Basic calc 3 problems, vectors and planes

[shemer77]

Homework Statement

1)show that the lines l1 and l2 intersect and find their point of intersection.
l1: x+1=4t y-3=t z-1=0
L2: x+13=12t y-1=6t z-2=3t

2)The plane through the points p1(-2,1,4),p2(1,0,3) that is perpendicular to the plane 4x-y+3z=2

3)Find the distance between the given skew lines
x=3-t y=4+4t z=1+2t
x=t y=3 z=2t

The Attempt at a Solution

1) I find that these don't intersect, am I wrong?
2) my answer came out to -2x +(-13/2)y + .5z +.5=0, if this is wrong i can show my work...
3) I got 55/(4*sqrt(6))

 

[Mark44]

I looked at 1 and 4 only.

Homework Statement

1)show that the lines l1 and l2 intersect and find their point of intersection.
l1: x+1=4t y-3=t z-1=0
L2: x+13=12t y-1=6t z-2=3t

2)The plane through the points p1(-2,1,4),p2(1,0,3) that is perpendicular to the plane 4x-y+3z=2

3)Find the distance between the given skew lines
x=3-t y=4+4t z=1+2t
x=t y=3 z=2t

4)Show that the graph of r=t*sin(t) i + t*cos(t) j + t^2 k
lies on the paraboloid of z=x^2 +y^2

The Attempt at a Solution

1) I find that these don't intersect, am I wrong?

Yes. I found a point of intersection. Show us what you did, so we can figure out where you went wrong.

2) my answer came out to -2x +(-13/2)y + .5z +.5=0, if this is wrong i can show my work...
3) I got 55/(4*sqrt(6))
4)I set x to t*sin(t), y to t*cos(t) and z to t^2 and then i put it in the equation z=x^2 +y^2 and i got t^2=t^2. Sounds right??

t² = t² is not what you want to conclude, since that's obviously and trivially true - any number is always equal to itself. Here's a better way to do it.
The parametric form of the vector equation r(t) = t*sin(t) i + t*cos(t) j + t² k is:
x = t*sin(t)
y = t*cos(t)
z = t²

From these equations, we have
x² + y² = (t*sin(t))² + (t*cos(t))²
= t²sin²(t) + t²cos²(t)
= t²(sin²(t) + cos²(t))
= t² * 1
= z
This shows that the graph of the given vector function lies on the paraboloid = x² + y².

 

[shemer77]

ah ok thanks, i got 4

For 1, I rewrote it so
L1: x=4t-1 y=T+3 z=1+0T
L2: x=12T-13 y=6T+1 z=3T-2

To solve this i set the to x's equal to each other but i change the x in l2 to the letter k
4t-1=12k-13
t=(12/4)k-(12/4)

I make my second equation
T+3=6k+1 and plug in t so then i get
(12/4)k -(12/4) +3 = 6k+1
solving for k i get -(1/3) and i plug that into my equation for t to get t=-4

Now i have values of k and T and i plug them into my final equation
1+0T =3T-2 and solving that you get 1=3*(-1/3) -2 which comes out to 1=-3 which they obviously don't so they don't intersect

 

[Dick]

You could probably think about this more clearly if you realize that the two lines are parametric equations. The 't' in the first equation is not necessarily the same as the 't' in the second equation. Rewrite them as:
L1: x+1=4t,y-3=t,z-1=0
L2: x+13=12s,y-1=6s,z-2=3s

Now try and find a value of s and t such that the x, y and z values are equal.

 

[shemer77]

You could probably think about this more clearly if you realize that the two lines are parametric equations. The 't' in the first equation is not necessarily the same as the 't' in the second equation. Rewrite them as:
L1: x+1=4t,y-3=t,z-1=0
L2: x+13=12s,y-1=6s,z-2=3s

Now try and find a value of s and t such that the x, y and z values are equal.

already did that, look at what i did

 

[Dick]

already did that, look at what i did

What you did is more than a little confusing and the conclusion is wrong. Look at the z equations. What must s be?

 

[shemer77]

What you did is more than a little confusing and the conclusion is wrong. Look at the z equations. What must s be?

i did exactly what you said except i used k instead of s.

anyways does s have to be -1/3

 

[Mark44]

ah ok thanks, i got 4

If you're talking about your previous work, I don't think so. Concluding that t² = t² is silly and probably won't get you credit for the problem, or maybe only partial credit, at best.

 

[Mark44]

To solve this i set the to x's equal to each other but i change the x in l2 to the letter k
4t-1=12k-13
t=(12/4)k-(12/4)

i did exactly what you said except i used k instead of s.

No, according to what you wrote, you replaced x with k.

And if I remember correctly, I got s = -1/3.

 

[Dick]

i did exactly what you said except i used k instead of s.

anyways does s have to be -1/3

Ok, then s=(-1/3), t=(-4). Doesn't that make x, y and z the same for both systems?

 

[shemer77]

If you're talking about your previous work, I don't think so. Concluding that t² = t² is silly and probably won't get you credit for the problem, or maybe only partial credit, at best.

no i did something else, i got it though I am sure :)

Ok, then s=(-1/3), t=(-4). Doesn't that make x, y and z the same for both systems?

sorry not sure how you figured that?

 

[Dick]

no i did something else, i got it though I am sure :)



sorry not sure how you figured that?

I thought that was what you had figured out. I looked at the z equation and figured s=(-1/3). Then I look at either of the other two equations and figured t=(-4). Then I checked third equation to make sure it was also true.

 

[shemer77]

Wooowwww, sorry about that I don't know why I was having such a stupid moment, yea i totally get it. thanks for your help! Now i just hope 2 and 3 are right...

edit: i got 3, i just need help with number 2 now, thanks!

 

[HallsofIvy]

Homework Statement

1)show that the lines l1 and l2 intersect and find their point of intersection.
l1: x+1=4t y-3=t z-1=0
L2: x+13=12t y-1=6t z-2=3t

2)The plane through the points p1(-2,1,4),p2(1,0,3) that is perpendicular to the plane 4x-y+3z=2

So a vector perpendicular to the plane is <4, -1, 3> and must be parallel to any plane that is also perpendicular to the given plane. Another vector in the plane is the vector from p1 to p2, <3, -1, -1>. The cross product of those two vectors is $4\vec{i}+ 13\vec{j}- \vec{k}$. Taking (-2, 1, 4) as the "base point", the equation of the plane is 4(x+ 2)+ 13(y- 1)- (z- 4)= 4x+ 13y- z+ 8- 13+ 4= 0 or 4x+ 13y- z= 1.

3)Find the distance between the given skew lines
x=3-t y=4+4t z=1+2t
x=t y=3 z=2t

The Attempt at a Solution

1) I find that these don't intersect, am I wrong?
2) my answer came out to -2x +(-13/2)y + .5z +.5=0, if this is wrong i can show my work...

Well, I would be inclined to multiply through by -2 and move the constant to the other side of the equation!

3) I got 55/(4*sqrt(6))

 

[shemer77]

3) I got 55/(4*sqrt(6))

wait are you saying that's the right answer?? because I redid it and rechecked everything twice and i got sqrt(6)