# Thread: Intersection of Vector Lines

1. ## Intersection of Vector Lines

2. Originally Posted by Del

$\displaystyle r_1 = r_2$
(3,5,-9) + t(0,2,0) = (-1,4,-8)+t(-4,2,1)
(3,5,-9) - (-1,4,-8) = t(-4,2,1) - t(0,2,0)
(4,1,-1) = t(-4,0,1)

No real t can satisfy this. Hence the lines do not intersect.

3. Originally Posted by Del

set each of the coordinates equal to find values for t

x: $\displaystyle 3+0t=-1-4t \iff t=-1$

y: $\displaystyle 5+2t=4+2t \iff 1=0 \mbox{ never true}$

we don't need to check z becuase the lines do not ever intersect because the y coordinates are never equal.

4. The program that I'm inputting my answers to accepted as the correct answer.. I'm confused but thanks!

5. Originally Posted by Del
The program that I'm inputting my answers to accepted as the correct answer.. I'm confused but thanks!
Yes you are right.
I made a dumb mistake of assuming t must be same.Sorry

$\displaystyle (3,5,-9) + t_2(0,2,0) = (-1,4,-8)+t_1(-4,2,1)$

$\displaystyle (3,5,-9) - (-1,4,-8) = t_1(-4,2,1) - t_2(0,2,0)$

$\displaystyle (4,1,-1) = (-4t_1,2t_1 - 2t_2,t_1)$

So $\displaystyle -4t_1 = 4 \Rightarrow t_1 = -1$

And $\displaystyle 2t_1 - 2t_2 = 1 \Rightarrow t_2 = -\frac32$

Thus $\displaystyle (-1,4,-8)+t_1(-4,2,1) = (-1,4,-8) - (-4,2,1) = (3,2,-9)$ is the right answer.

6. I did the same thing sorry