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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.

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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