Lines and Planes in Space - HELP SOON

[justagirl]

I am very confused regarding a few problems in Calculus III. Any help
or
hints to any of these would be greatly appreciated!

1.) Are lines L1 and L2 parallel?

L1: (x-7)/6 = (y+5)/4 = -(z-9)/8;
L2: -(x-11)/9 = -(y-7)/6 = (z-13)/12;

The answer says that they are parallel, which I don`t understand. I
know 2
lines are parallel if their direction vectors are parallel, but in this
case V1 = <6, 4, -8>, and V2 = <9, -6, 12>. So they are not multiples
of
each other and thus I didn`t think they are parallel. What am I
missing?

2.) Write an equation of the plane that contains both the point P and
the
line L:

P(2,4,6);
L: x = 7-3t, y = 3+4t, z = 5 + 2t;

I know to write an equation of the plane you need a direction vector
and a
point. I tried using <-3,4,2> crossed with <2,4,6> as my normal vector
and
<2,4,6> as my (X0, Y0, Z0). But I got the wrong answer...

3.) Find an equation of the plane through P(3,3,1) that is
perpendicular
to the planes x+y = 2Z and 2X + z = 10. If I take the cross product of
the
second 2 planes that would give me a vector parallel to the equation
that
I want to find, but I need a normal vector. What to do?

4.) Find an equation of the plane that passes through the points
P(1,0,-1), Q(2,1,0) and is parallel to the line of intersection of the
planes x+y+z = 5 and 3x -y = 4.

5.) Prove that the lines x -1 = 1/2(y+1) = z-2 and x-2 = 1/3(y-2) =
1/2(z-4) intersect. Find an equation of the only plane that contains
them
both.

Sorry for so many problems. But any help would be great! Thanks!

 

[matt grime]

Look at L2. Are you sure you've got the right direction vector?

 

[justagirl]

okay, so it's <-9. -6. 12> Does it make a difference though?

 

[justagirl]

oh nm

oh nevermind... that was a stupid question. I see it now. Got any suggestions on the other problems? :)

 

[rgrig]

You get -3/2 when you divide corresponding coordinates, any corresponding coordinates.