Vector fields and line equations Problem

  • #1
RJLiberator
Gold Member
1,095
63

Homework Statement


We are giving to lines:
r1(t)=<1-t,4,5+2t>
r2(s)=<2,1+s,-s>

1. Find an equation perpendicular to the two lines and passing point P(1,1,1)
2. Find Coordinates of points of intersection of the line found in #1 with planes x=-1, xz-plane
3. Parametrize the line segment joining these two points.

Homework Equations




The Attempt at a Solution



Okay, so #1 should be relatively easy. It's merely taking the cross product of the two lines in terms of their coefficients of the slope.
My cross product of this came out to be <-2,-1,-1> So the perpendicular line is r3(w)=<1-2w,1-w,1-w)

#2 is where things get weird.
So I set 1-2w=-1 to solve for the plane x=-1. This results in w=1 which makes r3(1)=(-1,0,0)

For the xz-plane I set y=0 so I set 1-w=0. This means that w=1 and I have the same result of (-1,0,0).

Proceeding to #3, I then get a weird result of <-1,0,0>

My question is: Where did I mess up? I assume I messed up because why would they ask a question that results in this manner. I have a feeling I messed up in part #2 where I said the xz-plane must mean y=0. What could I have done differently here?

Or is my cross product miscalculated ? (I think I've checked on it multiple times..)

Thanks for any help
 
Last edited:

Answers and Replies

  • #2
1,013
70

Homework Statement


We are giving to lines:
r1(t)=<1-t,4,5+2t>
r2(s)=<2,1+2,-2>
Unfortunately, we cannot yet check your work, due to the missing variable of s in the parametrization of r2.
 
  • Like
Likes RJLiberator
  • #3
RJLiberator
Gold Member
1,095
63
Ah, I apologize. The last two 2's were supposed to be "s". I have now fixed it.

Thank you.
 
  • #4
1,013
70
Ah, I apologize. The last two 2's were supposed to be "s". I have now fixed it.

Thank you.
In that case, your work checks out fine. Perhaps the book has a typo, or it is a trick question where you really are supposed to "parametrize" the line between (-1, 0, 0) and itself with r(t) = (-1, 0, 0). :-)
 
  • #5
RJLiberator
Gold Member
1,095
63
Interesting.

So my understanding of the xz plane where y must be = 0 is correct in this case and they are equal to each other.

A trick question!
And therefore the parametrization is r(t)=<-1,0,0>
 
  • #6
RJLiberator
Gold Member
1,095
63
Since the parametrization is r(t)=<-1,0,0>
Does this line intersect the z-axis or x-axis? If so, can we find the point of intersection?

This is just a point that is on the x-axis at x=-1. Correct?
 
  • #7
1,013
70
Since the parametrization is r(t)=<-1,0,0>
Does this line intersect the z-axis or x-axis? If so, can we find the point of intersection?

This is just a point that is on the x-axis at x=-1. Correct?
Yes. It is a single point, and only "intersects" the x-axis. This isn't a remarkable feat particular to points; many lines in 3 dimensions don't intersect any axis.
 
  • Like
Likes RJLiberator
  • #8
RJLiberator
Gold Member
1,095
63
Excellent. Thank you kindly.
 
Top