# Homework Help: Finding angle between two lines

1. Nov 4, 2008

### jeffreylze

1. The problem statement, all variables and given/known data

Two lines

x-3=2-y , z=1

x=7, y-2=z-5

Find the angle between the lines

2. Relevant equations

3. The attempt at a solution

The thing is that, i have no idea how to turn that into either (x,y,z) coordinates, parametric equations or vector equation. So, I really am stuck here. =/

2. Nov 4, 2008

### Dick

WHY can't you express them in parametric form? Choose y to be the parameter. Express the other variables in terms of y.

3. Nov 4, 2008

### jeffreylze

But how do you express the z term in terms of y?

4. Nov 4, 2008

### Dick

z=1+0*y, if you want to write it that way. z doesn't have any y dependence. Nothing wrong with that.

5. Nov 4, 2008

### jeffreylze

oh, so it will be

x = 5-y
y
z = 1

x= 7
y
z= y+3

Hence, to find the angle, i will need two direction vectors, right? But how do i convert that into vector equation? r = ro + tv

6. Nov 4, 2008

### Dick

I would write the first one, for example, as x=5-t, y=t, z=1. It can be a little confusing if you use the same name for the parameter as you do for the coordinate. Now you want r=(x,y,z)=(constant vector)+t*(another constant vector). Why don't you have a guess what those constant vectors might be?

7. Nov 4, 2008

### jeffreylze

Oh, i see. So it will be r = (5,0,1) + t*(-1,1,0) , what if i let x = t ? That will give me, r = (0,5,1) + t*(1,-1,0) . Or it doesnt matter? Because the direction vector is what I am after? Am I right here, or am I completely off track?

8. Nov 4, 2008

### Dick

You're on track. Another possibility is r=(5,0,1)+t*(-2,2,0). I just changed t -> 2*t. It's still the same line. There are lots of ways to parametrize a line. And, yes, in any two different ways of doing it, the direction vectors will still be parallel.

9. Nov 4, 2008

### jeffreylze

I see. Now, it all makes sense. Thanks a lot =D