Intersection of two lines in 3-space

[Physics345]

Homework Statement

Given l→1=(6,−1,0)+t(3,1,−4) and l→2=(4,0,5)+s(−1,1,5), find the intersection of l→1 and l→2
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Homework Equations

The Attempt at a Solution

I'm fairly certain I did this correctly, but I just thought I'd double check to make sure I have a good understanding.

 

[Mark44]

Homework Statement

Given l→1=(6,−1,0)+t(3,1,−4) and l→2=(4,0,5)+s(−1,1,5), find the intersection of l→1 and l→2

Tip: Use LaTeX to write vectors.
$\vec {l_1}$ is a lot nicer than l→1.
My script for the above: $\vec {l_1}$

.

Homework Equations

The Attempt at a Solution

I'm fairly certain I did this correctly, but I just thought I'd double check to make sure I have a good understanding.

View attachment 222571

Your values for s and t are the same as I got, but you need to check that the lines actually intersect. Just substitute for s and t in your vector equations for the two lines.

 

[Physics345]

Tip: Use LaTeX to write vectors.
$\vec {l_1}$ is a lot nicer than l→1.
My script for the above: $\vec {l_1}$

Your values for s and t are the same as I got, but you need to check that the lines actually intersect. Just substitute for s and t in your vector equations for the two lines.

Considering the lines are skew and the values of (s) and (t) did not satisfy the parametric equations. Isn't that self explanatory? Meaning there is no intersection of the lines? Or could the lines be skew and still intersect?
Anyways, It's always awesome to learn a new way to verify math! So I'm going to give it a shot.

Attempt at verifying the solution:

$\vec {l_1}$ = (6,−1,0)+t(3,1,−4)
= (6,−1,0)+(-1/4)(3,1,−4)
= (6,-1,0)+(-3/4, -1/4, 1)
=(21/4, -5/4, 1 )
$\vec {l_2}$ = (4,0,5)+s(−1,1,5)
=(4,0,5)+(-5/4)(−1,1,5)
=(4,0,5)+(5/4,-5/4,-25/4)
=(5, -5/4, -45/4)
I guess they do intersect at (x=-5/4)
That's cool. I guess the lines can be skew and still intersect. Good to know, my textbook should explain that.
And I have applied your teachings Sensei =) by using LaTex and finding the point of intersection :)

 

[Mark44]

Considering the lines are skew and the values of (s) and (t) did not satisfy the parametric equations. Isn't that self explanatory? Meaning there is no intersection of the lines? Or could the lines be skew and still intersect?

Sorry -- I overlooked that part of your post where you said this. The lines are skew, which means that they don't intersect. It seems my comment led you astray.

Attempt at verifying the solution:

$\vec {l_1}$ = (6,−1,0)+t(3,1,−4)
= (6,−1,0)+(-1/4)(3,1,−4)
= (6,-1,0)+(-3/4, -1/4, 1)
=(21/4, -5/4, 1 )
$\vec {l_2}$ = (4,0,5)+s(−1,1,5)
=(4,0,5)+(-5/4)(−1,1,5)
=(4,0,5)+(5/4,-5/4,-25/4)
=(5, -5/4, -45/4)

It's good to check, but you have mistakes in your work. By substitution values of t and s, you get specific points on the lines. On the first line, with t = -1/4, the point is (21/4, -5/4, 1), which you have. With s = -5/4 on the second line, the point is (21/4, -5/4, -5/4), which is different from what you show. These are different points, so there is no point that is on both lines.

I guess they do intersect at (x=-5/4)

For two lines in space to intersect, all three coordinates have to be the same, not just a single coordinate

 

[Ray Vickson]

Considering the lines are skew and the values of (s) and (t) did not satisfy the parametric equations. Isn't that self explanatory? Meaning there is no intersection of the lines? Or could the lines be skew and still intersect?
Anyways, It's always awesome to learn a new way to verify math! So I'm going to give it a shot.

Attempt at verifying the solution:

$\vec {l_1}$ = (6,−1,0)+t(3,1,−4)
= (6,−1,0)+(-1/4)(3,1,−4)
= (6,-1,0)+(-3/4, -1/4, 1)
=(21/4, -5/4, 1 )
$\vec {l_2}$ = (4,0,5)+s(−1,1,5)
=(4,0,5)+(-5/4)(−1,1,5)
=(4,0,5)+(5/4,-5/4,-25/4)
=(5, -5/4, -45/4)
I guess they do intersect at (x=-5/4)
That's cool. I guess the lines can be skew and still intersect. Good to know, my textbook should explain that.
And I have applied your teachings Sensei =) by using LaTex and finding the point of intersection :)

Of course two skew lines in space can intersect: just take a line in space, then take a point on that line and draw another line through that point but in a different direction. You get two skew lines that definitely intersect. You can do that in any number of dimensions, 2,3,4,5,...,

 

[Physics345]

Sorry -- I overlooked that part of your post where you said this. The lines are skew, which means that they don't intersect. It seems my comment led you astray.It's good to check, but you have mistakes in your work. By substitution values of t and s, you get specific points on the lines. On the first line, with t = -1/4, the point is (21/4, -5/4, 1), which you have. With s = -5/4 on the second line, the point is (21/4, -5/4, -5/4), which is different from what you show. These are different points, so there is no point that is on both lines.

For two lines in space to intersect, all three coordinates have to be the same, not just a single coordinate

I did find it odd, but I didn't want to doubt your experience thanks for clarifying.
On a side note, I should have done the math on paper, I just did it quickly on the pc which for me is more prone to mistakes.

 

[Mark44]

Of course two skew lines in space can intersect

By definition, skew lines don't intersect.
From Merriam Webster dictionary

straight lines that do not intersect and are not in the same plane

There are any number of sites, including Wikipedia, that say the same.

 

[Physics345]

By definition, skew lines don't intersect.
From Merriam Webster dictionary

There are any number of sites, including Wikipedia, that say the same.

Thanks for clarifying that Mark, I was a bit confused since I did a report on this subject with lots of research, which says skew lines do not intersect. Which I thought was wrong considering your two statements. My information came from textbooks I should have been more supportive of my ideals which is nostalgic and makes me think of "Anyone who has never made a mistake has never tried anything new." Albert Einstein.
Here's a picture of the diagrams I created for the report: