# Basic calc 3 problems, vectors and planes

1. Sep 23, 2011

### shemer77

1. The problem statement, all variables and given/known data
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

3. The attempt at a solution
1) I find that these dont 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))

Last edited: Sep 23, 2011
2. Sep 23, 2011

### Staff: Mentor

I looked at 1 and 4 only.
Yes. I found a point of intersection. Show us what you did, so we can figure out where you went wrong.
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.

3. Sep 23, 2011

### 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 dont so they dont intersect

4. Sep 23, 2011

### 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.

5. Sep 23, 2011

### shemer77

already did that, look at what i did

6. Sep 23, 2011

### Dick

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

7. Sep 23, 2011

### shemer77

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

anyways does s have to be -1/3

8. Sep 23, 2011

### Staff: Mentor

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.

9. Sep 23, 2011

### Staff: Mentor

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

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

10. Sep 23, 2011

### Dick

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

11. Sep 23, 2011

### shemer77

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

sorry not sure how you figured that?

12. Sep 23, 2011

### Dick

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.

13. Sep 24, 2011

### shemer77

Wooowwww, sorry about that I dont 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!!

Last edited: Sep 24, 2011
14. Sep 24, 2011

### HallsofIvy

Staff Emeritus
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.

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

15. Sep 24, 2011

### shemer77

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