- #1

- 3

- 0

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

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter jonney
- Start date

- #1

- 3

- 0

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

- #2

- 52

- 0

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

- #3

- 360

- 0

- #4

- 12,121

- 160

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

- #5

- 3

- 0

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

Mark44

Mentor

- 34,885

- 6,624

Your general equations above are equations of planes, not lines. Your parametric equations represent the lines, though.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

BTW, in your original post you said

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

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

- 0

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

Mark44

Mentor

- 34,885

- 6,624

Share: