# Homework Help: Calc 2 Lines and planes 2

1. Apr 29, 2010

### somebodyelse5

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

Find the vector equation for the line of intersection of the planes 5x−4y+z=3 and 5x+z=1

2. Relevant equations

N/A

3. The attempt at a solution

I know this much.
r=<__,__,0>+t<-4,__,__>

Could someone explain to me what I'm actually doing? I understand how the two intersecting planes make a line, but I dont understand how to actually find it. Id really appreciate it if someone could give me the "super simple, average joe" definition I would really appreciate it.
Conceptually I get it, but mathematically I dont know what I need to do.

2. Apr 29, 2010

### LCKurtz

There are several approaches to finding the line. Here's one way. You have two equations in 3 unknowns, so you can solve for two in terms of the third. So put the z on the right side of both equations:

5x - 4y = 3-z
5x = 1 - z.

To find a point on the line of intersection, not that it must be on both planes. To find such a point, now let z be anything. Make it easy on yourself. For example in your second equation if you let z = 6 you get x = -1 and the first equation gives y = -1/2.

So you have (-1, -1/2, 6) on the line.

Now take another value for z and get another point. Then use those two points to write the equation of the line.

3. Apr 29, 2010

### somebodyelse5

OK, so I used the given value of z=0 to find the other two points, just having issues solving for the rest of the line. I have <1/5,-1/2,0>+t<-4,__,__) I cant help but look at this as a form of y=mx+b, or y=b+mx. is that correct? because then I would now be solving for the slope of the line. Am I on the right track? There is not an equation for the slop of a 3D line is there?

4. Apr 29, 2010

### LCKurtz

There is no concept of slope for a 3D line. It is replaced by the concept of a direction vector. Parametric equations of lines are not unique. I would quit worrying about that -4 you are given. Just check that your two points are on both planes and that your parametric equation goes through those two points. Then it has to be correct.

5. Apr 29, 2010

### somebodyelse5

So the second vector in the equation, is just another point on the line?
I cant figure out how to actually solve for that second vector by using to points on the line. All that came to mind was slope. delta x, delta y, delta z?

My t(__,__,__) portion of the equation has to have the x=4 value because of how webwork is set up, my y and z values need to correspond to that value.

6. Apr 29, 2010

### LCKurtz

Oh good grief! WebWork!! So you have to get their point and their direction vector. Well, I guess taking z = 0 gives you their point. You can always get a direction vector for the intersection line by crossing the two normals to the planes [why?]. So do that to get a direction vector D and scale it so the first component is -4. Try that.

7. Apr 29, 2010

### somebodyelse5

Awesome thank you! I did <5,-4,1>x<5,0,1> and it worked!

Thanks a TON

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