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(Vectors) Proving 2 non-parellel lines meet in a unique point

I've tried to prove that two non-parellel lines meet in a unique point, but my answer differs from the proof given in the markscheme and I was hoping someone could tell me if my way is valid or if I've made a mistake somewhere.

My proof:

Let L1 and L2 be two non-parellel lines.
L1=a+λb[br]L2=c+μd L_1 = {a + \lambda b}[br]L_2 = {c + \mu d}
b and d are non-zero and b is not a scalar multiple of d

Suppose the lines intersect at points x and y. Then,
x=a+λ1b=c+μ1d[br]andy=a+λ2b=c+μ2d[br][br]Now,x−y=(λ1−λ2)b=(μ1−μ2)d[br] x =a + \lambda_1 b = c + \mu_1 d[br]and y = a + \lambda_2 b = c + \mu_2 d[br][br]Now, x-y = (\lambda_1 - \lambda_2)b = (\mu_1 - \mu_2)d[br]

As b and d are non-zero and b is not a scalar multiple of d
λ1−λ2=0[br]μ1−μ2=0[br][br]So,λ1=λ2[br][br]Therefore,x=a+λ1b[br]=a+λ2b[br]=y[br][br] \lambda_1 - \lambda_2 = 0 [br] \mu_1 - \mu_2 = 0[br][br]So, \lambda_1 = \lambda_2[br][br]Therefore, x = a + \lambda_1 b[br] = a + \lambda_2 b[br] =y[br][br]
x = y so the 2 non-parellel lines meet in a unique point x.

Apologies for the not-so-good latex and the unbolded vectors, and thanks for taking the time to read this!
I'm not entirely sure about your methods validity but the way I have been taught is to get two vector equations:

r1=(7+3y)i + (-3-2y)j + (3+y)k
r2=(7-2u)i + (u-2)j + (4-u)k

put the coefficients of i and j equal to one another:

i: 7+3y = 7-2u
j: -3-2y = u-2

solve to find y and u

find k using both y and u, should both k values be equal the lines intersect


Does that help?
Numbers are just a random example I had in my book
y = lambda
(edited 10 years ago)
Reply 2
Avatar for Jooooshy
Jooooshy
17
I was extremely confused by your use of x and y.. I'm now guessing they are both points of intersection!

^Or because I can't read..^

You've already assumed that the lines intersect, and then gone on to prove that the intersection is unique for non-parallel lines.

I don't know if you're meant to be proving that they intersect or just that the point of intersection is unique.. But I think you fulfil the second!
(edited 10 years ago)
Reply 3
Avatar for -Ace
-Ace
OP
Original post by InfernoxCJC
I'm not entirely sure about your methods validity but the way I have been taught is to get two vector equations:

r1=(7+3y)i + (-3-2y)j + (3+y)k
r2=(7-2u)i + (u-2)j + (4-u)k

put the coefficients of i and j equal to one another:

i: 7+3y = 7-2u
j: -3-2y = u-2

solve to find y and u

find k using both y and u, should both k values be equal the lines intersect


Does that help?
Numbers are just a random example I had in my book
y = lambda


Hmm, I suppose I could use the idea that there is a unique solution for the scalars to show that the point of intersection of 2 lines is unique. I believe this is the argument used in the markscheme. Cheers for the help :smile:
Reply 4
Avatar for -Ace
-Ace
OP
Original post by Jooooshy
I was extremely confused by your use of x and y.. I'm now guessing they are both points of intersection!

^Or because I can't read..^

You've already assumed that the lines intersect, and then gone on to prove that the intersection is unique for non-parallel lines.

I don't know if you're meant to be proving that they intersect or just that the point of intersection is unique.. But I think you fulfil the second!


Sorry for that confusion!

I assumed that I was only supposed to prove that the point of intersection is unique, but I could have interpreted the question incorrectly, which is

"Prove Proposition 1.2 of the lectures: any two non-parallel lines meet in a unique point."
Would you suggest I first prove that two non-parellel lines do in fact meet?

Thanks for the help- it's much appreciated.

EDIT: forgot to mention before, I'm working with R2 \mathbb{R}^2 at the moment, not R3 \mathbb{R}^3 . Would this make a difference? What I mean is, could I just state that any 2 non-parellel lines will, in fact, intersect, without proving it?
(edited 10 years ago)
Original post by -Ace
Hmm, I suppose I could use the idea that there is a unique solution for the scalars to show that the point of intersection of 2 lines is unique. I believe this is the argument used in the markscheme. Cheers for the help :smile:



No worries, that is what we are all here for! \0/
Good luck
Reply 6
Avatar for james22
james22
16
Original post by -Ace
Sorry for that confusion!

I assumed that I was only supposed to prove that the point of intersection is unique, but I could have interpreted the question incorrectly, which is

"Prove Proposition 1.2 of the lectures: any two non-parallel lines meet in a unique point."
Would you suggest I first prove that two non-parellel lines do in fact meet?

Thanks for the help- it's much appreciated.


You have to prove that they do meet.
Reply 7
Avatar for -Ace
-Ace
OP
Original post by james22
You have to prove that they do meet.


forgot to mention before, I'm working with R2 \mathbb{R}^2 at the moment, not R3 \mathbb{R}^3 . Would this make a difference? What I mean is, could I just state that any 2 non-parellel lines will in fact intersect, without proving it in this case?
Reply 8
Avatar for Jooooshy
Jooooshy
17
Original post by -Ace
forgot to mention before, I'm working with R2 \mathbb{R}^2 at the moment, not R3 \mathbb{R}^3 . Would this make a difference? What I mean is, could I just state that any 2 non-parellel lines will in fact intersect, without proving it in this case?

You're probably going to be required to prove that they do, in fact, meet! I'm not sure if two non-parallel lines in 3D space definitely intersect..
Reply 9
Avatar for james22
james22
16
Original post by -Ace
forgot to mention before, I'm working with R2 \mathbb{R}^2 at the moment, not R3 \mathbb{R}^3 . Would this make a difference? What I mean is, could I just state that any 2 non-parellel lines will in fact intersect, without proving it in this case?


No, you need to prove it. The proof is not hard, just solve the 2 equations.
Reply 10
Avatar for BlueSam3
BlueSam3
17
Original post by -Ace
forgot to mention before, I'm working with R2 \mathbb{R}^2 at the moment, not R3 \mathbb{R}^3 . Would this make a difference? What I mean is, could I just state that any 2 non-parellel lines will in fact intersect, without proving it in this case?


It makes the fairly enormous difference that it's not true in R3\mathbb{R}^3. Pick up two pencils, and you will probably see this immediately.




Anyway, for your proof, your notation is rather inconsistent (lines are sets, but you have equated them to single points, and haven't defined anything). As mentioned above, you haven't considered the case where the two lines do not intersect. Additionally, you have assumed your conclusion. So far as I can tell, you obtain λ1−λ2=0\lambda_1-\lambda_2=0 from assuming that x−y=0x-y=0, which is exactly what you're trying to show. You also haven't considered lines that are vertical in your coordinates.



To actually do the question, you're on the right line from assuming that there are two points of intersection, but there are significantly easier ways of doing it. Think about what it means for something to be a "line", and quickly prove that two lines that share two common points must, in fact, be the same line. For the existence part, probably the easiest way is to define your origin to be on one of the lines, with a basis vector pointing along that line (this just avoids having to remember to add constants to the end of everything), and show that the other line is at some point above, and at some point below, that zero line, and hence must contain some point that is on the zero line (if you really want to, you could do this by parameterising it and invoking the mean value theorem, but that's probably unnecessary). At this point, you're done.
Reply 11
Avatar for -Ace
-Ace
OP
Original post by BlueSam3
It makes the fairly enormous difference that it's not true in R3\mathbb{R}^3. Pick up two pencils, and you will probably see this immediately.

I think I phrased my question poorly. I thought that because I was in R2\mathbb{R}^2 and not R3\mathbb{R}^3 I would be allowed to just state it. (I'm aware that non-parallel lines don't always meet in R3\mathbb{R}^3). I do realise now that I can't just state it and that it needs proof. I may not have actually explained this any more clearly than earlier- if that's the case, I'm sorry!



Anyway, for your proof, your notation is rather inconsistent (lines are sets, but you have equated them to single points, and haven't defined anything). As mentioned above, you haven't considered the case where the two lines do not intersect. Additionally, you have assumed your conclusion. So far as I can tell, you obtain λ1−λ2=0\lambda_1-\lambda_2=0 from assuming that x−y=0x-y=0, which is exactly what you're trying to show.


Thanks for pointing out the issue with my notation- I'll be a lot more careful in the future.

As for obtaining λ1−λ2=0\lambda_1-\lambda_2=0..
I got that from (λ1−λ2)b=(μ1−μ2)d(\lambda_1-\lambda_2)b = (\mu_1-\mu_2)d
I thought that because b and d aren't scalar multiples of each other (and they're non-zero), the only way that the equation could be true was if the scalars in front of the b and d were zero. Is this incorrect?


You also haven't considered lines that are vertical in your coordinates.

I know I'm being really stupid here, but I'm not sure how this will change things


I will have a think about the last paragraph.

Your post has been really helpful. Thank you :smile:
Reply 12
Avatar for -Ace
-Ace
OP
Original post by Jooooshy
You're probably going to be required to prove that they do, in fact, meet! I'm not sure if two non-parallel lines in 3D space definitely intersect..


Original post by james22
No, you need to prove it. The proof is not hard, just solve the 2 equations.


Cheers for pointing this out to me :smile: I'll have to be a lot more careful when I read questions in the future!
The fact that this is true in the plane, but not 3 (or more) dimensions implies that to get the existence of a point where they do intersect, you will need to use the fact there's only 2 coordinates at one point (i.e. solve some simultaneous equations in terms of the 2 coordinates).

Your proof only shows uniqueness. For this, it has the correct idea, but as people said before it isn't very neat and there is inconsistencies in notation.
Reply 14
Avatar for BlueSam3
BlueSam3
17
Original post by -Ace
I think I phrased my question poorly. I thought that because I was in R2\mathbb{R}^2 and not R3\mathbb{R}^3 I would be allowed to just state it. (I'm aware that non-parallel lines don't always meet in R3\mathbb{R}^3). I do realise now that I can't just state it and that it needs proof. I may not have actually explained this any more clearly than earlier- if that's the case, I'm sorry!


Ah, ok, that makes more sense.


Thanks for pointing out the issue with my notation- I'll be a lot more careful in the future.

As for obtaining λ1−λ2=0\lambda_1-\lambda_2=0..
I got that from (λ1−λ2)b=(μ1−μ2)d(\lambda_1-\lambda_2)b = (\mu_1-\mu_2)d
I thought that because b and d aren't scalar multiples of each other (and they're non-zero), the only way that the equation could be true was if the scalars in front of the b and d were zero. Is this incorrect?


No, that's OK.


I know I'm being really stupid here, but I'm not sure how this will change things

Not all lines can be expressed in the way you've expressed them. Specifically, vertical lines can't. You could just consider them separately, or you could do some kind of coordinate change so that you can express your two lines in your given form.


I will have a think about the last paragraph.

Your post has been really helpful. Thank you :smile:


No problem.
It's parallel not parellel. I'm not even sure how you made this mistake repeatedly because I had to override the spell-checker to type it!

Here's what can happen in three dimensions.

SKew.jpg24.5KB

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