# Lines in R^3 space

jonney
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

transphenomen
You cannot add 960 + t(0,1,-1) since you cannot add a scalar and a vector meaningfully.

spamiam
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.

Staff Emeritus
Homework Helper
Moderator's note:

Before offering more help, let's let the OP respond with an attempt at solving the problem.

jonney
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

Mentor
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.

jonney
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).

Mentor
You can check this for yourself. Does your line intersect the two given lines (the ones in your first post)?