Find the coordinate of two line segments that are perpendicularcalculating perpendicular and angular distance...
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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*
$begingroup$
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
$endgroup$
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
$endgroup$
$begingroup$
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
1 hour ago
$begingroup$
@amd yes p1 and p2 are the endpoints of a and b. I need their mutual endpoint
$endgroup$
– user1938107
1 hour 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
$endgroup$
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.
geometry trigonometry
geometry trigonometry
asked 2 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
1 hour ago
$begingroup$
@amd yes p1 and p2 are the endpoints of a and b. I need their mutual endpoint
$endgroup$
– user1938107
1 hour ago
add a comment |
$begingroup$
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
1 hour ago
$begingroup$
@amd yes p1 and p2 are the endpoints of a and b. I need their mutual endpoint
$endgroup$
– user1938107
1 hour 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
1 hour 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
1 hour ago
$begingroup$
@amd yes p1 and p2 are the endpoints of a and b. I need their mutual endpoint
$endgroup$
– user1938107
1 hour ago
$begingroup$
@amd yes p1 and p2 are the endpoints of a and b. I need their mutual endpoint
$endgroup$
– user1938107
1 hour ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Sometimes a figure is worth a 1000 words:
In three dimensions:
$endgroup$
add a comment |
$begingroup$
If you draw a circle with diameter $sqrt{a^2+b^2}$ then $p_3$ can be either of two points on that circle.
$endgroup$
$begingroup$
The lengths $a$ and $b$ are also fixed so this doesn't completely work.
$endgroup$
– AHusain
2 hours ago
$begingroup$
Sorry, this answer should have been a comment.
$endgroup$
– steven gregory
2 hours ago
$begingroup$
sure, but how can I find where that point is on the circle?
$endgroup$
– user1938107
1 hour ago
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 $sqrt{a^2 - h^2}$. So
$$p_4 = p_1 + frac{sqrt{a^2-h^2}}{ell} (p_2 - p_1)$$
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 $widehat{u}$ in a perpendicular direction.
Then $p_3 = p_4 + hcdot widehat{u}$, 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. $A = sqrt{s(s-a)(s-b)(s-ell)}$, where $s$ is half the perimeter $s = (a+b+ell)/2$. Once you have the area, you can calculate the height of the altitude: $A = frac{1}{2}text{base}timestext{height}$, 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.
$endgroup$
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Sometimes a figure is worth a 1000 words:
In three dimensions:
$endgroup$
add a comment |
$begingroup$
Sometimes a figure is worth a 1000 words:
In three dimensions:
$endgroup$
add a comment |
$begingroup$
Sometimes a figure is worth a 1000 words:
In three dimensions:
$endgroup$
Sometimes a figure is worth a 1000 words:
In three dimensions:
edited 23 mins ago
answered 1 hour ago
David G. StorkDavid G. Stork
12.3k41836
12.3k41836
add a comment |
add a comment |
$begingroup$
If you draw a circle with diameter $sqrt{a^2+b^2}$ then $p_3$ can be either of two points on that circle.
$endgroup$
$begingroup$
The lengths $a$ and $b$ are also fixed so this doesn't completely work.
$endgroup$
– AHusain
2 hours ago
$begingroup$
Sorry, this answer should have been a comment.
$endgroup$
– steven gregory
2 hours ago
$begingroup$
sure, but how can I find where that point is on the circle?
$endgroup$
– user1938107
1 hour ago
add a comment |
$begingroup$
If you draw a circle with diameter $sqrt{a^2+b^2}$ then $p_3$ can be either of two points on that circle.
$endgroup$
$begingroup$
The lengths $a$ and $b$ are also fixed so this doesn't completely work.
$endgroup$
– AHusain
2 hours ago
$begingroup$
Sorry, this answer should have been a comment.
$endgroup$
– steven gregory
2 hours ago
$begingroup$
sure, but how can I find where that point is on the circle?
$endgroup$
– user1938107
1 hour ago
add a comment |
$begingroup$
If you draw a circle with diameter $sqrt{a^2+b^2}$ then $p_3$ can be either of two points on that circle.
$endgroup$
If you draw a circle with diameter $sqrt{a^2+b^2}$ then $p_3$ can be either of two points on that circle.
edited 2 hours ago
answered 2 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
2 hours ago
$begingroup$
Sorry, this answer should have been a comment.
$endgroup$
– steven gregory
2 hours ago
$begingroup$
sure, but how can I find where that point is on the circle?
$endgroup$
– user1938107
1 hour ago
add a comment |
$begingroup$
The lengths $a$ and $b$ are also fixed so this doesn't completely work.
$endgroup$
– AHusain
2 hours ago
$begingroup$
Sorry, this answer should have been a comment.
$endgroup$
– steven gregory
2 hours ago
$begingroup$
sure, but how can I find where that point is on the circle?
$endgroup$
– user1938107
1 hour ago
$begingroup$
The lengths $a$ and $b$ are also fixed so this doesn't completely work.
$endgroup$
– AHusain
2 hours ago
$begingroup$
The lengths $a$ and $b$ are also fixed so this doesn't completely work.
$endgroup$
– AHusain
2 hours ago
$begingroup$
Sorry, this answer should have been a comment.
$endgroup$
– steven gregory
2 hours ago
$begingroup$
Sorry, this answer should have been a comment.
$endgroup$
– steven gregory
2 hours ago
$begingroup$
sure, but how can I find where that point is on the circle?
$endgroup$
– user1938107
1 hour ago
$begingroup$
sure, but how can I find where that point is on the circle?
$endgroup$
– user1938107
1 hour ago
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 $sqrt{a^2 - h^2}$. So
$$p_4 = p_1 + frac{sqrt{a^2-h^2}}{ell} (p_2 - p_1)$$
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 $widehat{u}$ in a perpendicular direction.
Then $p_3 = p_4 + hcdot widehat{u}$, 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. $A = sqrt{s(s-a)(s-b)(s-ell)}$, where $s$ is half the perimeter $s = (a+b+ell)/2$. Once you have the area, you can calculate the height of the altitude: $A = frac{1}{2}text{base}timestext{height}$, 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.
$endgroup$
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 $sqrt{a^2 - h^2}$. So
$$p_4 = p_1 + frac{sqrt{a^2-h^2}}{ell} (p_2 - p_1)$$
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 $widehat{u}$ in a perpendicular direction.
Then $p_3 = p_4 + hcdot widehat{u}$, 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. $A = sqrt{s(s-a)(s-b)(s-ell)}$, where $s$ is half the perimeter $s = (a+b+ell)/2$. Once you have the area, you can calculate the height of the altitude: $A = frac{1}{2}text{base}timestext{height}$, 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.
$endgroup$
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 $sqrt{a^2 - h^2}$. So
$$p_4 = p_1 + frac{sqrt{a^2-h^2}}{ell} (p_2 - p_1)$$
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 $widehat{u}$ in a perpendicular direction.
Then $p_3 = p_4 + hcdot widehat{u}$, 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. $A = sqrt{s(s-a)(s-b)(s-ell)}$, where $s$ is half the perimeter $s = (a+b+ell)/2$. Once you have the area, you can calculate the height of the altitude: $A = frac{1}{2}text{base}timestext{height}$, 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.
$endgroup$
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 $sqrt{a^2 - h^2}$. So
$$p_4 = p_1 + frac{sqrt{a^2-h^2}}{ell} (p_2 - p_1)$$
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 $widehat{u}$ in a perpendicular direction.
Then $p_3 = p_4 + hcdot widehat{u}$, 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. $A = sqrt{s(s-a)(s-b)(s-ell)}$, where $s$ is half the perimeter $s = (a+b+ell)/2$. Once you have the area, you can calculate the height of the altitude: $A = frac{1}{2}text{base}timestext{height}$, 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.
answered 1 hour ago
user326210user326210
9,462927
9,462927
add a comment |
add a comment |
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$begingroup$
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
1 hour ago
$begingroup$
@amd yes p1 and p2 are the endpoints of a and b. I need their mutual endpoint
$endgroup$
– user1938107
1 hour ago