<p>Here's a hint to get you started: Parallel lines include the line through the camera. </p>

<p>So really all you need is the direction from the camera to the vanishing point on the view plane. Then create lines parallel to that line.</p>

CC BY-SA 3.0
2016-05-08T13:58:57.273Z
I have tough about that, so if my camera in $(0,0,0)$ and it's looking to $(0,0,1)$, the view plane is $z=1$ and lets say that the vanishing point is $(10,0,1)$. Then the line from the camera to the vanishing point is $(0,0,0)+t(10,0,1)$. So all the parallel lines that are going to that vanishing point will be $(a_1,a_2,a_3) + t(10,0,1)$? And then all the lines in plane $y=y_0$ that have that vanishing point will be $(a_1,y_0,a_3) + t(10,0,1)$?
nrofis
CC BY-SA 3.0
2016-05-08T14:08:22.040Z
stop focusing on the $y = y_0$ plane, just having the vanishing point on the view plane and the camera position is enough.
ratchet freak
CC BY-SA 3.0
2016-05-08T14:09:50.400Z
OK, so $(a_1,a_2,a_3)+t(10,0,1)$ are all the lines that their vanishing point is $(10,0,1)$?
nrofis
CC BY-SA 3.0
2016-05-08T13:42:24.037Z
2016-05-08T13:42:24.037Z
ratchet freak
<p>In perspective projection, group of parallel lines have the same vanishing point. I am interesting about the reverse calculation: Getting the group of parallel lines equations that their vanishing point specific point.</p>
<p>Say I know that the camera is perspective camera at <span class="math-container">$(0,0,0)$</span> and it's direction <span class="math-container">$(0,0,1)$</span>, the view plane is <span class="math-container">$z = 1$</span> and I am interesting about the lines in plane <span class="math-container">$y= y_0$</span> that their vanishing point is <span class="math-container">$P = (p_x,p_y,p_z)$</span>.</p>
<p>I have tried to calculate the projection point of some point <span class="math-container">$(x,y_0,z)$</span> and get the equations:<br />
(i) <span class="math-container">$p_x = x(\frac1z)$</span><br />
(ii) <span class="math-container">$p_y = y_0(\frac1z)$</span><br />
(iii) <span class="math-container">$p_z = 1$</span></p>
<p>But it seems wrong because if the vanishing point is like <span class="math-container">$(x_0,0,*)$</span> then form (ii) we will get <span class="math-container">$z\rightarrow \infty$</span> but then (i) is wrong because <span class="math-container">$x(\frac1z)\rightarrow0$</span> but it need to be equals to <span class="math-container">$x_0$</span>.</p>
<p>So how can I get the group of parallel lines have the same vanishing point in these conditions?</p>

CC BY-SA 3.0
2016-05-08T13:06:10.150Z
The lines in plane $y = y_0$ are not all parallel.
ratchet freak
CC BY-SA 3.0
2016-05-08T13:07:41.200Z
Of course, I need a group of parallel lines in plane $y=y_0$ that their vanishing point is some point, say $(x_0,0,1)$ (the camera position and direction and the view plane are defined in the question)
nrofis
<3d><perspective>
Calculate vanishing point
nrofis
# Calculate vanishing point

(Asked by: username_0 on 2016-05-08)

In perspective projection, group of parallel lines have the same vanishing point. I am interesting about the reverse calculation: Getting the group of parallel lines equations that their vanishing point specific point.

Say I know that the camera is perspective camera at $(0,0,0)$ and it's direction $(0,0,1)$, the view plane is $z = 1$ and I am interesting about the lines in plane $y= y\_0$ that their vanishing point is $P = (p\_x,p\_y,p\_z)$.

I have tried to calculate the projection point of some point $(x,y\_0,z)$ and get the equations:  
(i) $p\_x = x(\frac1z)$  
(ii) $p\_y = y\_0(\frac1z)$  
(iii) $p\_z = 1$

But it seems wrong because if the vanishing point is like $(x\_0,0,\*)$ then form (ii) we will get $z\rightarrow \infty$ but then (i) is wrong because $x(\frac1z)\rightarrow0$ but it need to be equals to $x\_0$.

So how can I get the group of parallel lines have the same vanishing point in these conditions?

**Tags:** 3d, perspective

> **Comments:**
> **username_1**: The lines in plane $y = y_0$ are not all parallel.  
> **username_0**: Of course, I need a group of parallel lines in plane $y=y_0$ that their vanishing point is some point, say $(x_0,0,1)$ (the camera position and direction and the view plane are defined in the question)  

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## 1 answers

### Answer by username_1 on 2016-05-08. Score: 2 | Accepted answer

Here's a hint to get you started: Parallel lines include the line through the camera.

So really all you need is the direction from the camera to the vanishing point on the view plane. Then create lines parallel to that line.

> **Comments:**
> **username_0**: I have tough about that, so if my camera in $(0,0,0)$ and it's looking to $(0,0,1)$, the view plane is $z=1$ and lets say that the vanishing point is $(10,0,1)$. Then the line from the camera to the vanishing point is $(0,0,0)+t(10,0,1)$. So all the parallel lines that are going to that vanishing point will be $(a_1,a_2,a_3) + t(10,0,1)$? And then all the lines in plane $y=y_0$ that have that vanishing point will be $(a_1,y_0,a_3) + t(10,0,1)$?  
> **username_1**: stop focusing on the $y = y_0$ plane, just having the vanishing point on the view plane and the camera position is enough.  
> **username_0**: OK, so $(a_1,a_2,a_3)+t(10,0,1)$ are all the lines that their vanishing point is $(10,0,1)$?  

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