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I am having difficulty understanding what to do for these questions.  

What is the relationship of the given line
L: (x,y,z) = (-2,1,5) + t<2,-8,-6>
to each of the lines below? Is it coincident, parallel, intersecting or skew?
1. (1, -3, 1) + t<1,2,3>
2. (4,-5,-1) + t<-5, 20, 15>
3. (5,5,3) + t<3,-4,1>
4. (-2,11,21) + t<-2,3,-2>
5. (-1,-3,2) + t<-3,12,9>

I have gotten as answers:
1. skew
2. parallel
3. skew
4. intersecting
5. parallel
it says that one of my answers is not correct. Any help would be greatly appreciated.

Answer
Questioner:Rob
Country:Quebec, Canada
Category:Advanced Math
Private:No
Subject:Linear algebra
Question:

I am having difficulty understanding what to do for these questions.  

What is the relationship of the given line
L: (x,y,z) = (-2,1,5) + t<2,-8,-6>

******* changed to

L: (x,y,z) = (-2,1,5) + t<1,-4,-3>

by removing the factor of 2.


to each of the lines below? Is it coincident, parallel, intersecting or skew?
1. (1, -3, 1) + t<1,2,3>
2. (4,-5,-1) + t<-5, 20, 15>
*******
  (4,-5,-1) + t<1, -4, -3>  <<< remove -5 factor

3. (5,5,3) + t<3,-4,1>
4. (-2,11,21) + t<-2,3,-2>
5. (-1,-3,2) + t<-3,12,9>
********
   (-1,-3,2) + t<1,-4,-3>  <<<< remove -3

I have gotten as answers:
1. skew
2. parallel
3. skew
4. intersecting
5. parallel
it says that one of my answers is not correct. Any help would be greatly

appreciated.
............................................................
Each line has a direction vector (a DV) and a point on the line.

If two lines have 'different' DV's, they are either skew or intersecting.  If they have a common point, they intersect.
.........................................
1. L and L1 have different DV's.

Set L1 equal to L for the first.
L: (x,y,z) = (-2,1,5) + t<1,-4,-3>
L1: (x,y,z) = (1, -3, 1) + t<1,2,3>

These have different DV's.  Now is there a point on both?  There would have to be a parameter for each that gives the same point.

(-2,1,5) + t<1,-4,-3> = (1, -3, 1) + s<1,2,3>   <<<<< use 's' for the second.


Now solve these three equations for s,t:

 t - 2 = s + 1     (A)
-4t + 1 = 2s - 3    (B)
-3t + 5 = 3s + 1    (C)

If you solve for s,t using A,B, then test your solution in C.  If it works, say 'intersecting'; if not, say 'skew'.
.......................
You will do the same for 3,4.

............................................
If two lines have 'the same' DV, they are either parallel or the same (coincident).  If they have a common point, they coincide.

Clearly L2 and L5 have the same DV as L.  Do the same:

Using t,s as parameters, write your three equations A,B,C; solve A,B for t,s; test in C.


NOTE: After I see you have read this, I will consider your other question.

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