# Find the normal vector to the projection plane

In the context of perspective projection. Given focal length is $2.387$, the camera is at $(0.0.0)$ looking at $-z$ direction A rectangle lies on a plane tilted from view plane. Also given the projected point on the view plane is

$P_1=(0,2.8)$

$P_2=(1.8,5.1)$

$P_3=(0.9,6.7)$

$P_4=(-1,4.4)$

How do you find the normal to the world plane that this rectangle lies on.

My idea is to first find the real world coordinates of these four points by using focal length, of cause, $z$ coord is lost.

However, I think the $z$ coordinate relation of these four points can be found by using this tilt argument, so set $\theta$ be the angle from view plane to world plane, such that all $z$ coord of $4$ points can be written in a relation.

But thing get quite tricky, just wondering is there any easy way to do it?

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If all is well and these four points actually lie on a plane without being on a same line, you can find three of them such that, say, $\vec{P_1P_2}$ and $\vec{P_1P_3}$ are not collinear/parallel/proportional.
Then the cross product will do what you want: $$\vec{P_1P_2}\times \vec{P_1P_3}.$$