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3D point from a know point, distance, angle.

1

$begingroup$

Let's assume I have a point in 3D space A(x,y,z).
Two points of distance 'd' from that point A with angle $alpha$ (with XY plane), $beta$ (with XZ plane).What are those two points ?

geometry vectors 3d

share|cite|improve this question

edited Mar 26 at 22:06

Aretino

25.9k31546

asked Mar 26 at 17:51

Maifee Ul AsadMaifee Ul Asad

1177

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  not quite sure what you mean by point with angle
  $endgroup$
  – Vasya
  Mar 26 at 17:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I am adding an image, give me few seconds
  $endgroup$
  – Maifee Ul Asad
  Mar 26 at 17:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  i have edited,please see
  $endgroup$
  – Maifee Ul Asad
  Mar 26 at 18:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Haha, that truly is a "bad image" :)
  $endgroup$
  – Rohit Pandey
  Mar 26 at 18:08
  

  
  
  
  

add a comment | 

1

$begingroup$

Let's assume I have a point in 3D space A(x,y,z).
Two points of distance 'd' from that point A with angle $alpha$ (with XY plane), $beta$ (with XZ plane).What are those two points ?

geometry vectors 3d

share|cite|improve this question

edited Mar 26 at 22:06

Aretino

25.9k31546

asked Mar 26 at 17:51

Maifee Ul AsadMaifee Ul Asad

1177

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  not quite sure what you mean by point with angle
  $endgroup$
  – Vasya
  Mar 26 at 17:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I am adding an image, give me few seconds
  $endgroup$
  – Maifee Ul Asad
  Mar 26 at 17:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  i have edited,please see
  $endgroup$
  – Maifee Ul Asad
  Mar 26 at 18:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Haha, that truly is a "bad image" :)
  $endgroup$
  – Rohit Pandey
  Mar 26 at 18:08
  

  
  
  
  

add a comment | 

$begingroup$

Let's assume I have a point in 3D space A(x,y,z).
Two points of distance 'd' from that point A with angle $alpha$ (with XY plane), $beta$ (with XZ plane).What are those two points ?

geometry vectors 3d

share|cite|improve this question

edited Mar 26 at 22:06

Aretino

25.9k31546

asked Mar 26 at 17:51

Maifee Ul AsadMaifee Ul Asad

1177

$endgroup$

share|cite|improve this question

edited Mar 26 at 22:06

Aretino

25.9k31546

asked Mar 26 at 17:51

Maifee Ul AsadMaifee Ul Asad

1177

share|cite|improve this question

edited Mar 26 at 22:06

Aretino

25.9k31546

asked Mar 26 at 17:51

Maifee Ul AsadMaifee Ul Asad

1177

edited Mar 26 at 22:06

Aretino

25.9k31546

edited Mar 26 at 22:06

Aretino

25.9k31546

asked Mar 26 at 17:51

Maifee Ul AsadMaifee Ul Asad

1177

asked Mar 26 at 17:51

Maifee Ul AsadMaifee Ul Asad

1177

1 Answer
1

active

oldest

votes

2

$begingroup$

Let the point you're interested in be $(a,b,c)$. Now, this point is distance $d$ from $A$. You get the first equation:

$$(a-x)^2+(b-y)^2+(c-z)^2=d^2$$

Also, it seems the position vector of this point has an angle $alpha$ with the x-y plane, meaning $fracpi2-alpha$ with the z-axis. So the second equation:

$$tan(fracpi2-alpha) = fracca^2+b^2+c^2$$
$$=> a^2+b^2+c^2 = c tan(alpha)$$

Similarly, the third equation becomes:

$$tan(fracpi2-beta) = fracba^2+b^2+c^2$$
$$=>a^2+b^2+c^2=btan(beta)$$

This makes it three equations in three unknowns ($a,b,c$). It's a system of quadratic equations. One way to solve them would be to use Buchberger's algorithm, which can solve an arbitrary system of Polynomial equations. See section 2.7 here and the python package, sympy implements it. See here section on solving polynomial equations. Although I'd recommend the python library, I also implemented this algorithm in C# a while back. See here.

share|cite|improve this answer

edited Mar 26 at 18:23

answered Mar 26 at 18:02

Rohit PandeyRohit Pandey

1,7651024

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  thanks for this great help, you are just awesome. i can't tell how much this was for me.
  $endgroup$
  – Maifee Ul Asad
  Mar 26 at 18:40
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Glad to help :)
  $endgroup$
  – Rohit Pandey
  Mar 26 at 18:40
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  i can't up-vote, but i accepted it, i wish i could
  $endgroup$
  – Maifee Ul Asad
  Mar 26 at 18:41
  

  
  
  
  

add a comment | 

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2

$begingroup$

Let the point you're interested in be $(a,b,c)$. Now, this point is distance $d$ from $A$. You get the first equation:

$$(a-x)^2+(b-y)^2+(c-z)^2=d^2$$

Also, it seems the position vector of this point has an angle $alpha$ with the x-y plane, meaning $fracpi2-alpha$ with the z-axis. So the second equation:

$$tan(fracpi2-alpha) = fracca^2+b^2+c^2$$
$$=> a^2+b^2+c^2 = c tan(alpha)$$

Similarly, the third equation becomes:

$$tan(fracpi2-beta) = fracba^2+b^2+c^2$$
$$=>a^2+b^2+c^2=btan(beta)$$

This makes it three equations in three unknowns ($a,b,c$). It's a system of quadratic equations. One way to solve them would be to use Buchberger's algorithm, which can solve an arbitrary system of Polynomial equations. See section 2.7 here and the python package, sympy implements it. See here section on solving polynomial equations. Although I'd recommend the python library, I also implemented this algorithm in C# a while back. See here.

share|cite|improve this answer

edited Mar 26 at 18:23

answered Mar 26 at 18:02

Rohit PandeyRohit Pandey

1,7651024

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  thanks for this great help, you are just awesome. i can't tell how much this was for me.
  $endgroup$
  – Maifee Ul Asad
  Mar 26 at 18:40
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Glad to help :)
  $endgroup$
  – Rohit Pandey
  Mar 26 at 18:40
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  i can't up-vote, but i accepted it, i wish i could
  $endgroup$
  – Maifee Ul Asad
  Mar 26 at 18:41
  

  
  
  
  

add a comment | 

2

$begingroup$

Let the point you're interested in be $(a,b,c)$. Now, this point is distance $d$ from $A$. You get the first equation:

$$(a-x)^2+(b-y)^2+(c-z)^2=d^2$$

Also, it seems the position vector of this point has an angle $alpha$ with the x-y plane, meaning $fracpi2-alpha$ with the z-axis. So the second equation:

$$tan(fracpi2-alpha) = fracca^2+b^2+c^2$$
$$=> a^2+b^2+c^2 = c tan(alpha)$$

Similarly, the third equation becomes:

$$tan(fracpi2-beta) = fracba^2+b^2+c^2$$
$$=>a^2+b^2+c^2=btan(beta)$$

This makes it three equations in three unknowns ($a,b,c$). It's a system of quadratic equations. One way to solve them would be to use Buchberger's algorithm, which can solve an arbitrary system of Polynomial equations. See section 2.7 here and the python package, sympy implements it. See here section on solving polynomial equations. Although I'd recommend the python library, I also implemented this algorithm in C# a while back. See here.

share|cite|improve this answer

edited Mar 26 at 18:23

answered Mar 26 at 18:02

Rohit PandeyRohit Pandey

1,7651024

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  thanks for this great help, you are just awesome. i can't tell how much this was for me.
  $endgroup$
  – Maifee Ul Asad
  Mar 26 at 18:40
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Glad to help :)
  $endgroup$
  – Rohit Pandey
  Mar 26 at 18:40
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  i can't up-vote, but i accepted it, i wish i could
  $endgroup$
  – Maifee Ul Asad
  Mar 26 at 18:41
  

  
  
  
  

add a comment | 

$begingroup$

Let the point you're interested in be $(a,b,c)$. Now, this point is distance $d$ from $A$. You get the first equation:

$$(a-x)^2+(b-y)^2+(c-z)^2=d^2$$

Also, it seems the position vector of this point has an angle $alpha$ with the x-y plane, meaning $fracpi2-alpha$ with the z-axis. So the second equation:

$$tan(fracpi2-alpha) = fracca^2+b^2+c^2$$
$$=> a^2+b^2+c^2 = c tan(alpha)$$

Similarly, the third equation becomes:

$$tan(fracpi2-beta) = fracba^2+b^2+c^2$$
$$=>a^2+b^2+c^2=btan(beta)$$

This makes it three equations in three unknowns ($a,b,c$). It's a system of quadratic equations. One way to solve them would be to use Buchberger's algorithm, which can solve an arbitrary system of Polynomial equations. See section 2.7 here and the python package, sympy implements it. See here section on solving polynomial equations. Although I'd recommend the python library, I also implemented this algorithm in C# a while back. See here.

share|cite|improve this answer

edited Mar 26 at 18:23

answered Mar 26 at 18:02

Rohit PandeyRohit Pandey

1,7651024

$endgroup$

share|cite|improve this answer

edited Mar 26 at 18:23

answered Mar 26 at 18:02

Rohit PandeyRohit Pandey

1,7651024

answered Mar 26 at 18:02

Rohit PandeyRohit Pandey

1,7651024

answered Mar 26 at 18:02

Rohit PandeyRohit Pandey

1,7651024