Basic calc 3 problems, vectors and planes

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Homework Help Overview

The discussion revolves around problems related to vectors and planes in a calculus context. Participants are tasked with showing intersections of lines, determining equations of planes, and calculating distances between skew lines.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore whether two lines intersect and attempt to find their point of intersection, with some expressing uncertainty about their conclusions.
  • There are discussions about the equations of planes, with participants questioning their own results and seeking clarification on the correctness of their derived equations.
  • Attempts to calculate the distance between skew lines are shared, with participants verifying their answers and discussing potential errors in their calculations.

Discussion Status

Some participants have provided guidance on re-evaluating the intersection of lines by clarifying the use of parameters. Others have expressed doubts about their results and are seeking confirmation or correction of their approaches. The discussion reflects an ongoing exploration of the problems without a clear consensus on all points.

Contextual Notes

Participants are working under the constraints of homework rules, which may limit the information they can share or the methods they can use. There is a focus on ensuring that the reasoning behind their answers is sound, particularly in the context of showing intersections and deriving plane equations.

shemer77
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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))
 
Last edited:
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I looked at 1 and 4 only.
shemer77 said:

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.
shemer77 said:
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.
 
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
 
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.
 
Dick said:
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
 
shemer77 said:
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?
 
Dick said:
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
 
shemer77 said:
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.
 
shemer77 said:
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)

shemer77 said:
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.
 
  • #10
shemer77 said:
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?
 
  • #11
Mark44 said:
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?
 
  • #12
shemer77 said:
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.
 
  • #13
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!
 
Last edited:
  • #14
shemer77 said:

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))
 
  • #15
shemer77 said:
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)
 

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