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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.

geometry trigonometry

share|cite|improve this question

asked 3 hours ago

user1938107user1938107

15310

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  3 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
  

  
  
  
  

add a comment | 

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.

geometry trigonometry

share|cite|improve this question

asked 3 hours ago

user1938107user1938107

15310

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  3 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
  

  
  
  
  

add a comment | 

$begingroup$

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.

geometry trigonometry

share|cite|improve this question

asked 3 hours ago

user1938107user1938107

15310

$endgroup$

share|cite|improve this question

asked 3 hours ago

user1938107user1938107

15310

share|cite|improve this question

asked 3 hours ago

user1938107user1938107

15310

asked 3 hours ago

user1938107user1938107

15310

asked 3 hours ago

user1938107user1938107

15310

3 Answers
3

active

oldest

votes

3

$begingroup$

Sometimes a figure is worth a 1000 words:

In three dimensions:

share|cite|improve this answer

edited 1 hour ago

answered 2 hours ago

David G. StorkDavid G. Stork

12.3k41836

$endgroup$

add a comment | 

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

edited 3 hours ago

answered 3 hours ago

steven gregorysteven gregory

18.5k32459

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

add a comment | 

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

answered 2 hours ago

user326210user326210

9,462927

$endgroup$

add a comment | 

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3

$begingroup$

Sometimes a figure is worth a 1000 words:

In three dimensions:

share|cite|improve this answer

edited 1 hour ago

answered 2 hours ago

David G. StorkDavid G. Stork

12.3k41836

$endgroup$

add a comment | 

3

$begingroup$

Sometimes a figure is worth a 1000 words:

In three dimensions:

share|cite|improve this answer

edited 1 hour ago

answered 2 hours ago

David G. StorkDavid G. Stork

12.3k41836

$endgroup$

add a comment | 

$begingroup$

Sometimes a figure is worth a 1000 words:

In three dimensions:

share|cite|improve this answer

edited 1 hour ago

answered 2 hours ago

David G. StorkDavid G. Stork

12.3k41836

$endgroup$

share|cite|improve this answer

edited 1 hour ago

answered 2 hours ago

David G. StorkDavid G. Stork

12.3k41836

answered 2 hours ago

David G. StorkDavid G. Stork

12.3k41836

answered 2 hours ago

David G. StorkDavid G. Stork

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

edited 3 hours ago

answered 3 hours ago

steven gregorysteven gregory

18.5k32459

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

add a comment | 

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

edited 3 hours ago

answered 3 hours ago

steven gregorysteven gregory

18.5k32459

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

add a comment | 

$begingroup$

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

edited 3 hours ago

answered 3 hours ago

steven gregorysteven gregory

18.5k32459

$endgroup$

share|cite|improve this answer

edited 3 hours ago

answered 3 hours ago

steven gregorysteven gregory

18.5k32459

answered 3 hours ago

steven gregorysteven gregory

18.5k32459

answered 3 hours ago

steven gregorysteven gregory

18.5k32459

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

answered 2 hours ago

user326210user326210

9,462927

$endgroup$

add a comment | 

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

answered 2 hours ago

user326210user326210

9,462927

$endgroup$

add a comment | 

$begingroup$

• 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

answered 2 hours ago

user326210user326210

9,462927

$endgroup$

share|cite|improve this answer

answered 2 hours ago

user326210user326210

9,462927

answered 2 hours ago

user326210user326210

9,462927

answered 2 hours ago

user326210user326210

9,462927