Find point on a line a certain distance away from another point
0
$begingroup$
I have a line defined by two points within the line, $A(x_1, y_1, z_1)$ and $B(x_2, y_2, z_2)$.
I have a third point $P(x_3, y_3, z_3)$.
How do I find the coordinates of the points on the line $AB$ which are $d$ units away from $P$ ?
geometry
asked Dec 18 '18 at 23:25
theonlygustitheonlygusti
687615
$endgroup$
|
show 1 more comment
0
$begingroup$
I have a line defined by two points within the line, $A(x_1, y_1, z_1)$ and $B(x_2, y_2, z_2)$.
I have a third point $P(x_3, y_3, z_3)$.
How do I find the coordinates of the points on the line $AB$ which are $d$ units away from $P$ ?
geometry
asked Dec 18 '18 at 23:25
theonlygustitheonlygusti
687615
$endgroup$
$begingroup$
Find the perpendicular distance of $P$ from the line, then slide along it either way, through a distance you get from Pythagoras.
$endgroup$
– J.G.
Dec 18 '18 at 23:28
$begingroup$
@J.G. how do I "slide along" it?
$endgroup$
– theonlygusti
Dec 18 '18 at 23:28
$begingroup$
@RossMillikan I don't think that's a dupe. Looks like they asked for finding a point on the line. I have 3 points. 2 define a line, one is away from it. If it is a dupe, idk how to re-apply the answer from the other question to this one.
$endgroup$
– theonlygusti
Dec 18 '18 at 23:46
$begingroup$
Sorry, I misread the question. i have reopened it. J.G.'s hint is a good one to make it a duplicate. If the distance from $P$ to the closest point on the line $Q$ is $p$, you want a point on the line that is $sqrt {d^2-p^2} $ away from $p$. Now apply the other question
$endgroup$
– Ross Millikan
Dec 18 '18 at 23:48
$begingroup$
Intersect the line with a sphere centered at $P$.
$endgroup$
– amd
Dec 19 '18 at 0:14
|
show 1 more comment
0
0
0
$begingroup$
I have a line defined by two points within the line, $A(x_1, y_1, z_1)$ and $B(x_2, y_2, z_2)$.
I have a third point $P(x_3, y_3, z_3)$.
How do I find the coordinates of the points on the line $AB$ which are $d$ units away from $P$ ?
geometry
asked Dec 18 '18 at 23:25
theonlygustitheonlygusti
687615
$endgroup$
I have a line defined by two points within the line, $A(x_1, y_1, z_1)$ and $B(x_2, y_2, z_2)$.
I have a third point $P(x_3, y_3, z_3)$.
How do I find the coordinates of the points on the line $AB$ which are $d$ units away from $P$ ?
geometry
geometry
asked Dec 18 '18 at 23:25
theonlygustitheonlygusti
687615
asked Dec 18 '18 at 23:25
theonlygustitheonlygusti
687615
asked Dec 18 '18 at 23:25
theonlygustitheonlygusti
687615
asked Dec 18 '18 at 23:25
theonlygustitheonlygusti
687615
asked Dec 18 '18 at 23:25
theonlygustitheonlygusti
687615
687615
$begingroup$
Find the perpendicular distance of $P$ from the line, then slide along it either way, through a distance you get from Pythagoras.
$endgroup$
– J.G.
Dec 18 '18 at 23:28
$begingroup$
@J.G. how do I "slide along" it?
$endgroup$
– theonlygusti
Dec 18 '18 at 23:28
$begingroup$
@RossMillikan I don't think that's a dupe. Looks like they asked for finding a point on the line. I have 3 points. 2 define a line, one is away from it. If it is a dupe, idk how to re-apply the answer from the other question to this one.
$endgroup$
– theonlygusti
Dec 18 '18 at 23:46
$begingroup$
Sorry, I misread the question. i have reopened it. J.G.'s hint is a good one to make it a duplicate. If the distance from $P$ to the closest point on the line $Q$ is $p$, you want a point on the line that is $sqrt {d^2-p^2} $ away from $p$. Now apply the other question
$endgroup$
– Ross Millikan
Dec 18 '18 at 23:48
$begingroup$
Intersect the line with a sphere centered at $P$.
$endgroup$
– amd
Dec 19 '18 at 0:14
|
show 1 more comment
$begingroup$
Find the perpendicular distance of $P$ from the line, then slide along it either way, through a distance you get from Pythagoras.
$endgroup$
– J.G.
Dec 18 '18 at 23:28
$begingroup$
@J.G. how do I "slide along" it?
$endgroup$
– theonlygusti
Dec 18 '18 at 23:28
$begingroup$
@RossMillikan I don't think that's a dupe. Looks like they asked for finding a point on the line. I have 3 points. 2 define a line, one is away from it. If it is a dupe, idk how to re-apply the answer from the other question to this one.
$endgroup$
– theonlygusti
Dec 18 '18 at 23:46
$begingroup$
Sorry, I misread the question. i have reopened it. J.G.'s hint is a good one to make it a duplicate. If the distance from $P$ to the closest point on the line $Q$ is $p$, you want a point on the line that is $sqrt {d^2-p^2} $ away from $p$. Now apply the other question
$endgroup$
– Ross Millikan
Dec 18 '18 at 23:48
$begingroup$
Intersect the line with a sphere centered at $P$.
$endgroup$
– amd
Dec 19 '18 at 0:14
$begingroup$
Find the perpendicular distance of $P$ from the line, then slide along it either way, through a distance you get from Pythagoras.
$endgroup$
– J.G.
Dec 18 '18 at 23:28
$begingroup$
Find the perpendicular distance of $P$ from the line, then slide along it either way, through a distance you get from Pythagoras.
$endgroup$
– J.G.
Dec 18 '18 at 23:28
$begingroup$
@J.G. how do I "slide along" it?
$endgroup$
– theonlygusti
Dec 18 '18 at 23:28
$begingroup$
@J.G. how do I "slide along" it?
$endgroup$
– theonlygusti
Dec 18 '18 at 23:28
$begingroup$
@RossMillikan I don't think that's a dupe. Looks like they asked for finding a point on the line. I have 3 points. 2 define a line, one is away from it. If it is a dupe, idk how to re-apply the answer from the other question to this one.
$endgroup$
– theonlygusti
Dec 18 '18 at 23:46
$begingroup$
@RossMillikan I don't think that's a dupe. Looks like they asked for finding a point on the line. I have 3 points. 2 define a line, one is away from it. If it is a dupe, idk how to re-apply the answer from the other question to this one.
$endgroup$
– theonlygusti
Dec 18 '18 at 23:46
$begingroup$
Sorry, I misread the question. i have reopened it. J.G.'s hint is a good one to make it a duplicate. If the distance from $P$ to the closest point on the line $Q$ is $p$, you want a point on the line that is $sqrt {d^2-p^2} $ away from $p$. Now apply the other question
$endgroup$
– Ross Millikan
Dec 18 '18 at 23:48
$begingroup$
Sorry, I misread the question. i have reopened it. J.G.'s hint is a good one to make it a duplicate. If the distance from $P$ to the closest point on the line $Q$ is $p$, you want a point on the line that is $sqrt {d^2-p^2} $ away from $p$. Now apply the other question
$endgroup$
– Ross Millikan
Dec 18 '18 at 23:48
$begingroup$
Intersect the line with a sphere centered at $P$.
$endgroup$
– amd
Dec 19 '18 at 0:14
$begingroup$
Intersect the line with a sphere centered at $P$.
$endgroup$
– amd
Dec 19 '18 at 0:14
|
show 1 more comment
1 Answer
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If you write out $|A+alpha(B-A)-P|^2=d^2$ you get a quadratic equation in $alpha$. You can solve for $alpha$ with standard methods.
answered Dec 19 '18 at 0:35
SmileyCraftSmileyCraft
3,611517
$endgroup$
add a comment |
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$begingroup$
If you write out $|A+alpha(B-A)-P|^2=d^2$ you get a quadratic equation in $alpha$. You can solve for $alpha$ with standard methods.
answered Dec 19 '18 at 0:35
SmileyCraftSmileyCraft
3,611517
$endgroup$
add a comment |
0
$begingroup$
If you write out $|A+alpha(B-A)-P|^2=d^2$ you get a quadratic equation in $alpha$. You can solve for $alpha$ with standard methods.
answered Dec 19 '18 at 0:35
SmileyCraftSmileyCraft
3,611517
$endgroup$
add a comment |
0
0
0
$begingroup$
If you write out $|A+alpha(B-A)-P|^2=d^2$ you get a quadratic equation in $alpha$. You can solve for $alpha$ with standard methods.
answered Dec 19 '18 at 0:35
SmileyCraftSmileyCraft
3,611517
$endgroup$
If you write out $|A+alpha(B-A)-P|^2=d^2$ you get a quadratic equation in $alpha$. You can solve for $alpha$ with standard methods.
answered Dec 19 '18 at 0:35
SmileyCraftSmileyCraft
3,611517
answered Dec 19 '18 at 0:35
SmileyCraftSmileyCraft
3,611517
answered Dec 19 '18 at 0:35
SmileyCraftSmileyCraft
3,611517
answered Dec 19 '18 at 0:35
SmileyCraftSmileyCraft
3,611517
3,611517
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$begingroup$
Find the perpendicular distance of $P$ from the line, then slide along it either way, through a distance you get from Pythagoras.
$endgroup$
– J.G.
Dec 18 '18 at 23:28
$begingroup$
@J.G. how do I "slide along" it?
$endgroup$
– theonlygusti
Dec 18 '18 at 23:28
$begingroup$
@RossMillikan I don't think that's a dupe. Looks like they asked for finding a point on the line. I have 3 points. 2 define a line, one is away from it. If it is a dupe, idk how to re-apply the answer from the other question to this one.
$endgroup$
– theonlygusti
Dec 18 '18 at 23:46
$begingroup$
Sorry, I misread the question. i have reopened it. J.G.'s hint is a good one to make it a duplicate. If the distance from $P$ to the closest point on the line $Q$ is $p$, you want a point on the line that is $sqrt {d^2-p^2} $ away from $p$. Now apply the other question
$endgroup$
– Ross Millikan
Dec 18 '18 at 23:48
$begingroup$
Intersect the line with a sphere centered at $P$.
$endgroup$
– amd
Dec 19 '18 at 0:14