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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?

John
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  • Be careful. The two $t$s are dummy parameters and do not represent the same value in both lines. When you formulate the distance between points on the two lines, use two different variables, otherwise you’ll end up with a bogus solution. – amd May 20 '17 at 00:53
  • But this is the exact form in which the question was given. The second part even asks to find the value of $t$ at which the curves are closest. How I am supposed to interpret this when the question doesn't seem to hint that the parameters are different at all. It seems very odd to me because, as you say, we usually change the letter of the parameter so as avoid confusion when defining the lines. The differentiation idea seems like the only one that makes sense, although I've never actually seen it used in answering these kinds of distance questions. – John May 20 '17 at 01:23
  • If the parameters are indeed coupled, then differentiating the distance function is the way to go. – amd May 20 '17 at 04:23

3 Answers3

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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.

Isaac Browne
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  • The projection approach yields $D$ = $6/\sqrt5$. Yet the differentiation approach yields a minimum $t=23/29$, which when you plug into the equations, solve for what are presumably closest points, and then use the distance formula, you arrive at $D$ = $\sqrt{3075}/29$. Why the difference? Or must I be making some computation error? Also, the projection vector I obtained was $\langle 6/5, 12/5, 0\rangle$, but if I try and solve for the closest points from this vector I cannot find a unique value for $t$ which would give me two appropriate points. How do you make sense of this? Thanks. – John May 19 '17 at 15:02
  • I think it is just a computation error, for these methods are correct. – Isaac Browne May 19 '17 at 16:42
  • Note: the easiest way to find the perpendicular vector that @IsaacBrowne talks about is to take two vectors along these two lines and calculate their cross product. – fractal1729 May 19 '17 at 19:43
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    @John See my comment to your question above. The $t$s in the two parametric equations are different and you need to use different parameters when formulating the distance between points on those lines. – amd May 20 '17 at 00:54
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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

hamam_Abdallah
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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?

Michael Hoppe
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  • Yes, but the issue here is with the parameters. If the first curve were defined in terms of $t$ and the second in terms of $s$, this would be a familiar problem and one could use the standard projection method yielding $D = 6/\sqrt{5}$. But since they are both defined in terms of $t$ I chose to minimize the distance function instead, finding a minimum value for $t$ of $23/29$. Now if I plug this value of $t$ into the two curves to obtain two points and calculate the distance between them, the distance is smaller at $\sqrt{3075}/29$. I'm asking why, and whether it is related to the parameters. – John May 19 '17 at 21:18
  • @John I believe that, unless you have information to the contrary, there’s no reason to think that the $t$s in the two parameterizations are the same. Indeed, equating them is a common error that you can find in many questions here. On the other hand, if these were motions of equation and they were synchronized via a time parameter $t$, that would be a different story. – amd May 20 '17 at 00:58
  • @John But as usual the parameter in each equation are private. – Michael Hoppe May 20 '17 at 09:28