1
$\begingroup$

Given two skew lines defined by 2 points lying on them as $(\vec{x}_1,\vec{x}_2)$ and $(\vec{x}_3,\vec{x}_4)$. What are the vectors for the two points on the corrwsponding lines, distance between which is minimum? That distance is thus but what are the points where it is achieved?

$\endgroup$
  • 1
    $\begingroup$ The title is misleading or confusing, as skew lines do not intersect :) You're asking about the distance between the skew lines, or the points on the respective skew lines that are closest to one another. $\endgroup$ – Ted Shifrin Jun 29 '15 at 22:01
  • $\begingroup$ @TedShifrin I meant the points on the respective skew lines that are closest to one another. $\endgroup$ – user_1_1_1 Jun 29 '15 at 22:03
  • 1
    $\begingroup$ Yup, much better title; thanks. See if my hint helps. $\endgroup$ – Ted Shifrin Jun 29 '15 at 22:04
  • $\begingroup$ See math.stackexchange.com/questions/1414285/…. $\endgroup$ – Ujjwal Rajput Mar 18 '16 at 10:45
  • $\begingroup$ See math.stackexchange.com/questions/1414285/…. $\endgroup$ – Ujjwal Rajput Mar 18 '16 at 10:53
2
$\begingroup$

HINT: You're looking for points $P$ and $Q$ on the respective lines for which the vector $\overrightarrow{PQ}$ will be orthogonal (perpendicular) to both lines. (To understand why, think about the hypotenuse of a right triangle.) So, for starters, you want a vector orthogonal to both $\vec x_2-\vec x_1$ and $\vec x_4-\vec x_3$.

$\endgroup$
  • $\begingroup$ Yup that hint completely solves my problem thanks, but i will wait to see if someone else actually solves and answers $P,Q$ in terms of $(\vec{x}_1,\vec{x}_2)$ and $(\vec{x}_3,\vec{x}_4)$? $\endgroup$ – user_1_1_1 Jun 29 '15 at 22:19
  • 1
    $\begingroup$ Well, work it out! Parametrize the lines by $\vec\alpha(t)=\vec x_1+t(\vec x_2-\vec x_1)$ and $\vec\beta(s)$ analogous;y, and solve for the values of $s,t$ for which $\vec\alpha(t)-\vec\beta(s)$ is orthogonal to both $\vec x_2-\vec x_1$ and $\vec x_4-\vec x_3$. With numbers this won't be too bad. With all these letters, a bit less fun. $\endgroup$ – Ted Shifrin Jun 29 '15 at 22:22

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.