It looks like you're new here. If you want to get involved, click one of these buttons!

- 141.6K All Categories
- 104.7K Programming Languages
- 6.4K Assembler Developer
- 1.9K Basic
- 39.8K C and C++
- 4.3K C#
- 7.9K Delphi and Kylix
- 4 Haskell
- 9.6K Java
- 4.1K Pascal
- 1.3K Perl
- 2K PHP
- 518 Python
- 37 Ruby
- 4.3K VB.NET
- 1.6K VBA
- 20.8K Visual Basic
- 2.6K Game programming
- 312 Console programming
- 89 DirectX Game dev
- 1 Minecraft
- 110 Newbie Game Programmers
- 2 Oculus Rift
- 8.9K Applications
- 1.8K Computer Graphics
- 730 Computer Hardware
- 3.4K Database & SQL
- 522 Electronics development
- 1.6K Matlab
- 628 Sound & Music
- 257 XML Development
- 3.3K Classifieds
- 196 Co-operative Projects
- 185 For sale
- 189 FreeLance Software City
- 1.9K Jobs Available
- 600 Jobs Wanted
- 201 Wanted
- 2.9K Microsoft .NET
- 1.7K ASP.NET
- 1.1K .NET General
- 3.3K Miscellaneous
- 4 Join the Team
- 0 User Profiles
- 353 Comments on this site
- 60 Computer Emulators
- 2.1K General programming
- 183 New programming languages
- 604 Off topic board
- 170 Mobile & Wireless
- 44 Android
- 124 Palm Pilot
- 335 Multimedia
- 151 Demo programming
- 184 MP3 programming
- 0 Bash scripts
- 19 Cloud Computing
- 53 FreeBSD
- 1.7K LINUX programming
- 367 MS-DOS
- 0 Shell scripting
- 320 Windows CE & Pocket PC
- 4.1K Windows programming
- 895 Software Development
- 408 Algorithms
- 68 Object Orientation
- 89 Project Management
- 90 Quality & Testing
- 239 Security
- 7.6K WEB-Development
- 1.8K Active Server Pages
- 61 AJAX
- 2 Bootstrap Themes
- 55 CGI Development
- 19 ColdFusion
- 222 Flash development
- 1.4K HTML & WEB-Design
- 1.4K Internet Development
- 2.2K JavaScript
- 34 JQuery
- 286 WEB Servers
- 151 WEB-Services / SOAP

Shkaboinka
Posts: **79**Member

in 3D Graphics

I need to know how to find the line perendicular to a plane in 3D space. More specifically, I need to do this: I need find point D, where line AD is perpendiculer to plane ABC, and the distance from A to D is 1 (A,B,C, and D are all 3D {x,y,z} points).

The point of this is that when D is further away than A, the plane will not be drawn because it is not facing the "camera". I am using MY OWN 3D graphics routines through 2D graphics provided in Java; don't tell me how I should be using something to do this for me like OpenGL; I KNOW it's all much better...I just want to do this on my own.

The point of this is that when D is further away than A, the plane will not be drawn because it is not facing the "camera". I am using MY OWN 3D graphics routines through 2D graphics provided in Java; don't tell me how I should be using something to do this for me like OpenGL; I KNOW it's all much better...I just want to do this on my own.

Terms of use / Privacy statement / Publisher: Lars Hagelin

Programmers Heaven articles / Programmers Heaven files / Programmers Heaven uploaded content / Programmers Heaven C Sharp ebook / Operated by CommunityHeaven LLC

© 1997-2015 Programmersheaven.com - All rights reserved.

## Comments

22Member: I need to know how to find the line perendicular to a plane in 3D space. More specifically, I need to do this: I need find point D, where line AD is perpendiculer to plane ABC, and the distance from A to D is 1 (A,B,C, and D are all 3D {x,y,z} points).

:

: The point of this is that when D is further away than A, the plane will not be drawn because it is not facing the "camera". I am using MY OWN 3D graphics routines through 2D graphics provided in Java; don't tell me how I should be using something to do this for me like OpenGL; I KNOW it's all much better...I just want to do this on my own.

:

Well, you can always find a vector normal (perpendicular) to a plane by taking the cross product of two vectors lying within the plane...

Let me explain...

You can always create two vectors lying within a plane if you know three points on the plane. Taking A as our base point, the vector from point A to point B would be

, and the vector from point A to point C would be (just subtract the points). Then take the cross product of those two vectors (I'll call them P and Q respectively), which can be defined as the determinant of the matrix

[code]

|- -|

| i j k |

| Px Py Pz |

| Qx Qy Qz |

|_ _|

[/code]

where i = <1,0,0>, j = <0,1,0>, and k = <0,0,1>. The result will be a new vector that is perpendicular to the plane. Then all you have to do is normalize the new vector (divide each component by the vector's magnitude), and add it to your base point, A. This will give you a new point D a distance 1 from your plane along a line normal to the plane.

Q.E.D. (well, not really, but you know what I mean)

Now test the distance of point D to see if is further away than point A.

You could also use the dot product to find the angle between a vector from the camera to the base point and your normal vector. If the angle is < 90, the plane is facing away from the camera.

Note: you do have to be careful about which vector is placed in what row in the matrix, because reversing the order will still give you a vector normal to the plane, but it will be facing in the opposite direction. So you still have to figure out which vector to place in the second row to give you a vector. The vectors should be arranged in counter-clockwise order to get a normal facing you, and clockwise order to get a normal facing away.

Hope this helps!

Zaffire

P.S. Hey, how much math have you had? You should really take a linear algebra class or a course in vector mechanics.