- #1
deanchhsw
- 14
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* 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.
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.