How to find the equations of the plane that contains the parallel lines...?

Question

The parametric equations of the two parallel lines are the following...

{x=-2+3t, y=1+4t, z=2-t}, {x=2-2t, y=3-4t, z=1+t}

and I do have the right answer...

which is supposed to be, x+y+6z=11

Answer 1

{x=-2+3t, y=1+4t, z=2-t}
{x=2-2t, y=3-4t, z=1+t}

Let's rewrite the equations of the lines in a more calculation friendly form.

Line 1: L1 = <-2,1,2> + t<3,4,-1>
Line 2: L2 = <2,3,1> + s<-2,-4,1>

First of all, the two lines are not parallel, because the directional vectors are not the same (or a multiple of each other). If the two lines intersect they define a plane. If they do not intersect but are skew, no plane can contain them both.

To see if they intersect solve for s and t.

-2 + 3t = 2 - 2s
1 + 4t = 3 - 4s
2 - t = 1 + s

Let's start by looking at the second and third equations. Add four times the third equation to the second equation.

9 = 7

This is inconsistent. Therefore there is no intersection. The lines are skew and no plane contains them both.

Answer 2

The given airplane has a common N=( 2,4,8).- A parallel airplane has an identical direction , so widely used would be n= (2,4,8) includes the line , so at t=0 a place vector is P(4,0,7) The airplane is (R-P) .N=0 2(x-4)+4(y-0)+8(z-7) =0 Simplify by ability of two (x-4)+2(y-0)+4(z-7)=0 x+2y+4z-32=0