# Homework Help: Intersection of Lines in Space

1. Nov 7, 2004

### Windwaker2004

Hi, I need some help with this question:

Find all values of $$\ k$$ for which the lines do not intersect.

$$\ (x-2,y+1,z-3) = (r,0,3r)\ and\ (x,y,z) = (2,1,4)\ +\ s(2,k,6)$$

I put the first equation in vector form:

$$\ (x,y,z) = (2,-1,3)\ +\ r(1,0,3)$$

Now I know that if the direction vectors are scalar multiples of one another, they are parallel lines and therefore do no intersect...

$$\ d_1 = (1,0,3)\ and\ d_2 = (2,k,6) \ \ d_1 = t(d_2)\ therefore...\ (1,0,3) = t(2,k,6)\ since\ 1 = t2,\ t = 1/2 \ then\ 0 = 1/2(k),\ therefore\ k=0$$

The second direction vector is a scalar multiple of direction vector 1 at any scalar k?

Last edited: Nov 7, 2004
2. Nov 7, 2004

### Windwaker2004

*bump*........