# Calculate vanishing point

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 is $(0,0,1)$, 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?

• The lines in plane $y = y_0$ are not all parallel. – ratchet freak May 8 '16 at 13:06
• 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 May 8 '16 at 13:07

• 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 May 8 '16 at 13:58
• 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 May 8 '16 at 14:08
• 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 May 8 '16 at 14:09