Vector geometry - Intersection of lines

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Homework Help Overview

The discussion revolves around finding the intersection of two parametric vector equations representing lines in three-dimensional space. The original poster presents the equations and seeks to determine the coordinates at which the lines intersect.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to equate the components of the two vector equations to find the intersection point. They express uncertainty about their method and results, suggesting a potential misunderstanding of the approach.

Discussion Status

Participants have provided guidance on the notation used for the parameters of the lines, indicating that the parameters do not need to be the same for the lines to intersect. There is an ongoing exploration of how to equate the components and solve for the parameters.

Contextual Notes

The original poster expresses confusion regarding the method used to find the intersection and has noted discrepancies between their calculations and the expected answer. There is an indication of a lack of consensus on the approach to take.

Keshroom
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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
 
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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.
 

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