Question

help?

Answer 1

I don't remember vectors very well.
Lemme see if this makes sense...

Answer 2

" its parallel for sure. To see if it's coincident, set the x equal to each other and then the y. Then set the z to each other to see if it's equal. "

-One of my classmates

Answer 3

If we examine t=0 and t=1,\[\large\rm <1,2,0>\qquad\qquad <-5,~11,-3>\]We get these two vectors pointing to our parameterized line.
(Is that a word? Parameterized? Probably not XD lol)

Let's examine the change in the vectors,\[\large\rm m=\frac{<-5,~11,-3>-<1,2,0>}{1-0}=<-6,9,-3>\]And our starting point is t=0, seems legit.
So line 1 can be written as\[\large\rm \ell_1=<-6,9,-3>t+<1,2,0>\]

Answer 4

Hmm ok, then maybe my way is a little bit more tedious.

Answer 5

whatever works :P

Answer 6

If you do that same process for line 2,
you would get this equation,\[\large\rm \ell_2=<2,-3,~1>t+<2,3,0>\]

Answer 7

And since \(\large\rm <-6,9,-3>=-3<2,-3,1>\)
That tells us that ummmm

Answer 8

Nah I dunno what I'm doing -_-

Answer 9

haha no worries zep, you tried. Once I understand how to do this stuff ill explain it to you so you can teach other people

Answer 10

haha nice XD

Answer 11

Ok ok I guess the way you're supposed to do it is by looking at the "direction vectors". We get these values from the coefficients of our parameterizations.\[\large\rm x=1\color{orangered}{-6}t,\qquad y=2+\color{orangered}{9}t,\qquad z=\color{orangered}{-3}t\]So the direction vector for this first line is \(\large\rm <-6,9,-3>\) ya?

Answer 12

The other being \(\large\rm <2,-3,1>\)

Answer 13

One is a scalar multiple of the other.
Since our direction vectors are proportional,
the lines are parallel.

Ok ok ok I think this is making a little sense...

Answer 14

If the lines intersected,
then there would be some x value for which they are the same,
\(\large\rm 1-6t=2+2u\)
there would be some place where the y's are the same as well,
\(\large\rm 2+9t=3-3u\)
and also some place where the z's are equivalent,
\(\large\rm -3t=u\)

Solve the system for t and u.
If no solution exists, then they are not coincident.

Answer 15

@legomyego180 do it! >:O lol

Answer 16

@zepdrix, sorry! fell asleep after I thought you had given up last night lol. Looking back over it now...

Answer 17

Makes sense, just set them equal to each other and solve for the respective variables.
If they are equal they are parallel and coincident, and if not then they are something else.