Intersection of Lines in Space

[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?

 

[Windwaker2004]

*bump*...

 

[wais]

Yes, that is correct. In order for the lines to not intersect, the direction vectors must be parallel, which means they must be scalar multiples of each other. In this case, the second direction vector is a scalar multiple of the first direction vector at any value of k. Therefore, any value of k will result in parallel lines and the lines will never intersect.