# Vector question- tough one

1. Jun 3, 2004

### mathrocks

Vector question--"tough one"

My teacher said this question is really tough and doesn't really expect anyonre in the class to solve it so I thought I might give it a shot in here...

Find parametric equations for the line through the point (0,1,2) that is perpendicular to the line x=1+t, y=1-t, z=2t and intersects this line.

Any help would be much appreciated.

2. Jun 3, 2004

### Hurkyl

Staff Emeritus
Ugh... it's been a long while since I couldn't think of any hint to give that doesn't spoil the problem.

Well, I guess there's always the hard way: you could start by writing down the general set of parametric equations for an arbitrary line through (0, 1, 2), then solve for the parameters...

3. Jun 3, 2004

Edit: Think I was being too nice... So here's a hint instead:

What direction is a line between P_1(0,1,2) and P_2(1+t,1-t,2t) going? What do we know about this direction?

4. Jun 3, 2004

### mathrocks

x=at
y=1+bt
z=2+ct

How exactly do I go about solving for the parameters? Do I just set each of them equal to the corresponding givens?

5. Jun 3, 2004

### Hurkyl

Staff Emeritus
You're given two properties of the line you just wrote; it has to intersect another line, and it has to be perpendicular to that line.

You just need to translate these properties into equations.

6. Jun 3, 2004

### TALewis

Another hint: the dot product of two perpendicular vectors is equal to zero.

7. Jun 3, 2004

### mathrocks

Hmm..
In order for something to be perpendicular, the dot product has to be equal to zero...but in this case would my two vectors be <1,-1.2> and <a,b,c>? Or am I suppose to find a,b,c first?
And I'm really confused on how to go about finding the line that intersects with it also. I guess this problem is just really hard for the knowledge I have

8. Jun 3, 2004

### TALewis

You're doing fine. Everything you've said so far has been correct.

You're trying to build a system of equations to use to solve for a, b, and c. The dot product condition gives you another equation to use. Let's list what we know.

You gave:

\begin{align*} x &= 1+t\\ y &= 1-t\\ z &= 2t \end{align*}

And the line we're looking for is given by:

\begin{align*} x &= as\\ y &= 1 + bs\\ z &= 2 + cs \end{align}

Notice on the last set of equations I used "s" instead of "t" for the parameter. Two lines intersect when their equations are equal. Although we want the two lines to intersect at some point, it's quite possible that they could intersect at "different times." Therefore, we need to use a different parameter for each line when we set them equal (intersecting).

Finally, the dot product condition gives us:

\begin{align*} \langle 1,-1,2 \rangle \cdot \langle a,b,c \rangle &= 0\\ a - b + 2c &= 0 \end{align}

Set the two lines equal to each other (x = x, y = y, z = z). This satisfies the "intersection" condition. The dot product result satisfies the "perpendicular" condition. With those two conditions, you should have four equations and five unknowns (a, b, c, t, s). More unknowns than available equations -- how can we solve this?

You must realize that there is no unique solution for a, b, and c. Why? Consider your original line. We said its direction vector was <1,-1,2>. Couldn't we use the direction vector <2,-2,4> to describe the same line? Or <6,-6,12>? They all point in the same direction; they just have different magnitudes.

My point is, you can pick an arbitrary value for a, b, or c (say, let c = 2) and then you have a system of four equations and four unknowns, which you can solve to give you the equation of your line, along with the s and t where they intersect.

Last edited: Jun 3, 2004
9. Jun 3, 2004

### mathrocks

I think im getting stuck on finding how they intersect and are perpendicular...the intersection part is getting me confused.

10. Jun 3, 2004

### mathrocks

Oh wow...thank you so much. I guess I just couldnt connect the two things..but I understand now. Awesome, thanks again!

11. Jun 3, 2004

### mathrocks

Ok, I want to make sure I'm doing this correctly. My equations are:

1+t=as
1-t=bs
2t=2t+cs
a-b+2c=0

Those are my four equations. Now to solve them, I just pick a point? So if I used the point that you mentioned as an example, c=2. I get...

1+t=as
1-t=1+bs
2t=2t+2s
a-b=-4

But if I try to solve it I get s=0 and t has two different values when s=0 so wouldnt that mean it's skewed? I'm sure I'm doing something horribly wrong somewhere...

12. Jun 3, 2004

### mathrocks

OH man, I see my mistake I think. One of my equations I have 2t=2t+cs..when it should be 2t=2+cs.
Hopefully that's the mistake.

13. Jun 3, 2004

### TALewis

You have the right idea, but you have either made some small errors in setting the lines equal to each other, or your post has a few typos. The system should be:

\begin{align*} a - b + 2c &= 0\\ 1 + t &= as\\ 1 - t &= 1 + bs\\ 2t &= 2 + cs \end{align}

Pick a value for a, b, or c (c = 2 is a good choice), solve that system carefully, and you should get the answer you're looking for.

14. Jun 3, 2004

### mathrocks

is s=(-1/2), t=(1/2) a=-3, b=1, and c=2?

And will my final final answer be :

x=-3s
y=1+s
z=2+2s

Last edited: Jun 3, 2004
15. Jun 3, 2004

### TALewis

That's what I got. Check it to see that your conditions are satisfied:

Your new line must pass through the point (0,1,2).
It must intersect with the original line.
It must be perpendicular to the original line.

16. Jun 3, 2004

### mathrocks

Yup, all conditions are met. Yay!