point of intersection

[cutepiglet]

Determine if the lines r1(t)=<3,0,2>+t<1,2,-2> and r2(s)=<0,1,-1>+s<4,1,1> intersect, and if they do at what point.

I have worked on the problem for a long time and I believe this is the right answer. I am taking cal 3 and it has been 7 years since my last class. I do not remember 1 and 2 calculus and had a very confusing teacher.

This is what i have, please help me if i completely missed it.

<3+t,2t,2-2t>=<4s,1+s,-1+s>
3+t=4s
2t=1+s
2-2t=-1+s


used the first equation
3+t=4s
t=4s-3

inserted the equation in the second
2(4s-3)=1+s
8s-6=1+s
8s=7+s
7s=7 (this is the part I am not sure i can legally do in math)
s=1
Plugged in the third
2-2t=-1+1
-2t=-2
t=1

So they do intersect. But I do not understand the question of where they intersect. It is more than likely a easy answer that I am completely missing because my brain is scrambled.

 

[Deleted member 4993]

Determine if the lines r1(t)=<3,0,2>+t<1,2,-2> and r2(s)=<0,1,-1>+s<4,1,1> intersect, and if they do at what point.

I have worked on the problem for a long time and I believe this is the right answer. I am taking cal 3 and it has been 7 years since my last class. I do not remember 1 and 2 calculus and had a very confusing teacher.

This is what i have, please help me if i completely missed it.

<3+t,2t,2-2t>=<4s,1+s,-1+s>
3+t=4s
2t=1+s
2-2t=-1+s


used the first equation
3+t=4s
t=4s-3

inserted the equation in the second
2(4s-3)=1+s
8s-6=1+s
8s=7+s
7s=7 (this is the part I am not sure i can legally do in math)
s=1
Plugged in the third
2-2t=-1+1
-2t=-2
t=1

So they do intersect. But I do not understand the question of where they intersect. It is more than likely a easy answer that I am completely missing because my brain is scrambled.

At the point of intersection

r1(t)=<3,0,2>+t<1,2,-2> → r1(1)=<3,0,2>+<1,2,-2> = <4,2,0>

r2(s)=<0,1,-1>+s<4,1,1> → r2(1)=<0,1,-1>+<4,1,1> = <4,2,0>

Now where do you think the point of intersection is.....