Show two lines are parallel & find equation for their plane

[jheld]

Homework Statement

Show that the two lines:
L1: x = t - 3; y = 1 - 2t; z = 2t + 5..
L2: x = 4 - 2s; y = 4s + 3; z = 6 - 4s... (I changed the t's to s's for L2)
are parallel, and find an equation for the plane that contains them.

Homework Equations

A(x - x0) + B(y - y0) + C(z - z0) = 0 : equation of a plane;
Set each variable (x, y, z) equal for each line, and if the equations don't match up with the found s's and t's, then the two lines are not going to intersect, or just look to see if the directional vectors (i.e. t's and s's) are scalar multiples of each other.

The Attempt at a Solution

I found that the lines directional vectors were scalar multiples of each other, thus proving that they are parallel. I could show this a couple of other ways, but this satisfies the first question.

As far as finding the equation for the plane containing the two lines, here is where I get stuck.

When I picture it in my head, I see each of the normal vectors pointing to the other line, respectively, which makes perfect sense. But from there, I don't see the connection that I need to make in order to find both 1)the vector for the equation and 2) the initial x,y,z point.

 

[rootX]

L1: x = t - 3; y = 1 - 2t; z = 2t + 5..
L2: x = 4 - 2s; y = 4s + 3; z = 6 - 4s

One vector = <1,-2,2>
Other vector can be P2-P1.

There is no unique normal vector for one line. You need two non-parallel vectors to get a unique normal for them.

 

[Дьявол]

The first parallel vector to the line L1 is <1,-2,2> and the vector parallel to the line L2 is <-2,4,-4>. The lines are parallel if the first vector is scalar multiple of the other, i.e:
λ(1,-2,2)=(-2,4,-4)
As we can see λ=-2

For the second one, find three points from the lines and you will find the plane.