Question: Given the vector equations of 2 lines $$r_1 =(-1,3,0)+ k(2,1,-3)$$ and $r_2 =(4,0,1)+ t(-4,-2,6)$. Show that $r_1$ and $r_2$ are distinct parallel lines.
Answer: It is parallel because $(-4,-2,6) = -2 (2,1,3).$
BUT if I find the $k$ and $t$, they are all different.
$k$ is $5/2, -3$ and $-1/3$, while $t$ is $-3/4, -3/2$ and $-1/6.$ None of them match up so does this mean that the lines are parallel but not distinct?
Thank you very much!
Yes, that is exactly what it means. If they were not distinct, then you can choose any value of one of the parameters, say $k=0$, and the resulting vector $r_1(0)=\pmatrix{-1 \\ 3 \\ 0}$ lies on the other line for some (single!) value of $t$, i.e. $r_1(0)=r_2(t)$ should have a unique solution in $t$. Since this is not the case, the two lines are distinct.