Given two lines: r 1 ( t ) = ( a 1 , b 1 , c 1 ) + t ( p 1 , q 1 , r 1 ) r 2 ( s ) = ( a 2 , b 2 , c 2 ) + s ( p 2 , q 2 , r 2 )I have calculated the cross product of the direction vectors to get the vector perpendicular to both lines: n → = ( p 1 , q 1 , r 1 ) × ( p 2 , q 2 , r 2 )Which gives the vector n → .I then got a vector V → between a point on each line, and calculated the minimum distance d between both lines by finding the scalar projection of V → onto n → d = | V → ⋅ n → | | n → | I now need the points on both lines where this minimum distance is achieved, how would I go about doing that?

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Given two lines:      r    1    (  t  )  =  (      a    1    ,      b    1    ,      c    1    )  +  t  (      p    1    ,      q    1    ,      r    1    )      r    2    (  s  )  =  (      a    2    ,      b    2    ,      c    2    )  +  s  (      p    2    ,      q    2    ,      r    2    )I have calculated the cross product of the direction vectors to get the vector perpendicular to both lines:            n      →        =  (      p    1    ,      q    1    ,      r    1    )  ×  (      p    2    ,      q    2    ,      r    2    )Which gives the vector             n      →       .I then got a vector             V      →        between a point on each line, and calculated the minimum distance d between both lines by finding the scalar projection of             V      →       onto             n      →        d  =                    |                              V          →                    ⋅                        n          →                            |                            |                              n          →                            |            I now need the points on both lines where this minimum distance is achieved, how would I go about doing that?

Jakob Chavez · Accepted Answer

L    1    (  t  )  =  (      a    1    ,      b    1    ,      c    1    )  +  t  (      p    1    ,      q    1    ,      r    1    )      L    2    (  s  )  =  (      a    2    ,      b    2    ,      c    2    )  +  s  (      p    2    ,      q    2    ,      r    2    )If two points on lines       L    1   and       L    2   are A and B resp.,                    A        B            →        =  (      a    1    −      a    2    +  t      p    1    −  s      p    2    ,        b    1    −      b    2    +  t      q    1    −  s      q    2    ,        c    1    −      c    2    +  t      r    1    −  s      r    2    )to find two points on lines with minimal distance, the vector                     A        B            →       should be perpendicular to both lines. So the dot product of                     A        B            →       to the directional vectors of both lines be zero.                    A        B            →        ⋅  (      p    1    ,      q    1    ,      r    1    )  =  0                    A        B            →        ⋅  (      p    2    ,      q    2    ,      r    r    )  =  0Solve for s and t and that should give you two points. You can also find the minimum distance from there.

lexi13xoxla · Accepted Answer

Hint.Calling      {                                        r            1                    =                      r                          01                                +          t                                                    v                →                                      1                                                            r            2                    =                      r                          02                                +          s                                                    v                →                                      2                                  we have      d    2    (  t  ,  s  )  =      |        |        r    1    −      r    2        |              |        2    =      t    2        |        |                      v        →              1        |              |        2    +      s    2        |        |                      v        →              2        |              |        2    +  2  (      r          01        −      r          02        )  ⋅                    v        →              1    t  −  2  (      r          01        −      r          02        )  ⋅                    v        →              2    s  −  2  s  t                    v        →              1    ⋅                    v        →              2  now the minimum to this quadratic form is attained at the stationary point given by the solution to  ∇      d    2    (  t  ,  s  )  =  0

Finding the points on two lines where the minimum distance is achieved Given two lines: r_1(t)=(a_1,b_1,c_1)+t(p_1,q_1,r_1) r_2(s)=(a_2,b_2,c_2)+s(p_2,q_2,r_2)

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