline{PQ}[/MATH] be a vector from one line to the other. Let [MATH]\vec N = \vec D_1 \times \vec D_2[/MATH] be the cross product of direction vectors for the two lines, and [MATH]\hat N[/MATH] be a unit vector in the direction of [MATH]\vec N[/MATH]. Then the distance between the non-parallel lines is [MATH]d = |\vec V \cdot \hat N|[/MATH].
 
LCKurtz said:
Let [MATH]P[/MATH] and [MATH]Q[/MATH] be points on the two lines and [MATH]\vec V = \overline{PQ}[/MATH] be a vector from one line to the other. Let [MATH]\vec N = \vec D_1 \times \vec D_2[/MATH] be the cross product of direction vectors for the two lines, and [MATH]\hat N[/MATH] be a unit vector in the direction of [MATH]\vec N[/MATH]. Then the distance between the non-parallel lines is [MATH]d = |\vec V \cdot \hat N|[/MATH].
Click to expand...

Does this give the minimum distance between the lines?
 
rtareen said:
Does this give the minimum distance between the lines?
Click to expand...
If you do not know about skew lines that is on p43 of Smith&Minton.
Given \(\displaystyle \ell_1(t): P+t\vec{D}~\&~\ell_2(t): Q+t\vec{E}\) are two skew lines Then the distance between the lines is
\(\displaystyle \mathcal{D}(\ell_1,\ell_2)=\dfrac{|\overrightarrow {PQ}\cdot(\vec{D}\times\vec{E})|}{\|\vec{D}\times\vec{E}\|}\).
The fact is, given any pair of skew lines then there is a unique line segment that is perpendicular to each line of the pair.
The length of that line segment is the distance between the skew lines.
On page 34 of S&M is the rule for finding is distance from a point to a line.
 
rtareen said:
Determining if two lines in R3 are parallel is easy. If they are parallel, then one of their slope vectors should be a scalar multiple of the other. In the linked problem, the two lines are not parallel.
What I don’t actually understand is determining if the lines intersect or not. What is the general method to determine if two lines intersect, and what is the intuition behind it?
Click to expand...
Given that each of \(\displaystyle \ell_1(t): P+t\vec{D}~\&~\ell_2(t): Q+t\vec{E}\) is a line then:
1) If Given \(\displaystyle \vec{D} ||\vec{E}\) (i.e. parallel) then the lines are parallel.
2) If Given \(\displaystyle \vec{D} \cancel {||}\vec{E}\) (i.e. not parallel) then the lines are not parallel.
 
pka said:
If you do not know about skew lines that is on p43 of Smith&Minton.
Given \(\displaystyle \ell_1(t): P+t\vec{D}~\&~\ell_2(t): Q+t\vec{E}\) are two skew lines Then the distance between the lines is
\(\displaystyle \mathcal{D}(\ell_1,\ell_2)=\dfrac{|\overrightarrow {PQ}\cdot(\vec{D}\times\vec{E})|}{\|\vec{D}\times\vec{E}\|}\).
The fact is, given any pair of skew lines then there is a unique line segment that is perpendicular to each line of the pair.
The length of that line segment is the distance between the skew lines.
On page 34 of S&M is the rule for finding is distance from a point to a line.
Click to expand...

I think I have a different version of smith and Minton than you. My book starts at page 688. View attachment 15877
 
Here is a practice problem for you. Are these two lines parallel, intersecting, or skew?
\(\displaystyle \ell_1 (t) = \left\{ \begin{array}{l}x(t) = 1 - 2t\\y(t) = 2 t\\z(t) = 5-t\end{array} \right.\) \(\displaystyle \ell_2 (t) = \left\{ \begin{array}{l}x(t) = 3 + 2s\\y(t) = -2\\z(t) = 3+2s\end{array} \right.\)
 
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