Vector geometry - Intersection of lines

Click For Summary
To find the intersection of the two parametric vector equations r(t) and s(u), the components of the vectors must be equated: 2+t = 1+2u, -4-3t = -1-u, and 4+4t = u. Solving the first equation for t yields t = -1 + 2u. Substituting this value into the other equations allows for solving for u, which can then be used to find the corresponding t. The intersection point is confirmed to be (1, -1, 0), indicating that the lines intersect at this coordinate.
Keshroom
Messages
25
Reaction score
0

Homework Statement


I have 2 parametric vector equations (of a line)

r(t) = (2,-4,4) + t(1,-3,4)
s(t) = (1,-1,0) + t(2,-1,1)

how do i find the coordinates for which they intersect each other?
The answers is (1,-1,0)


Homework Equations



x=a+λv, for some λ in ℝ (parametric vector form of line)

The Attempt at a Solution


As in high school, with the form y=mx+b i would make the 2 equations equal to each other, solve for x, then substitute back into either equations to find y.

I've tried making the (x,y,z) components equal to each other, solve for 't' and substitute back in but i can't get the answer in the back of the book

parametic equations
for r(t): x = 2+t, y= -4-3t, z= 4+4t
for s(t): x = 1+2t, y= -1-t, z= t

Now i did
2+t = 1+2t
t = 1

substituting back into x=2+t: x = 3

i did this for also y and x components and got (3, 1/2, -4/3)
hmmmm
I have a feeling that this method isn't correct :s
 
Physics news on Phys.org
Change the notation to
r(t) = (2,-4,4) + t(1,-3,4)
s(u) = (1,-1,0) + u(2,-1,1)
the parameters t and u don't have to be the same for the lines to intersect each other.
 
Dick said:
Change the notation to
r(t) = (2,-4,4) + t(1,-3,4)
s(u) = (1,-1,0) + u(2,-1,1)
the parameters t and u don't have to be the same for the lines to intersect each other.

alright. Now how do i solve it?
 
Keshroom said:
alright. Now how do i solve it?

The same way you tried before. Equate components of the vectors and solve them. Try it. The first component gives you 2+t=1+2u. Solve that for t and substitute into the rest.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

Similar threads

  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
12
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
2
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
6
Views
2K
  • · Replies 21 ·
Replies
21
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
7
Views
1K