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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


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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

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
Then you must rotate the plane using standard rotation matrices, and project onto the resulting plane. – Emily Sep 11 '12 at 15:04

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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