# Finding the equation of a plane given two lines

## Homework Statement

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>

## 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!

## Answers and Replies

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

Your answer simplifies to x + y = 2.

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

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

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

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

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

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.

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

Dick
Science Advisor
Homework Helper
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

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

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