two dots on a line that are the same distance from plane
$begingroup$
Find two dots on a line that is determined by plane cross:
$5x + 3y - 1 = 0$
$2x + 3z + 5 = 0$
That are the same length from two planes $3x + 3y - 2 = 0$ and $4x + y + z + 4 = 0$.
I found a line equation:
$x = -1 + 3t$
$y = 2 - 5t$
$z = -1 - 2t$
Now I don't know how to find two dots that are the same length from those two planes.
My another question is how it is possible that there are two dots on a line that is the same distance from two planes. I thought there is only one.
linear-algebra
edited Dec 28 '18 at 19:38
David G. Stork
11.1k41432
asked Dec 28 '18 at 15:39
PetarPetar
317
$endgroup$
add a comment |
$begingroup$
Find two dots on a line that is determined by plane cross:
$5x + 3y - 1 = 0$
$2x + 3z + 5 = 0$
That are the same length from two planes $3x + 3y - 2 = 0$ and $4x + y + z + 4 = 0$.
I found a line equation:
$x = -1 + 3t$
$y = 2 - 5t$
$z = -1 - 2t$
Now I don't know how to find two dots that are the same length from those two planes.
My another question is how it is possible that there are two dots on a line that is the same distance from two planes. I thought there is only one.
linear-algebra
edited Dec 28 '18 at 19:38
David G. Stork
11.1k41432
asked Dec 28 '18 at 15:39
PetarPetar
317
$endgroup$
$begingroup$
You're working in 3d so you can try to visualize it in real life. Get a piece of paper, and pick a direction and point that way. See if you can find two points which are equally close to the plane. You're right that there is a single closest point, but for any other distance to the plane, hopefully you'll be able to see there are two points which are both this distance to the plane.
$endgroup$
– tch
Dec 28 '18 at 16:10
$begingroup$
okay thank you! But do you know how to find those two dots?
$endgroup$
– Petar
Dec 28 '18 at 17:58
$begingroup$
I would start by coming up with an equation which gives the distance from any point on the line to the plane. I.e. a function $d(t)$ which gives the distance from $(x(t),y(t),z(t))$ to the plane. Now, for a given distance $d_0$, try to solve $d(t) = d_0$ for $t$.
$endgroup$
– tch
Dec 28 '18 at 18:01
$begingroup$
It seems to me that there is only one point on the line that is equidistant from the two planes.
$endgroup$
– Doug M
Dec 28 '18 at 19:57
$begingroup$
@DougM The line isn’t parallel to either angle bisector, so there are two.
$endgroup$
– amd
Feb 1 at 0:45
add a comment |
$begingroup$
Find two dots on a line that is determined by plane cross:
$5x + 3y - 1 = 0$
$2x + 3z + 5 = 0$
That are the same length from two planes $3x + 3y - 2 = 0$ and $4x + y + z + 4 = 0$.
I found a line equation:
$x = -1 + 3t$
$y = 2 - 5t$
$z = -1 - 2t$
Now I don't know how to find two dots that are the same length from those two planes.
My another question is how it is possible that there are two dots on a line that is the same distance from two planes. I thought there is only one.
linear-algebra
edited Dec 28 '18 at 19:38
David G. Stork
11.1k41432
asked Dec 28 '18 at 15:39
PetarPetar
317
$endgroup$
Find two dots on a line that is determined by plane cross:
$5x + 3y - 1 = 0$
$2x + 3z + 5 = 0$
That are the same length from two planes $3x + 3y - 2 = 0$ and $4x + y + z + 4 = 0$.
I found a line equation:
$x = -1 + 3t$
$y = 2 - 5t$
$z = -1 - 2t$
Now I don't know how to find two dots that are the same length from those two planes.
My another question is how it is possible that there are two dots on a line that is the same distance from two planes. I thought there is only one.
linear-algebra
linear-algebra
edited Dec 28 '18 at 19:38
David G. Stork
11.1k41432
asked Dec 28 '18 at 15:39
PetarPetar
317
edited Dec 28 '18 at 19:38
David G. Stork
11.1k41432
asked Dec 28 '18 at 15:39
PetarPetar
317
edited Dec 28 '18 at 19:38
David G. Stork
11.1k41432
edited Dec 28 '18 at 19:38
David G. Stork
11.1k41432
edited Dec 28 '18 at 19:38
David G. Stork
11.1k41432
11.1k41432
asked Dec 28 '18 at 15:39
PetarPetar
317
asked Dec 28 '18 at 15:39
PetarPetar
317
asked Dec 28 '18 at 15:39
PetarPetar
317
317
$begingroup$
You're working in 3d so you can try to visualize it in real life. Get a piece of paper, and pick a direction and point that way. See if you can find two points which are equally close to the plane. You're right that there is a single closest point, but for any other distance to the plane, hopefully you'll be able to see there are two points which are both this distance to the plane.
$endgroup$
– tch
Dec 28 '18 at 16:10
$begingroup$
okay thank you! But do you know how to find those two dots?
$endgroup$
– Petar
Dec 28 '18 at 17:58
$begingroup$
I would start by coming up with an equation which gives the distance from any point on the line to the plane. I.e. a function $d(t)$ which gives the distance from $(x(t),y(t),z(t))$ to the plane. Now, for a given distance $d_0$, try to solve $d(t) = d_0$ for $t$.
$endgroup$
– tch
Dec 28 '18 at 18:01
$begingroup$
It seems to me that there is only one point on the line that is equidistant from the two planes.
$endgroup$
– Doug M
Dec 28 '18 at 19:57
$begingroup$
@DougM The line isn’t parallel to either angle bisector, so there are two.
$endgroup$
– amd
Feb 1 at 0:45
add a comment |
$begingroup$
You're working in 3d so you can try to visualize it in real life. Get a piece of paper, and pick a direction and point that way. See if you can find two points which are equally close to the plane. You're right that there is a single closest point, but for any other distance to the plane, hopefully you'll be able to see there are two points which are both this distance to the plane.
$endgroup$
– tch
Dec 28 '18 at 16:10
$begingroup$
okay thank you! But do you know how to find those two dots?
$endgroup$
– Petar
Dec 28 '18 at 17:58
$begingroup$
I would start by coming up with an equation which gives the distance from any point on the line to the plane. I.e. a function $d(t)$ which gives the distance from $(x(t),y(t),z(t))$ to the plane. Now, for a given distance $d_0$, try to solve $d(t) = d_0$ for $t$.
$endgroup$
– tch
Dec 28 '18 at 18:01
$begingroup$
It seems to me that there is only one point on the line that is equidistant from the two planes.
$endgroup$
– Doug M
Dec 28 '18 at 19:57
$begingroup$
@DougM The line isn’t parallel to either angle bisector, so there are two.
$endgroup$
– amd
Feb 1 at 0:45
$begingroup$
You're working in 3d so you can try to visualize it in real life. Get a piece of paper, and pick a direction and point that way. See if you can find two points which are equally close to the plane. You're right that there is a single closest point, but for any other distance to the plane, hopefully you'll be able to see there are two points which are both this distance to the plane.
$endgroup$
– tch
Dec 28 '18 at 16:10
$begingroup$
You're working in 3d so you can try to visualize it in real life. Get a piece of paper, and pick a direction and point that way. See if you can find two points which are equally close to the plane. You're right that there is a single closest point, but for any other distance to the plane, hopefully you'll be able to see there are two points which are both this distance to the plane.
$endgroup$
– tch
Dec 28 '18 at 16:10
$begingroup$
okay thank you! But do you know how to find those two dots?
$endgroup$
– Petar
Dec 28 '18 at 17:58
$begingroup$
okay thank you! But do you know how to find those two dots?
$endgroup$
– Petar
Dec 28 '18 at 17:58
$begingroup$
I would start by coming up with an equation which gives the distance from any point on the line to the plane. I.e. a function $d(t)$ which gives the distance from $(x(t),y(t),z(t))$ to the plane. Now, for a given distance $d_0$, try to solve $d(t) = d_0$ for $t$.
$endgroup$
– tch
Dec 28 '18 at 18:01
$begingroup$
I would start by coming up with an equation which gives the distance from any point on the line to the plane. I.e. a function $d(t)$ which gives the distance from $(x(t),y(t),z(t))$ to the plane. Now, for a given distance $d_0$, try to solve $d(t) = d_0$ for $t$.
$endgroup$
– tch
Dec 28 '18 at 18:01
$begingroup$
It seems to me that there is only one point on the line that is equidistant from the two planes.
$endgroup$
– Doug M
Dec 28 '18 at 19:57
$begingroup$
It seems to me that there is only one point on the line that is equidistant from the two planes.
$endgroup$
– Doug M
Dec 28 '18 at 19:57
$begingroup$
@DougM The line isn’t parallel to either angle bisector, so there are two.
$endgroup$
– amd
Feb 1 at 0:45
$begingroup$
@DougM The line isn’t parallel to either angle bisector, so there are two.
$endgroup$
– amd
Feb 1 at 0:45
add a comment |
1 Answer
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$begingroup$
Points that are equidistant from a pair of non-parallel planes lie on their two angle bisectors, which are themselves planes. Work out their equations and then find the intersection of the given line with the bisectors.
answered Dec 28 '18 at 19:17
amdamd
31k21051
$endgroup$
add a comment |
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$begingroup$
Points that are equidistant from a pair of non-parallel planes lie on their two angle bisectors, which are themselves planes. Work out their equations and then find the intersection of the given line with the bisectors.
answered Dec 28 '18 at 19:17
amdamd
31k21051
$endgroup$
add a comment |
$begingroup$
Points that are equidistant from a pair of non-parallel planes lie on their two angle bisectors, which are themselves planes. Work out their equations and then find the intersection of the given line with the bisectors.
answered Dec 28 '18 at 19:17
amdamd
31k21051
$endgroup$
add a comment |
$begingroup$
Points that are equidistant from a pair of non-parallel planes lie on their two angle bisectors, which are themselves planes. Work out their equations and then find the intersection of the given line with the bisectors.
answered Dec 28 '18 at 19:17
amdamd
31k21051
$endgroup$
Points that are equidistant from a pair of non-parallel planes lie on their two angle bisectors, which are themselves planes. Work out their equations and then find the intersection of the given line with the bisectors.
answered Dec 28 '18 at 19:17
amdamd
31k21051
answered Dec 28 '18 at 19:17
amdamd
31k21051
answered Dec 28 '18 at 19:17
amdamd
31k21051
answered Dec 28 '18 at 19:17
amdamd
31k21051
31k21051
add a comment |
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$begingroup$
You're working in 3d so you can try to visualize it in real life. Get a piece of paper, and pick a direction and point that way. See if you can find two points which are equally close to the plane. You're right that there is a single closest point, but for any other distance to the plane, hopefully you'll be able to see there are two points which are both this distance to the plane.
$endgroup$
– tch
Dec 28 '18 at 16:10
$begingroup$
okay thank you! But do you know how to find those two dots?
$endgroup$
– Petar
Dec 28 '18 at 17:58
$begingroup$
I would start by coming up with an equation which gives the distance from any point on the line to the plane. I.e. a function $d(t)$ which gives the distance from $(x(t),y(t),z(t))$ to the plane. Now, for a given distance $d_0$, try to solve $d(t) = d_0$ for $t$.
$endgroup$
– tch
Dec 28 '18 at 18:01
$begingroup$
It seems to me that there is only one point on the line that is equidistant from the two planes.
$endgroup$
– Doug M
Dec 28 '18 at 19:57
$begingroup$
@DougM The line isn’t parallel to either angle bisector, so there are two.
$endgroup$
– amd
Feb 1 at 0:45