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Lines in R^3 space

  • Thread starter jonney
  • Start date
  • #1
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given two lines x=s(1,2,1) and x=(960)+t(0,1,-1)
how would i go about finding a vector that intersects both lines and then finding another line parallel to that vector? help would be much appreciated
 

Answers and Replies

  • #2
You cannot add 960 + t(0,1,-1) since you cannot add a scalar and a vector meaningfully.
 
  • #3
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I have a feeling that that is supposed to be (9, 6, 0). In any case, to find a vector that intersects both lines, just pick a point on each line, and then find the line connecting the two.
 
  • #4
Redbelly98
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Moderator's note:

Before offering more help, let's let the OP respond with an attempt at solving the problem.
 
  • #5
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here is what i have come up with is this right

general equation for line 1 x-y+z=0, parametric equation x=s y=2s z=s
general equation for line 2 x+y+z=15, parametric equation x=9 y=6+t z=-t

then created a line with the vector (1,-1,1) going through the origin.
x =(0,0,0) +v(1,-1,1)
therefor general equation is x+2y+z=0, parametric x=v y=-v z=v

crossing of line 1 and line 3
s+4s+s=0
6s=0
s=0
therefore x=0 y=0 z=0

crossing of line 2 and line 3
v-v+v=15
v=15
therefore x=15 y=-15 z=15

so the third line crosses line one at (0,0,0) and line 2 and (15,-15,15)
is this correct???
appreciate all the help
 
  • #6
33,513
5,195
here is what i have come up with is this right

general equation for line 1 x-y+z=0, parametric equation x=s y=2s z=s
general equation for line 2 x+y+z=15, parametric equation x=9 y=6+t z=-t
Your general equations above are equations of planes, not lines. Your parametric equations represent the lines, though.

BTW, in your original post you said
jonney said:
... x=(960)+t(0,1,-1)
Without commas, the 960 part was confusing to several posters.


then created a line with the vector (1,-1,1) going through the origin.
x =(0,0,0) +v(1,-1,1)
therefor general equation is x+2y+z=0, parametric x=v y=-v z=v

crossing of line 1 and line 3
s+4s+s=0
6s=0
s=0
therefore x=0 y=0 z=0

crossing of line 2 and line 3
v-v+v=15
v=15
therefore x=15 y=-15 z=15

so the third line crosses line one at (0,0,0) and line 2 and (15,-15,15)
is this correct???
appreciate all the help
All you need to do is find one point on each line, and then construct a vector from one point to the other. By inspection, (0, 0, 0) is a point on line 1, and (9, 6, 0) is a point on line 2.

Now, form a vector from one point to the other, and then find the equation of the line with this direction that goes through, say, (0, 0, 0). That's what I would do.
 
  • #7
3
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Hey thanks heaps for that, just one more question how do i know that this line is parallel to the vector.
I got x = (0,0,0)+t(9,6,0) for the vector equation of the line does this look right.
also sorry about the confusion on the (960) when it should have been (9,6,0).
 
  • #8
33,513
5,195
You can check this for yourself. Does your line intersect the two given lines (the ones in your first post)?
 

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