Let $s \equiv$ $x=t, y=\left(\frac{t}{3}\right)+2$ and $z=\left(\frac{bt}{3}\right)+1$ $$$$ $r \equiv x=2bu-4b+1, y=u, z=2u-4$
Question: determine if there exists any value(s) of $b$ s.t. $r$ and $s$ are skew. If such value(s) exist then find the distance between $r$ and $s$.
Attempt: I did the first part and found that $r$ and $s$ are only co-planar if $b=1$ which means they are skewed when $b\neq 1$ $$$$ The direction vectors of $r$ & $s$ are $$d_s=(1,\left(\frac{1}{3}\right),\left(\frac{b}{3}\right))$$ and $$d_r=(2b,1,2)$$ respectively. So I worked out the normal vector and got $$n=(\left(\frac{2-b}{3}\right),\left(\frac{2b^2-6}{3}\right),\left(\frac{3-2b}{3}\right))$$ with magnitude $\mid n \mid=\left(\frac{\sqrt{4b^4-19b^2-16b+49}}{3}\right)$
Since the shortest distance is the perpendicular distance than the line with $n$ as a directional vector will cross both $r$ & $s$ at some points and the length of that point will be the shortest distance.
Thi is where I keep getting stuck. If I use the general points $P_s=(t,\left(\frac{t}{3}\right)+2,\left(\frac{bt}{3}\right)+1)$ and $P_r=(2bu-4b+1,u,2u-4)$ then things get very messy with $u$ and $t$.
$\dot{\vec{P_sP_r}}=(2bu-4b-1-t,u-\left(\frac{t}{3}\right)-2,2u-4-\left(\frac{bt}{3}\right)-1)$
Do I just multiply $\dot{\vec{P_sP_r}}$ by $\left(\frac{n}{\mid n \mid}\right)$ even though I'll still have $u$ and $t$ everywhere or is there a simplification that I am missing? $$$$Note: I know that this is similar to two other questions here but none have the case where there is variable so this is not a duplicate.
