Find the points that give the shortest distance between the line $(2,3,1)+s(1,2,-1)$ and $(1,2,0)+t(2,-3,5)$ using lagrange multipliers

The problem is that these lines aren't giving in the form like $x+y=2$, or something like $x^2+y^2=4$

They aren't defined "via an equation", and so I am unsure how to do this problem.

All the questions on MSE have lagrange multipliers of equations, something like Find points that give the shortest distance between $y = x^2$ and $y-x+2=0$ using Lagrange multipliers

So I don't even know where to start here?

We know the distance formula squared is like $(x_\text{on line 1}-x_\text{on line 2})+(y_\text{on line 1}-y_\text{on line 2})+(z_\text{on line 1}-z_\text{on line 2})$

  • $\begingroup$ The problem's been answered, so I just want to comment that the problem is actually easier given in parametric form like this. You don't need to use Lagrange multipliers here. $\endgroup$ – Dylan Nov 7 '17 at 4:41

$d^2 = ((2+s) - (1+2t))^2 + ((3+2s) - (2-3t))^2 + ((1-s) - (5t))^2\\ d^2 = (1+s-2t)^2 + (1+2s+3t)^2 + (1-s - 5t)^2\\ d^2 = 38t^2 + 6s^2 + 3 -8t +4s+ 18 st$

$s, t$ are unconstrained, so there is no further constraint to apply.

Take the partials with respect to $s, t$ and set them equal to $0.$

$76 t + 18 s - 8 = 0\\ 12 s + 18 t + 4 = 0$

and that is a system of linear equations.

| cite | improve this answer | |
  • $\begingroup$ Wouldnt the constraint be that the point must be one the line 1 and line 2 repsectively? $\endgroup$ – K Split X Nov 7 '17 at 4:35
  • $\begingroup$ $(2+s, 3+2s, 1-s)$ is the set of points on line 1, and $(1+2t, 2-3t, -5t)$ is the set of points on line 2. The distance formula applied gives the distance between any two points on the respective lines. The lines are infinite, so there are no further constraints. $\endgroup$ – Doug M Nov 7 '17 at 4:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.