# Finding the equation of a plane given two lines

1. Sep 19, 2011

### navalava

1. The problem statement, all variables and given/known data
Find an equation of the plane that contains these lines:
r=<1,1,0>+t<1,-1,2>
r=<2,0,2>+s<-1,1,0>

2. Relevant equations

3. The attempt at a solution
I took the cross product of <1,-1,2> and <-1,1,0> to get <-2,-2,0>.
I used the point (1,1,0) to get the equation of the plane:
-2(x-1)-2(y-1)=0

But the correct answer is supposed to be x+y=2.

I'd really appreciate it if someone could show me what I'm doing wrong. Thanks!

2. Sep 19, 2011

### Dick

-2(x-1)-2(y-1)=0 and x+y=2 are the same plane. Can you figure out a way to show that?

3. Sep 19, 2011

### glebovg

Your answer simplifies to x + y = 2.

4. Sep 19, 2011

### glebovg

As with equations of lines in three dimensions, it should be noted that there is not a unique equation for a given plane.

5. Sep 19, 2011

### flyingpig

I thought that is only true for planes in the form of

a(x - x_0) + b(y - y_0) + c(z - z0) = 0

but not for a'x + b'y + c'z = d

6. Sep 19, 2011

### glebovg

There are various ways of describing a plane. The two equations you provided are essentially equivalent.

7. Sep 19, 2011

### flyingpig

No, but the first one, you can choose any intercept and it can still be reduced to my second one. By just observation the first one "isn't unique", even though the answer will be the same.

8. Sep 19, 2011

### glebovg

Since there are infinitely many points on the plane you may choose any point as long as it is on the plane.

9. Sep 19, 2011

### Dick

For the second form you can still multiply by any constant and it's the same plane. That's not quite unique.

10. Sep 19, 2011

### glebovg

If you multiply it by a constant what you obtain is a different equation of the plane.