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I have points in 3D system like this

$$p1=(2,3,4)$$ $$p2=(3,5,5)$$

Here I would like draw point $p1$ and $p2$ in $2D$ view.

Project type = orthographic. Coordinate system = Cartesian

X- axis, min = 2, max=9 y-axis min=2, max=12 z-axis min=1, max=10

Basically I would like draw $3D$ points in $2D$ view.(using Cartesian coordinate system)

Please help me find an answer for the following question.

1) how can convert $3D$ points $(p1, p2)$ to $2D$ points. What is the formula for this?

I cann't upload images yet, as I need at least 10 reputation as per the forum rules.

Any idea

Thanks

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If you need of 10 of reputation I will give one vote. –  user29999 Sep 11 '12 at 14:56
    
Answer would be great help than Vote! –  flex Sep 11 '12 at 15:01
    
You need to specify which plane you want the points projected onto. –  copper.hat Sep 11 '12 at 15:17
    
To do an orthographic projection you need to specify a plane onto which the original points are projected. The plane is usually described by a 3D vector. A projection onto the x-y plane would use $(0,0,1)$, another 'perspective-like' projection would be $(1,1,1)$. –  copper.hat Sep 11 '12 at 16:26
    
when it is rotatable, it is difficult identify which plane is in the eye view(visible). am telling in perspective of programming –  flex Sep 11 '12 at 16:59
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2 Answers

You also need to specify the orientation of the plane you are projecting onto. The easiest examples are planes perpendicular to one of the axes. So if you project onto a plane perpendicular to $z$, your get $p1=(2,3), p2=(3,5)$

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What if rectangle is rotatable? –  flex Sep 11 '12 at 15:01
1  
Then you must rotate the plane using standard rotation matrices, and project onto the resulting plane. –  Arkamis Sep 11 '12 at 15:04
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If I understand correctly, orthographic is parallel projection.

Pick a normalized direction $h \in \mathbb{R}^3$ (ie, $\|h\| = 1$). Then you will be projecting onto the plane $\{x \in \mathbb{R}^3 | h^T x = 0\}$. An example would be $h = (0,0,1)$ which would be a plan view.

Then the projection $P: \mathbb{R}^3 \to \mathbb{R}^3$ is given by $P(x) = (I - h h^T) x = x - \langle h, x \rangle h$.

For example, if $h = (0,0,1)$, then $P((x,y,z)) = (x,y,0)$. (This is essentially the same as Ross' example.)

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Help me to solve the equation by using above points. –  flex Sep 11 '12 at 15:10
    
@flex: You haven't given enough information to uniquely solve the problem. –  copper.hat Sep 11 '12 at 15:23
    
could u pleas tell me, What other information is missing? –  flex Sep 11 '12 at 16:20
    
@flex: the missing information is the plane you want to project onto. It could be the $xy$ plane, as in my example. It could be the $yz$ plane. It could be the plane perpendicular to $(1,1,1)$ or any other. Think of looking at your box from various angles. We need to specify the viewpoint. –  Ross Millikan Sep 12 '12 at 13:09
    
Ok, lets Assume XY plane –  flex Sep 12 '12 at 13:24
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