
I have a homework question and I dare not put it in general. Thanks in advance if you help me out! I need to find the point where two lines intersect in space, and then find the plane that they determine. Here are the lines: x = t y = t + 2 z = t + 1 and x = 2s + 2 y = s + 3 z = 5s + 6
I tried making making the equations equal but it didn't work: x: t = s + 2 y: t + 2 = 2s + 3 z: t + 1 = 5s + 6
I tried solving for s and t (since s and t should be the same for all the eqations at their point of intersection), but I get different s's and t's for z than i did for x and y. help me please!
EDIT: Nevermind I wrote down the wrong info. Man I feel stupid! =(
EDIT 2: I put the right info in there, I still can't figure it out.

Unless we're both making calculation errors or the info is wrong those lines do not intersect. They are not parallel since their direction vectors (check out the coefficients of t/s) are (1, 1, 1) and (1, 1, 5) are not scaler multiples of eachother. So they are skew lines, and furthermore do not define a plane.
Trick question maybe? Good luck!

fuck i messed it up again! do you think my approach is correct though?

I get the same result. I'm pretty sure they don't intersect or your correction was wrong?
On January 30 2008 13:18 Meta wrote: fuck i messed it up again! do you think my approach is correct though?
Ya it's right for the first part at least.

If I understand what you're doing then yes.
What I did goes like this:
t=s+2 and t+2=s+3 so from this we can solve for s and t as its 2 equations and 2 unknowns. So I got t=3/2 and s=1/2. Then I subbed those values into the third equation t+1=5s+6 and got that 3/2=3/2, which is gibberish clearly. So the lines don't intersect. It looks like you did something similar ^^
Edit: looking at what you entered for the info there is an inconsistancy, though it doesn't change the outcome heh. You have: x=t and x=2s+2 and then later you do t=s+2, missing that 2s.

for the plane they determine get your direction numbers for each line <1,1,1> and <2,1,5> take the cross product of that and plug it into the equation where (xo, yo, zo) is the point of intersection and <a,b,c> is the cross product of the two vectors a(xxo)+b(yyo)+c(zzo)=0

okay, solving the system of equations, s = 1, t = 0 plugging it in x=0, y=(0)+2, z=(0)+1, you get the point (0,2,1) so that is the point of intersection
the cross product of the direction numbers <1,1,1> x <2,1,5> is <6,3,3> (check this... i hate doing cross products and always make stupid errors on them) edit: yup... stupid error...
so the equation of your plane would be 6x3(y2)+3(z1)=0 or 6x3y+3z=3

Nvm I didn't read properly XD

Actually it does, I dunno what I was doing. x=t=2s+2=0=2(1)+2 s= 1, t=0 works. Well I'm a fool ^^

yay pwn but seriously... where are you taking this i'm taking it in high school right now and i thought we were moving really slow some of my college friends are around 3 chapters ahead of me and we did this last semester...

I did this in highschool in my last year, but again in first year university, theres a lot of repetition I find. Thought I know you meant him
So here is what I got along the way... Finish the problem first! + Show Spoiler +point of intersection=(0, 2, 1) normal vector=(6, 3, 3) plane: 6x3y+3z+3=0

On January 30 2008 13:33 Zortch wrote: Actually it does, I dunno what I was doing. x=t=2s+2=0=2(1)+2 s= 1, t=0 works. Well I'm a fool ^^
I think you did the same thing as me and looked at the equations he formed rather than the information given at the start.

On January 30 2008 13:41 Zortch wrote:I did this in highschool in my last year, but again in first year university, theres a lot of repetition I find. Thought I know you meant him So here is what I got along the way... Finish the problem first! + Show Spoiler +point of intersection=(0, 2, 1) normal vector=(6, 3, 3) plane: 6x3y+3z+3=0
if you look up, i did do the problem... how did you get your normal vector

I just used the cross product. It is really easy to check, simply take the dot product with each of the two line vectors and make sure they are both zero. Dot product is nice and easy ^^ sum of the product of corresponding components.

On January 30 2008 13:58 Zortch wrote: I just used the cross product. It is really easy to check, simply take the dot product with each of the two line vectors and make sure they are both zero. Dot product is nice and easy ^^ sum of the product of corresponding components. and inevitably, i made a stupid mistake on the cross product... _ god i hate taking cross products

On January 30 2008 13:37 Saracen wrote:yay pwn but seriously... where are you taking this i'm taking it in high school right now and i thought we were moving really slow some of my college friends are around 3 chapters ahead of me and we did this last semester...
i'm a second semester freshman in college. we covered this stuff in our second week of class.. my high school didn't offer past calc 1



