How to determine whether two lines are parallel or perpendicular to eachother

[Nickg140143]

Homework Statement

Let l₁:x=3+t, y=1-t, z=2t and l₂:x=-1+s, y=2s, z=1+kt (not sure if this one is just a typo, in which t should actually be s, or whether this is fundamental to the problem) be two lines in R³.
a) Find all value(s) of k, (if any) for which l₁ and l₂ are parallel. If not possible then explain why.
a) Find all value(s) of k, (if any) for which l₁ and l₂ are perpendicular. If not possible then explain why.

Homework Equations

These two lines are given in parametric forms, from which I should be able to get the directional vector, and the point associated with each one.

The Attempt at a Solution

After noticing that I could grab a little information about each line, I decided to take a look at their directional vectors.
l_1: <1,-1,2>
l_2: <1,2,k>
I'm assuming I'm supposed to be solving for k in this problem?

after looking at these, I can't see them being proportional to each other, so I don't think they are parallel, granted, I'm not sure how to necessarily show this mathematically.

My biggest problem is finding all k in which these may be perpendicular, the only idea I've been playing around with would be that if I took the Dot Product between the directional vectors of line 1 and line 2, it would need to equal 0 in order for these two lines to be perpendicular, but I'm not sure if this is the right track of thinking, since I'm not sure what to do about k.

Is my thinking somewhere along the right lines, or have I made a horrible mistake?

Any help would be great.

 

[flyingpig]

Yes, they are orthogonal if the dot product is 0 and parallel if the dot product is 1

 

[micromass]

parallel if the dot product is 1

Not true at all. They are parallel if the directional vectors are a nonzero multiple of each other.

To the OP: your track of thinking is great, continue like that.

 

[flyingpig]

Not true at all. They are parallel if the directional vectors are a nonzero multiple of each other.

To the OP: your track of thinking is great, continue like that.

Oh wait I meant the cosine between them is 1.

 

[micromass]

Oh wait I meant the cosine between them is 1.

Not true either. It could be -1 as well.

 

[flyingpig]

Not true either. It could be -1 as well.

But that's anti-parallel

 

[micromass]

But that's anti-parallel

There is no such thing.

 

[flyingpig]

There is no such thing.

Yeah when you are 180 degrees, i mean the vectors, not the lines...

 

[icystrike]

You can take the absolute value of its dot product and the do what is stated by the replies above.Otherwise, you can easily find a vector that is perpendicular to the first vector and "dot" with the second.

 

[micromass]

Yeah when you are 180 degrees, i mean the vectors, not the lines...

If vectors make an angle of 180 degrees, then they are parallel.

 

[flyingpig]

If vectors make an angle of 180 degrees, then they are parallel.

BUt they point in opposite directions, so they are ANTI- parallel!

 

[Nickg140143]

Alright, good to know my approach isn't a bust. So I'll try taking the dot product and solve for k.

<1,-1,2> . <1,2,k> = 0
1(1)+2(-1)+2(k) = 0
1-2+2k = 0
-1+2k = 0
2k = 1
k = \frac{1}{2}

how does this look?

 

[micromass]

Alright, good to know my approach isn't a bust. So I'll try taking the dot product and solve for k.

<1,-1,2> . <1,2,k> = 0
1(1)+2(-1)+2(k) = 0
1-2+2k = 0
-1+2k = 0
2k = 1
k = \frac{1}{2}

how does this look?

That's good!

BUt they point in opposite directions, so they are ANTI- parallel!

No, they're not.

 

[Nickg140143]

Alright, thanks for the tips. Just one last thing, I wanted to know if my statement about these lines not being parallel is valid.

Since I can't see the directional vector of l1 and l2 being proportional for any value of k, they can't be parallel. This is because if their proportional, that would mean that they both point in the same direction, but have different magnitudes correct?

And also, what if these lines WERE parallel for some value of k, is there a way to mathematically solve for these values of k?

Thanks for all the help so far everyone.

 

[micromass]

Alright, thanks for the tips. Just one last thing, I wanted to know if my statement about these lines not being parallel is valid.

Since I can't see the directional vector of l1 and l2 being proportional for any value of k, they can't be parallel. This is because if their proportional, that would mean that they both point in the same direction, but have different magnitudes correct?

And also, what if these lines WERE parallel for some value of k, is there a way to mathematically solve for these values of k?

Thanks for all the help so far everyone.

Yes, your solution for checking parallelism is ok.

But, let's say that our exercise was actually the following:
(1,-1,2)
(2,-2,k)

Then the lines are parallel for a certain k. Indeed, they are a multiple of each other if there is a t such that

1=2t
-1=-2t
2=kt

The first and second equation give us that t=1/2. Plugging in in the second gives us k=4. So this is when the lines are parallel!

 

[Nickg140143]

Yes, your solution for checking parallelism is ok.

But, let's say that our exercise was actually the following:
(1,-1,2)
(2,-2,k)

Then the lines are parallel for a certain k. Indeed, they are a multiple of each other if there is a t such that

1=2t
-1=-2t
2=kt

The first and second equation give us that t=1/2. Plugging in in the second gives us k=4. So this is when the lines are parallel!

Beautiful

Thanks, this helped a lot.