Hi! I am working on some online homework for my calc2 class and I am having trouble with this problem. I first set $r_1$ and $r_2$ equal to one another to get $(-1-4t, 2+2t, -14+2t)=(-13+4t, 8-2t, -12+2t)$. I then equated the corresponding components to get $-1-4t=-13+4t$, $2+2t=8-2t$, and $-14+2t=-12+2t$. I then solved for $t$ and calculated a value of $1.5$. Once I had $t=1.5$ I plugged it back into the right hand side of $(-1-4t, 2+2t, -14+2t)=(-13+4t, 8-2t, -12+2t)$ to get $(-13+4(1.5), 8-2(1.5), -12+2(1.5))=(-7,5,-9)$. I thought that the pint of intersection would be the terminal point of the vector which would be $(-7,5,-9)$, but that is coming up as incorrect. I am not quite sure what I did wrong, but if someone can help me figure out where I went wrong I would really appreciate it.
You (both) got confused by notation. Instead, consider $(−1−4t,2+2t,−14+2t)=(−13+4s,8−2s,−12+2s)$
It's nominally the same $t$ in both equations, but the lines may (or may not) intersect at different values of their parameters.
So, solve the system above for a pair of $(t,s)$.