R3 Line Problem: Intersection of 3 lines, find eqn of L

[ace68]

Intersection of 3 Lines in Space

Homework Statement

L0 contains point P(1,0,2) and it meets

L1: x=y=z+2
L2: x+3=-y/2=z/3

Find equation of Line L0.

The Attempt at a Solution

So we know that l1 and l2 are skewed.

Based on the given info:

L1) x=t
y=t
z=t-2

L2) x =t-3
y = -2t
z = 3t

these parametric eqns are the same eqns in vector form

r1 = (0,0,-2) + t(1,1,1)
r2 = (-3,0,0) + t (1,-2,3)

with P(1,0,2) for L0, i deduce

L0 = (1,0,2) + t(x,y,z)

im stuck now, with no directional vector since L0 is not necessarily perpendicular to both lines so i can't even use cross product of directional vectors of L1 & L2.

Thanks for the help!

 

[rootX]

"L0 contains point P(1,0,2) and it meets

L1: x=y=z+2
L2: x+3=-y/2=z/3

Find equation of Line L0."

Rewording your question:
finding line l0 that passes through (1,0,2) and common point between L1 and L2.What if L1 and L2 don't intersect; you get a plane. Try to visualize the thing.

P.S. I think there must be some other condition: meeting those two lines at 90 deg ...

 

[ace68]

Im sorry I don't quite understand what you mean "common point between L1 and L2"

L0 does pass thru (1,0,2).

if that other condition is meeting it at 90deg...the cross product of the directional vectors of L1 and L2 is

n = (5, -2, -3)

with that...i get r3 = (1,0,2) + n(5,-2,-3)

written in parametric form and equating corresponding coordinates, i don't end up with
LS = RS, and not have a common point. =/

 

[rootX]

Im sorry I don't quite understand what you mean "common point between L1 and L2"

L0 does pass thru (1,0,2).

if that other condition is meeting it at 90deg...the cross product of the directional vectors of L1 and L2 is

n = (5, -2, -3)

with that...i get r3 = (1,0,2) + n(5,-2,-3)

written in parametric form and equating corresponding coordinates, i don't end up with
LS = RS, and not have a common point. =/

r3 does not passes through both lines. So that means 90 deg cannot be other condition.
From common point, I meant L0 passes through (1,0,2) and the point where L1 and L2 intersect.
I think there's something wrong with the question -- you must be given some additional information. Hopefully, someone else would be able to figure out (so either wait or ask your prof/ but don't spend too much time on this question, alone :D)

P.S. do you have the answer?

 

[ace68]

well my polish teacher does word his questions weirdly so I'm not surprised at all!
and unfortunately i don't have the answer
thanks anyways!

 

[Bertrandkis]

I think what we want to know is this: What is the relationship betwen L1, L2 and L0? Is
L0 perpendicular to these 2 lines, or is L0 parallel to these two lines? Does L0 intersects
L1 and L2 at all?

 

[HallsofIvy]

I don't see anything weird about the way the question is phrased- you are given two skew lines, L1 and L2 and asked to find the unique line that passes through (1, 0, 2) and both given lines. Since the lines are skew there is no "common point" between L1 and L2- also "What if L1 and L2 don't intersect; you get a plane" makes no sense. The one case in which L1 and L2 don't define at least one plane is when they are skew and don't intersect.

This can be done directly from the basic definitions- my preferred method! Any line passing through (1, 0, 2) has parametric equations x= At+ 1, y= Bt, z= Ct+ 2 where <A, B, C> is the "direction vector". As you say, L1 can be written in parametric equation as x= s, y= s, z= s-2. For those two line to intersect there must be values of s and t such that x= s= At+1, y= s= Bt, z= s- 2= Ct+ 2. From the second equation s= Bt so, replacing s by Bt in the first equation, Bt= At+ 1 or (B-A)t= 1 so t= 1/(B-A) and s= B/(B-a). Putting those into the third equation, s-2= B/(B-A)- 2= C/(B-A)+ 2 or, multiplying the entire equation by B-A, B= C+ 4(B-A). In order that the two line intersect we must have 3B- 4A+ C= 0 (and B not equal to A).

Do the same with L2 and you will get another equation involving A, B, C. You can solve those two equations for two of A, B, C, in terms of the third: say solve for A and B in terms of C. That leaves C but that doesn't matter: the "direction vector" can have any length. Take C to be any convenient value and write down the equation of the line.

 

[ace68]

I was actually able to solve this question already. Here is the image...
Thanks again for the input and
help!

http:/\/img134.imageshack.us/img134/5236/booyahmg5.jpg