Calculate X, Y, Z of the 4 points in the 3D space.
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I want to find the $x, y, z$ coordinates for $4$ points in a 3D space. Point $A$ is my origin $(X, Y, Z = 0,0,0)$ and other points $B, C, D$ are with reference to point A. I know all six distances from $A to B, Ato C, Ato D, Bto C, Bto D$, and $Cto D$.
Point $B$ lies on axis $X$, point $C$ lies in $XY$ plane, and lastly, point $D$ can be anywhere.
Coordinates of the points $A, B, C, D$ are as follows:
$A = (0, 0, 0)$
$B = (X_B, 0, 0)$
$C = (X_C, Y_C, 0)$
$D = (X_D, Y_D, Z_D)$
It is straightforward to determine $X_B$, $X_C$, and $Y_C$. I am having difficulty finding X, Y, Z coordinate of the point D with respect to the point A.
To determine Xd, Yd, Zd, I am using the following equations
$(X_D-X_A)^2 + (Y_D-Y_A)^2 + (Z_D-Z_A)^2 = AD^2$
$(X_D-X_B)^2 + (Y_D-Y_B)^2 + (Z_D-Z_B)^2 = BD^2$
$(X_D-X_C)^2 + (Y_D-Y_C)^2 + (Z_D-Z_C)^2 = CD^2$
After doing some expansion and eliminating the 2nd order variables I came up with the following:
$Ax = b$
$x = (X_D, Y_D, Z_D)$
after doing the pseudo inverse on both sides I can solve for x. I am not sure if my method of solving $X_D$, $Y_D$, and $Z_D$ is correct.
4 points diagram
linear-algebra trigonometry plane-geometry
edited Nov 16 at 20:23
Vasya
3,2791515
asked Nov 16 at 19:49
Bhavin
61
add a comment |
up vote
0
down vote
favorite
I want to find the $x, y, z$ coordinates for $4$ points in a 3D space. Point $A$ is my origin $(X, Y, Z = 0,0,0)$ and other points $B, C, D$ are with reference to point A. I know all six distances from $A to B, Ato C, Ato D, Bto C, Bto D$, and $Cto D$.
Point $B$ lies on axis $X$, point $C$ lies in $XY$ plane, and lastly, point $D$ can be anywhere.
Coordinates of the points $A, B, C, D$ are as follows:
$A = (0, 0, 0)$
$B = (X_B, 0, 0)$
$C = (X_C, Y_C, 0)$
$D = (X_D, Y_D, Z_D)$
It is straightforward to determine $X_B$, $X_C$, and $Y_C$. I am having difficulty finding X, Y, Z coordinate of the point D with respect to the point A.
To determine Xd, Yd, Zd, I am using the following equations
$(X_D-X_A)^2 + (Y_D-Y_A)^2 + (Z_D-Z_A)^2 = AD^2$
$(X_D-X_B)^2 + (Y_D-Y_B)^2 + (Z_D-Z_B)^2 = BD^2$
$(X_D-X_C)^2 + (Y_D-Y_C)^2 + (Z_D-Z_C)^2 = CD^2$
After doing some expansion and eliminating the 2nd order variables I came up with the following:
$Ax = b$
$x = (X_D, Y_D, Z_D)$
after doing the pseudo inverse on both sides I can solve for x. I am not sure if my method of solving $X_D$, $Y_D$, and $Z_D$ is correct.
4 points diagram
linear-algebra trigonometry plane-geometry
edited Nov 16 at 20:23
Vasya
3,2791515
asked Nov 16 at 19:49
Bhavin
61
your method is correct, just be aware that you can have multiple solutions (for example, point $B'$ can have coordinates $(-X_B,0,0)$, etc.). Also sometimes there is no solution (triangle inequality).
– Vasya
Nov 16 at 20:15
Yes you are correct, it can have multiple solutions but for the Points B and C, I am assuming that corresponding planes are positive.
– Bhavin
Nov 16 at 20:38
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I want to find the $x, y, z$ coordinates for $4$ points in a 3D space. Point $A$ is my origin $(X, Y, Z = 0,0,0)$ and other points $B, C, D$ are with reference to point A. I know all six distances from $A to B, Ato C, Ato D, Bto C, Bto D$, and $Cto D$.
Point $B$ lies on axis $X$, point $C$ lies in $XY$ plane, and lastly, point $D$ can be anywhere.
Coordinates of the points $A, B, C, D$ are as follows:
$A = (0, 0, 0)$
$B = (X_B, 0, 0)$
$C = (X_C, Y_C, 0)$
$D = (X_D, Y_D, Z_D)$
It is straightforward to determine $X_B$, $X_C$, and $Y_C$. I am having difficulty finding X, Y, Z coordinate of the point D with respect to the point A.
To determine Xd, Yd, Zd, I am using the following equations
$(X_D-X_A)^2 + (Y_D-Y_A)^2 + (Z_D-Z_A)^2 = AD^2$
$(X_D-X_B)^2 + (Y_D-Y_B)^2 + (Z_D-Z_B)^2 = BD^2$
$(X_D-X_C)^2 + (Y_D-Y_C)^2 + (Z_D-Z_C)^2 = CD^2$
After doing some expansion and eliminating the 2nd order variables I came up with the following:
$Ax = b$
$x = (X_D, Y_D, Z_D)$
after doing the pseudo inverse on both sides I can solve for x. I am not sure if my method of solving $X_D$, $Y_D$, and $Z_D$ is correct.
4 points diagram
linear-algebra trigonometry plane-geometry
edited Nov 16 at 20:23
Vasya
3,2791515
asked Nov 16 at 19:49
Bhavin
61
I want to find the $x, y, z$ coordinates for $4$ points in a 3D space. Point $A$ is my origin $(X, Y, Z = 0,0,0)$ and other points $B, C, D$ are with reference to point A. I know all six distances from $A to B, Ato C, Ato D, Bto C, Bto D$, and $Cto D$.
Point $B$ lies on axis $X$, point $C$ lies in $XY$ plane, and lastly, point $D$ can be anywhere.
Coordinates of the points $A, B, C, D$ are as follows:
$A = (0, 0, 0)$
$B = (X_B, 0, 0)$
$C = (X_C, Y_C, 0)$
$D = (X_D, Y_D, Z_D)$
It is straightforward to determine $X_B$, $X_C$, and $Y_C$. I am having difficulty finding X, Y, Z coordinate of the point D with respect to the point A.
To determine Xd, Yd, Zd, I am using the following equations
$(X_D-X_A)^2 + (Y_D-Y_A)^2 + (Z_D-Z_A)^2 = AD^2$
$(X_D-X_B)^2 + (Y_D-Y_B)^2 + (Z_D-Z_B)^2 = BD^2$
$(X_D-X_C)^2 + (Y_D-Y_C)^2 + (Z_D-Z_C)^2 = CD^2$
After doing some expansion and eliminating the 2nd order variables I came up with the following:
$Ax = b$
$x = (X_D, Y_D, Z_D)$
after doing the pseudo inverse on both sides I can solve for x. I am not sure if my method of solving $X_D$, $Y_D$, and $Z_D$ is correct.
4 points diagram
linear-algebra trigonometry plane-geometry
linear-algebra trigonometry plane-geometry
edited Nov 16 at 20:23
Vasya
3,2791515
asked Nov 16 at 19:49
Bhavin
61
edited Nov 16 at 20:23
Vasya
3,2791515
asked Nov 16 at 19:49
Bhavin
61
edited Nov 16 at 20:23
Vasya
3,2791515
edited Nov 16 at 20:23
Vasya
3,2791515
edited Nov 16 at 20:23
Vasya
3,2791515
3,2791515
asked Nov 16 at 19:49
Bhavin
61
asked Nov 16 at 19:49
Bhavin
61
asked Nov 16 at 19:49
Bhavin
61
61
your method is correct, just be aware that you can have multiple solutions (for example, point $B'$ can have coordinates $(-X_B,0,0)$, etc.). Also sometimes there is no solution (triangle inequality).
– Vasya
Nov 16 at 20:15
Yes you are correct, it can have multiple solutions but for the Points B and C, I am assuming that corresponding planes are positive.
– Bhavin
Nov 16 at 20:38
add a comment |
your method is correct, just be aware that you can have multiple solutions (for example, point $B'$ can have coordinates $(-X_B,0,0)$, etc.). Also sometimes there is no solution (triangle inequality).
– Vasya
Nov 16 at 20:15
Yes you are correct, it can have multiple solutions but for the Points B and C, I am assuming that corresponding planes are positive.
– Bhavin
Nov 16 at 20:38
your method is correct, just be aware that you can have multiple solutions (for example, point $B'$ can have coordinates $(-X_B,0,0)$, etc.). Also sometimes there is no solution (triangle inequality).
– Vasya
Nov 16 at 20:15
your method is correct, just be aware that you can have multiple solutions (for example, point $B'$ can have coordinates $(-X_B,0,0)$, etc.). Also sometimes there is no solution (triangle inequality).
– Vasya
Nov 16 at 20:15
Yes you are correct, it can have multiple solutions but for the Points B and C, I am assuming that corresponding planes are positive.
– Bhavin
Nov 16 at 20:38
Yes you are correct, it can have multiple solutions but for the Points B and C, I am assuming that corresponding planes are positive.
– Bhavin
Nov 16 at 20:38
add a comment |
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your method is correct, just be aware that you can have multiple solutions (for example, point $B'$ can have coordinates $(-X_B,0,0)$, etc.). Also sometimes there is no solution (triangle inequality).
– Vasya
Nov 16 at 20:15
Yes you are correct, it can have multiple solutions but for the Points B and C, I am assuming that corresponding planes are positive.
– Bhavin
Nov 16 at 20:38