# Parallel Vectors Problem

1. Mar 6, 2005

### deanchhsw

* This post was moved from the general math section by the poster himself

There are two lines:

L1 : x = 1 + 2t ; y = 2 - t ; z = -1 + 3t.
L2: x = 2 - 3m ; y = 2m ; z = 1 - m.

The problem states to find a general point on A on L1 and a general point B on L2, and then find the vector AB from those points.

Hence,

Vector AB = ( -3m - 2t - 1 , 2m + t - 2, -m -3t +2).

Then, the problem states to find specific points A and B such that vector AB is parallel to the product of direction vectors of lines L1 and L2.

using determinants, the product of direction vectors come out to be (-5,-7,1). I'm pretty sure it is. This also means that direction vector of vector AB has to be either (-5, -7, 1) or scalar multiples of it. But when I equate

-3m-2t-1 = -5
2m + t - 2 = -7
-m -3t +2 = 1

and do simultaneous equation, the value for m and t doesn't come out to be right. if it works for x and y, it doesnt' work for z, and so on.

…..additional info after not getting the first one responded

Here's what I meant by product of direction vectors.
direction vector L1 = (2,-1,3)
direction vector L2 = (-3,2,-1)

Thus D.V. L1 * D.V. L2 = det( i j k ) = (-5,-7,1)
......................................( 2 -1 3 )
..................................... ( -3 2 -1 )

forgive me for using parenthesis where abstract value sign should be.
At any rate, that's what I meant by product.

if you pictured it correctly that it's the "cross product", that's what I originally figured: that since product of direction vector is perpendicular to the lines connecting the two "general points", it cannot be parallel. but I can't imagine that the problem is flawed because it is coming straight out of the infamous International Baccalaureate internal assessment sheet! So, here's the dilemma. I think I'm interpreting the problem in a wrong direction.

Since I really can't see what to do, I'll just present the full problem here:

Consider the two lines:

L1 : x = 1 + 2t ; y = 2 - t ; z = -1 + 3t.
L2: x = 2 - 3m ; y = 2m ; z = 1 - m.

4) Given that l-1 and l-2 are direction vectors for lines L1 and L2, find the vector product l-1 x l-2.

5) Taking a general point A on L1 and a general point B on L2, find the vector AB.

6) Find points A and B such that AB is parallel to l-1 x l-2.

7) Find the magnitude of vector AB.

....

As noted before, vector product, if I did things correctly, should be (-5,-7,1).
Number 5's answer, I believed, was
Vector AB = ( -3m - 2t - 1 , 2m + t - 2, -m -3t +2).
As I stated before in the question.

Number 6 is the point where I have trouble with- the way it is worded, it sounds as if two points each in Line L1 and L2 are supposed to be parallel to their direction vector products. I stared at the problem for 20 minutes, and yelled out, "THIS DOES NOT MAKE SENSE."

I'm really sorry that I had to resort to actually presenting a problem, but
please understand that I tried my best and even more.

Thank you so much.

2. Mar 7, 2005

### HallsofIvy

Staff Emeritus
Yes, it was posted in general math- and I answered it there- but I answered it wrong! I assumed that the two lines intersected and asserted that any such vector AB must lie in the plane defined by the two lines and so could not be parallel to the cross product which is perpendicular to that plane.

HOWEVER, in this case the two lines are skew. They do not intersect nor are they parallel and so they do not define a plane.

You are correct that the cross product vector is -5i- 7j+ k.
You are also correct that a vector between two points A and B on each of the lines must be of the form (-3m - 2t - 1)i+ (2m + t - 2)j+ (-m -3t +2)k.

Now, in order for two vectors two be parallel one must be a multiple of the other. In other words, you must have -3m- 2t- 1= -5p, 2m+ t- 2= -7p, -m-3t+ 2= p for some number p. That is 3 equations for the 3 variables m, t, p. You don't really need to know p but m and t will give you the points A and B.

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