Distance between two parametric lines

Question

Compute the shortest distance between the following two parametric curves,

$\vec{r_1(t)}$ = $\langle -1+2t, 4-t, 2\rangle$

$\vec{r_2(t)}$ = $\langle 3-2t, 5+t, -1+3t\rangle$

I think this can be done by simply minimizing the distance function and solving for $t$, but I was wondering if it is any way possible to do this by projection? I've only ever used the projection method for computing the distance between two skew lines given by two independent parameters, so I don't know if this would be completely wrong in this case, where both curves are given in terms of $t$. If so, why? And is there any other approach to this problem other than direct differentiation?

Answer 1

You can find the perpendicular vector, then project any line from a point on $r_1$ to a point on $r_2$ onto it. The length of this projection will be the minimum distance between the two parametric.

Yet another approach is to find the parallel planes that the parametrics lie on, then find the distance between these planes.

Answer 2

Just a hint

The square of the distance between the point $A (t) $of curve one and the point $B (u) $ of curve two is $$f (t,u)=( 2 (t+u)-4 )^2+(-(t+u) -1)^2+(3-3u )^2$$

make $f'_t=f'_u=0$ to find the minimum

Answer 3

Standard procedure for two lines $\vec r_1(t)=\vec a +t\vec v$ and $\vec r_2(s)=\vec b+s \vec v$ is to calculate the absolute value of $$\frac{\langle \vec a-\vec b,\vec u\times \vec v\rangle}{\|\vec u\times \vec v\|}.$$ Do you see why?