Find the coordinate of two line segments that are perpendicularcalculating perpendicular and angular distance between line segments in 3dPerpendicular line passing through the midpoint of another lineFind points on perpendicular lineAngle between two line segmentsFind two points on two lines in the plane where the line between the two points go through a third point and are equidistant from that pointModelling the difference between intersections of two lines on the circumference of a circleUsing vector math to get point on perpendicular line from a point with the same YFind the point of intersection between a line segment $AC$ and a perpendicular line going through a point $B$ not on $AC$How to find the coordinates of points on a line perpendicular to a given planeShortest distance between skew line *segments*

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Find the coordinate of two line segments that are perpendicular

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Find the coordinate of two line segments that are perpendicular


calculating perpendicular and angular distance between line segments in 3dPerpendicular line passing through the midpoint of another lineFind points on perpendicular lineAngle between two line segmentsFind two points on two lines in the plane where the line between the two points go through a third point and are equidistant from that pointModelling the difference between intersections of two lines on the circumference of a circleUsing vector math to get point on perpendicular line from a point with the same YFind the point of intersection between a line segment $AC$ and a perpendicular line going through a point $B$ not on $AC$How to find the coordinates of points on a line perpendicular to a given planeShortest distance between skew line *segments*













1












$begingroup$


How can the x,y position (actually will need in 3d but for simplicity asking in 2d) of a point be found if it is the intersection of two perpendicular line segments.



I have two points, $p_1$ and $p_2$ that are known $x,y$ locations. I also have line segments with known lengths $a$ and $b$. I made a diagram below to illustrate the problem.



enter image description here










share|cite|improve this question









$endgroup$











  • $begingroup$
    If the lengths $a$ and $b$ are fixed, then you must also have $lVert p_1-p_2rVert^2=a^2+b^2$ for there to be any solution at all. It this condition is meth, then there can be up to four solutions, two if $p_1$ must be an endpoint of $a$ and $p_2$ must be an endpoint of $b$.
    $endgroup$
    – amd
    2 hours ago










  • $begingroup$
    @amd yes p1 and p2 are the endpoints of a and b. I need their mutual endpoint
    $endgroup$
    – user1938107
    2 hours ago















1












$begingroup$


How can the x,y position (actually will need in 3d but for simplicity asking in 2d) of a point be found if it is the intersection of two perpendicular line segments.



I have two points, $p_1$ and $p_2$ that are known $x,y$ locations. I also have line segments with known lengths $a$ and $b$. I made a diagram below to illustrate the problem.



enter image description here










share|cite|improve this question









$endgroup$











  • $begingroup$
    If the lengths $a$ and $b$ are fixed, then you must also have $lVert p_1-p_2rVert^2=a^2+b^2$ for there to be any solution at all. It this condition is meth, then there can be up to four solutions, two if $p_1$ must be an endpoint of $a$ and $p_2$ must be an endpoint of $b$.
    $endgroup$
    – amd
    2 hours ago










  • $begingroup$
    @amd yes p1 and p2 are the endpoints of a and b. I need their mutual endpoint
    $endgroup$
    – user1938107
    2 hours ago













1












1








1





$begingroup$


How can the x,y position (actually will need in 3d but for simplicity asking in 2d) of a point be found if it is the intersection of two perpendicular line segments.



I have two points, $p_1$ and $p_2$ that are known $x,y$ locations. I also have line segments with known lengths $a$ and $b$. I made a diagram below to illustrate the problem.



enter image description here










share|cite|improve this question









$endgroup$




How can the x,y position (actually will need in 3d but for simplicity asking in 2d) of a point be found if it is the intersection of two perpendicular line segments.



I have two points, $p_1$ and $p_2$ that are known $x,y$ locations. I also have line segments with known lengths $a$ and $b$. I made a diagram below to illustrate the problem.



enter image description here







geometry trigonometry






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 3 hours ago









user1938107user1938107

15310




15310











  • $begingroup$
    If the lengths $a$ and $b$ are fixed, then you must also have $lVert p_1-p_2rVert^2=a^2+b^2$ for there to be any solution at all. It this condition is meth, then there can be up to four solutions, two if $p_1$ must be an endpoint of $a$ and $p_2$ must be an endpoint of $b$.
    $endgroup$
    – amd
    2 hours ago










  • $begingroup$
    @amd yes p1 and p2 are the endpoints of a and b. I need their mutual endpoint
    $endgroup$
    – user1938107
    2 hours ago
















  • $begingroup$
    If the lengths $a$ and $b$ are fixed, then you must also have $lVert p_1-p_2rVert^2=a^2+b^2$ for there to be any solution at all. It this condition is meth, then there can be up to four solutions, two if $p_1$ must be an endpoint of $a$ and $p_2$ must be an endpoint of $b$.
    $endgroup$
    – amd
    2 hours ago










  • $begingroup$
    @amd yes p1 and p2 are the endpoints of a and b. I need their mutual endpoint
    $endgroup$
    – user1938107
    2 hours ago















$begingroup$
If the lengths $a$ and $b$ are fixed, then you must also have $lVert p_1-p_2rVert^2=a^2+b^2$ for there to be any solution at all. It this condition is meth, then there can be up to four solutions, two if $p_1$ must be an endpoint of $a$ and $p_2$ must be an endpoint of $b$.
$endgroup$
– amd
2 hours ago




$begingroup$
If the lengths $a$ and $b$ are fixed, then you must also have $lVert p_1-p_2rVert^2=a^2+b^2$ for there to be any solution at all. It this condition is meth, then there can be up to four solutions, two if $p_1$ must be an endpoint of $a$ and $p_2$ must be an endpoint of $b$.
$endgroup$
– amd
2 hours ago












$begingroup$
@amd yes p1 and p2 are the endpoints of a and b. I need their mutual endpoint
$endgroup$
– user1938107
2 hours ago




$begingroup$
@amd yes p1 and p2 are the endpoints of a and b. I need their mutual endpoint
$endgroup$
– user1938107
2 hours ago










3 Answers
3






active

oldest

votes


















2












$begingroup$

Sometimes a figure is worth a 1000 words:



enter image description here



In three dimensions:



enter image description here






share|cite|improve this answer











$endgroup$




















    1












    $begingroup$

    If you draw a circle with diameter $sqrta^2+b^2$ then $p_3$ can be either of two points on that circle.






    share|cite|improve this answer











    $endgroup$












    • $begingroup$
      The lengths $a$ and $b$ are also fixed so this doesn't completely work.
      $endgroup$
      – AHusain
      3 hours ago










    • $begingroup$
      Sorry, this answer should have been a comment.
      $endgroup$
      – steven gregory
      3 hours ago











    • $begingroup$
      sure, but how can I find where that point is on the circle?
      $endgroup$
      – user1938107
      3 hours ago


















    1












    $begingroup$

    • Let $ell$ be the distance between $p_1$ and $p_2$, and let $p_3$ be the point where the two line segments will meet.


    • The three points $p_1, p_2, p_3$ form a triangle whose side lengths we know: $a, b, ell$.


    • We can find out where $p_3$ is by measuring properties of this triangle. Draw a perpendicular altitude from $p_3$ down to the side of the triangle between $p_1$ and $p_2$. Call the intersection point $p_4$.



    • If we knew the length $h$ of this altitude, we would know where to put $p_3$. First, we could find out where $p_4$ is. By the Pythagorean theorem, the distance between $p_1$ and $p_4$ is $sqrta^2 - h^2$. So
      $$p_4 = p_1 + fracsqrta^2-h^2ell (p_2 - p_1)$$



      Next, we would use $p_4$ to find out where to put $p_3$. We start from $p_4$ and travel a distance $h$ in a direction perpendicular to the $p_1$ $p_2$ base of the triangle.



      In two dimensions, we have two choices of perpendicular direction. In three dimensions, we actually have an infinite number of perpendicular directions[*], so you'll have to pick one. Pick a unit vector $widehatu$ in a perpendicular direction.



      Then $p_3 = p_4 + hcdot widehatu$, which gives you the answer you want.



    • But how do we find $h$? We can find $h$ if we know the area of the triangle. Heron's formula lets you calculate the area of the triangle when you know the length of all the sides. $A = sqrts(s-a)(s-b)(s-ell)$, where $s$ is half the perimeter $s = (a+b+ell)/2$. Once you have the area, you can calculate the height of the altitude: $A = frac12textbasetimestextheight$, so $h = 2A/ell$.


    [*] In 3D, you have an infinite number of perpendicular directions. Imagine $p_1=langle 0,0rangle$ and $p_2 = langle 1,0rangle$, and $a$ and $b$ are some numbers. There's a satisfactory point $p_3$ in the x-y plane. But if you spin $p_3$ around the $x$-axis, you find an infinite number of other satisfactory points too.






    share|cite|improve this answer









    $endgroup$













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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      Sometimes a figure is worth a 1000 words:



      enter image description here



      In three dimensions:



      enter image description here






      share|cite|improve this answer











      $endgroup$

















        2












        $begingroup$

        Sometimes a figure is worth a 1000 words:



        enter image description here



        In three dimensions:



        enter image description here






        share|cite|improve this answer











        $endgroup$















          2












          2








          2





          $begingroup$

          Sometimes a figure is worth a 1000 words:



          enter image description here



          In three dimensions:



          enter image description here






          share|cite|improve this answer











          $endgroup$



          Sometimes a figure is worth a 1000 words:



          enter image description here



          In three dimensions:



          enter image description here







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 1 hour ago

























          answered 2 hours ago









          David G. StorkDavid G. Stork

          12.3k41836




          12.3k41836





















              1












              $begingroup$

              If you draw a circle with diameter $sqrta^2+b^2$ then $p_3$ can be either of two points on that circle.






              share|cite|improve this answer











              $endgroup$












              • $begingroup$
                The lengths $a$ and $b$ are also fixed so this doesn't completely work.
                $endgroup$
                – AHusain
                3 hours ago










              • $begingroup$
                Sorry, this answer should have been a comment.
                $endgroup$
                – steven gregory
                3 hours ago











              • $begingroup$
                sure, but how can I find where that point is on the circle?
                $endgroup$
                – user1938107
                3 hours ago















              1












              $begingroup$

              If you draw a circle with diameter $sqrta^2+b^2$ then $p_3$ can be either of two points on that circle.






              share|cite|improve this answer











              $endgroup$












              • $begingroup$
                The lengths $a$ and $b$ are also fixed so this doesn't completely work.
                $endgroup$
                – AHusain
                3 hours ago










              • $begingroup$
                Sorry, this answer should have been a comment.
                $endgroup$
                – steven gregory
                3 hours ago











              • $begingroup$
                sure, but how can I find where that point is on the circle?
                $endgroup$
                – user1938107
                3 hours ago













              1












              1








              1





              $begingroup$

              If you draw a circle with diameter $sqrta^2+b^2$ then $p_3$ can be either of two points on that circle.






              share|cite|improve this answer











              $endgroup$



              If you draw a circle with diameter $sqrta^2+b^2$ then $p_3$ can be either of two points on that circle.







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited 3 hours ago

























              answered 3 hours ago









              steven gregorysteven gregory

              18.5k32459




              18.5k32459











              • $begingroup$
                The lengths $a$ and $b$ are also fixed so this doesn't completely work.
                $endgroup$
                – AHusain
                3 hours ago










              • $begingroup$
                Sorry, this answer should have been a comment.
                $endgroup$
                – steven gregory
                3 hours ago











              • $begingroup$
                sure, but how can I find where that point is on the circle?
                $endgroup$
                – user1938107
                3 hours ago
















              • $begingroup$
                The lengths $a$ and $b$ are also fixed so this doesn't completely work.
                $endgroup$
                – AHusain
                3 hours ago










              • $begingroup$
                Sorry, this answer should have been a comment.
                $endgroup$
                – steven gregory
                3 hours ago











              • $begingroup$
                sure, but how can I find where that point is on the circle?
                $endgroup$
                – user1938107
                3 hours ago















              $begingroup$
              The lengths $a$ and $b$ are also fixed so this doesn't completely work.
              $endgroup$
              – AHusain
              3 hours ago




              $begingroup$
              The lengths $a$ and $b$ are also fixed so this doesn't completely work.
              $endgroup$
              – AHusain
              3 hours ago












              $begingroup$
              Sorry, this answer should have been a comment.
              $endgroup$
              – steven gregory
              3 hours ago





              $begingroup$
              Sorry, this answer should have been a comment.
              $endgroup$
              – steven gregory
              3 hours ago













              $begingroup$
              sure, but how can I find where that point is on the circle?
              $endgroup$
              – user1938107
              3 hours ago




              $begingroup$
              sure, but how can I find where that point is on the circle?
              $endgroup$
              – user1938107
              3 hours ago











              1












              $begingroup$

              • Let $ell$ be the distance between $p_1$ and $p_2$, and let $p_3$ be the point where the two line segments will meet.


              • The three points $p_1, p_2, p_3$ form a triangle whose side lengths we know: $a, b, ell$.


              • We can find out where $p_3$ is by measuring properties of this triangle. Draw a perpendicular altitude from $p_3$ down to the side of the triangle between $p_1$ and $p_2$. Call the intersection point $p_4$.



              • If we knew the length $h$ of this altitude, we would know where to put $p_3$. First, we could find out where $p_4$ is. By the Pythagorean theorem, the distance between $p_1$ and $p_4$ is $sqrta^2 - h^2$. So
                $$p_4 = p_1 + fracsqrta^2-h^2ell (p_2 - p_1)$$



                Next, we would use $p_4$ to find out where to put $p_3$. We start from $p_4$ and travel a distance $h$ in a direction perpendicular to the $p_1$ $p_2$ base of the triangle.



                In two dimensions, we have two choices of perpendicular direction. In three dimensions, we actually have an infinite number of perpendicular directions[*], so you'll have to pick one. Pick a unit vector $widehatu$ in a perpendicular direction.



                Then $p_3 = p_4 + hcdot widehatu$, which gives you the answer you want.



              • But how do we find $h$? We can find $h$ if we know the area of the triangle. Heron's formula lets you calculate the area of the triangle when you know the length of all the sides. $A = sqrts(s-a)(s-b)(s-ell)$, where $s$ is half the perimeter $s = (a+b+ell)/2$. Once you have the area, you can calculate the height of the altitude: $A = frac12textbasetimestextheight$, so $h = 2A/ell$.


              [*] In 3D, you have an infinite number of perpendicular directions. Imagine $p_1=langle 0,0rangle$ and $p_2 = langle 1,0rangle$, and $a$ and $b$ are some numbers. There's a satisfactory point $p_3$ in the x-y plane. But if you spin $p_3$ around the $x$-axis, you find an infinite number of other satisfactory points too.






              share|cite|improve this answer









              $endgroup$

















                1












                $begingroup$

                • Let $ell$ be the distance between $p_1$ and $p_2$, and let $p_3$ be the point where the two line segments will meet.


                • The three points $p_1, p_2, p_3$ form a triangle whose side lengths we know: $a, b, ell$.


                • We can find out where $p_3$ is by measuring properties of this triangle. Draw a perpendicular altitude from $p_3$ down to the side of the triangle between $p_1$ and $p_2$. Call the intersection point $p_4$.



                • If we knew the length $h$ of this altitude, we would know where to put $p_3$. First, we could find out where $p_4$ is. By the Pythagorean theorem, the distance between $p_1$ and $p_4$ is $sqrta^2 - h^2$. So
                  $$p_4 = p_1 + fracsqrta^2-h^2ell (p_2 - p_1)$$



                  Next, we would use $p_4$ to find out where to put $p_3$. We start from $p_4$ and travel a distance $h$ in a direction perpendicular to the $p_1$ $p_2$ base of the triangle.



                  In two dimensions, we have two choices of perpendicular direction. In three dimensions, we actually have an infinite number of perpendicular directions[*], so you'll have to pick one. Pick a unit vector $widehatu$ in a perpendicular direction.



                  Then $p_3 = p_4 + hcdot widehatu$, which gives you the answer you want.



                • But how do we find $h$? We can find $h$ if we know the area of the triangle. Heron's formula lets you calculate the area of the triangle when you know the length of all the sides. $A = sqrts(s-a)(s-b)(s-ell)$, where $s$ is half the perimeter $s = (a+b+ell)/2$. Once you have the area, you can calculate the height of the altitude: $A = frac12textbasetimestextheight$, so $h = 2A/ell$.


                [*] In 3D, you have an infinite number of perpendicular directions. Imagine $p_1=langle 0,0rangle$ and $p_2 = langle 1,0rangle$, and $a$ and $b$ are some numbers. There's a satisfactory point $p_3$ in the x-y plane. But if you spin $p_3$ around the $x$-axis, you find an infinite number of other satisfactory points too.






                share|cite|improve this answer









                $endgroup$















                  1












                  1








                  1





                  $begingroup$

                  • Let $ell$ be the distance between $p_1$ and $p_2$, and let $p_3$ be the point where the two line segments will meet.


                  • The three points $p_1, p_2, p_3$ form a triangle whose side lengths we know: $a, b, ell$.


                  • We can find out where $p_3$ is by measuring properties of this triangle. Draw a perpendicular altitude from $p_3$ down to the side of the triangle between $p_1$ and $p_2$. Call the intersection point $p_4$.



                  • If we knew the length $h$ of this altitude, we would know where to put $p_3$. First, we could find out where $p_4$ is. By the Pythagorean theorem, the distance between $p_1$ and $p_4$ is $sqrta^2 - h^2$. So
                    $$p_4 = p_1 + fracsqrta^2-h^2ell (p_2 - p_1)$$



                    Next, we would use $p_4$ to find out where to put $p_3$. We start from $p_4$ and travel a distance $h$ in a direction perpendicular to the $p_1$ $p_2$ base of the triangle.



                    In two dimensions, we have two choices of perpendicular direction. In three dimensions, we actually have an infinite number of perpendicular directions[*], so you'll have to pick one. Pick a unit vector $widehatu$ in a perpendicular direction.



                    Then $p_3 = p_4 + hcdot widehatu$, which gives you the answer you want.



                  • But how do we find $h$? We can find $h$ if we know the area of the triangle. Heron's formula lets you calculate the area of the triangle when you know the length of all the sides. $A = sqrts(s-a)(s-b)(s-ell)$, where $s$ is half the perimeter $s = (a+b+ell)/2$. Once you have the area, you can calculate the height of the altitude: $A = frac12textbasetimestextheight$, so $h = 2A/ell$.


                  [*] In 3D, you have an infinite number of perpendicular directions. Imagine $p_1=langle 0,0rangle$ and $p_2 = langle 1,0rangle$, and $a$ and $b$ are some numbers. There's a satisfactory point $p_3$ in the x-y plane. But if you spin $p_3$ around the $x$-axis, you find an infinite number of other satisfactory points too.






                  share|cite|improve this answer









                  $endgroup$



                  • Let $ell$ be the distance between $p_1$ and $p_2$, and let $p_3$ be the point where the two line segments will meet.


                  • The three points $p_1, p_2, p_3$ form a triangle whose side lengths we know: $a, b, ell$.


                  • We can find out where $p_3$ is by measuring properties of this triangle. Draw a perpendicular altitude from $p_3$ down to the side of the triangle between $p_1$ and $p_2$. Call the intersection point $p_4$.



                  • If we knew the length $h$ of this altitude, we would know where to put $p_3$. First, we could find out where $p_4$ is. By the Pythagorean theorem, the distance between $p_1$ and $p_4$ is $sqrta^2 - h^2$. So
                    $$p_4 = p_1 + fracsqrta^2-h^2ell (p_2 - p_1)$$



                    Next, we would use $p_4$ to find out where to put $p_3$. We start from $p_4$ and travel a distance $h$ in a direction perpendicular to the $p_1$ $p_2$ base of the triangle.



                    In two dimensions, we have two choices of perpendicular direction. In three dimensions, we actually have an infinite number of perpendicular directions[*], so you'll have to pick one. Pick a unit vector $widehatu$ in a perpendicular direction.



                    Then $p_3 = p_4 + hcdot widehatu$, which gives you the answer you want.



                  • But how do we find $h$? We can find $h$ if we know the area of the triangle. Heron's formula lets you calculate the area of the triangle when you know the length of all the sides. $A = sqrts(s-a)(s-b)(s-ell)$, where $s$ is half the perimeter $s = (a+b+ell)/2$. Once you have the area, you can calculate the height of the altitude: $A = frac12textbasetimestextheight$, so $h = 2A/ell$.


                  [*] In 3D, you have an infinite number of perpendicular directions. Imagine $p_1=langle 0,0rangle$ and $p_2 = langle 1,0rangle$, and $a$ and $b$ are some numbers. There's a satisfactory point $p_3$ in the x-y plane. But if you spin $p_3$ around the $x$-axis, you find an infinite number of other satisfactory points too.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 2 hours ago









                  user326210user326210

                  9,462927




                  9,462927



























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