I am very confused regarding a few problems in Calculus III. Any help(adsbygoogle = window.adsbygoogle || []).push({});

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!

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