# Finding line of intersection of two planes

1. Aug 25, 2012

### Spacec0wboy

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

Find parametric equations for the line of intersection of the
planes x + y + z = 1 and
r = (1, 0, 0) + $\lambda$(2, 1, 0) + $\mu$(0, 1, 1) where $\lambda$, $\mu$ $\in$ R

2. Relevant equations

3. The attempt at a solution

I attempted to convert the 2nd plane equation to scalar form by finding the normal: (2, -1, 0) cross product with (0, 1, -1). That ended up being i + 2j + 2k, meaning that the plane had equation x + 2y + 2z = d, subbing in the point in the vector equation (1, 0, 0) gives d =1 therefore x + 2y + 2z = 1. I'm not sure what to do from here or if what i've done is either correct or necessary.

2. Aug 25, 2012

### kushan

the normal vector would be (-1,2,-2).
And Cartesian equation would be x-2y+2z=1

3. Aug 25, 2012

### HallsofIvy

Staff Emeritus
The parametric equations for the second plane are saying that $x= 1+ 2\lambda+ 0\mu$, $y= 0+ 1\lambda+ 1\mu$, and $z= 0+ 0\lambda+ 1\mu$. Putting those into the equation of the first plane, x+ y+ z= 1 gives $(1+ 2\lambda)+ (\lambda+ \mu)+ (\mu)= 1+ 3\lambda+ 2\mu= 1$ or $3\lambda+ 2\mu= 0$. Solve that for either $\lambda$ or $\mu$ in terms of the other and replace in the parametric equation, getting linar equations in a single parameter- a line.

I don't know if it was just a typo or copying mistake but the normal to the second plane is NOT the cross product of (2, -1, 0) and (0, 1, -1). I'm not sure where you got those vectors. Two vectors in the plane are (1, 0, 0)+ (2, 1, 0)= (3, 1, 0) (taking $\lambda= 1$ and $\mu= 0$ and another is (1, 0, 0)+ (0, 1, 1)= (1, 1, 1) (taking $\lambda= 0$ and $\mu= 1$. The cross product of those vectors is normal to the plane.

4. Aug 25, 2012

### kushan

HallsofIvy
I dont think cross product of 3,1,0 and 1,1,1 would be normal to the 2nd plane
as the cross product of above is 1,-3,2 , but the normal vector should be 1,-2,2

Last edited: Aug 25, 2012
5. Aug 25, 2012

### ehild

(2,1,0) and (0,1,1) are vectors which lay in the plane. You mixed the signs a bit.

Kushan is right, (-1, 2,-2) is a normal vector of the plane.

ehild

6. Aug 25, 2012

### kushan

ehild would it be -1,2,-2 or 1,-2,2 ?

7. Aug 25, 2012

### ehild

The normal of a plane is the cross product of two vectors laying in the plane, your vectors are position vectors of two points in the plane.

ehild

Last edited: Aug 25, 2012
8. Aug 25, 2012

### ehild

They are parallel vectors, both normal to the plane, one is -1 times the other.

ehild

9. Aug 25, 2012

### kushan

Direction doesn't matter?

10. Aug 25, 2012

### Bacle2

Not for normality: if a vector v is normal to the surface, then so is the vector -v.

11. Aug 25, 2012

### kushan

Ok I used to imagine like the one originating from center and piercing though the plane perpendicularly
So it is both way .
Thank Ehild and Bacle2

12. Aug 25, 2012

### Bacle2

Actually,

maybe thinking of it this way would help: imagine you just have , say the vector

1,-2,2 on the plane. Now, give the plane a 180 turn,and you get the vector

-1,2,-2 --which is the same as 1,-2,2, but with opposite orientation.

Give 99% of the tanks to Ehild, he did most of the work.

13. Aug 25, 2012

### kushan

1/100 thanks Bacle

14. Aug 25, 2012

### Spacec0wboy

Wow yep I can't even read the question properly, the normal is indeed (-1, 2,-2). So if I cross product (-1, 2, -2) with (1, 1, 1) I get 4i + 3j - 3k, which is the direction vector of the line. So that's half of what I need for the parametric equations, the other half is a point on the line, how do I get that?

15. Aug 25, 2012

### kushan

find the equation of line Spacec , using the equation of planes .

16. Aug 25, 2012

### ehild

You do not need that cross product really. You need to give the parametric equation of the line. Follow HallsofIvy's hint.

ehild

17. Aug 25, 2012

### Spacec0wboy

Ok so using HallsofIvy's method I get:

λ = -2/3*μ

x = 1 - 4/3*μ
y = 5/3*μ
z = -μ

Is this right? Thanks for helping me out and that method was pretty easy but I would still like to know how to do it using the direction vector of the line and finding a point on it, just so I know another way to do it.

18. Aug 25, 2012

### kushan

You mean other method for solving this?

19. Aug 25, 2012

### Spacec0wboy

Yea, by finding the vector equation of the line then putting it into parametric form. We got the direction vector of the line before from the cross product of the 2 plane normals. Now we need a point on the line but I'm not sure how to get that.

20. Aug 25, 2012

### kushan

put some x=0 in both plane then solve the two equation .