- #1
justagirl
- 29
- 0
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!
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!
