In a Cartesian 3D coordinate system defined by XYZ, there is a known point $\textbf{P}$ defined with it coordinates $(x_0, y_0, z_0)$ in X-Y plane. Following is the diagram:
$P$ in X-Y plane is vertically (along Z axis) backprojected onto the light blue plane" />
There is a blue plane defined by its norm vector $\textbf{n}$ in this XYZ coordinate system. The blue plane can be obtained by rotating X-Y plane around XYZ axis.
My question is if the point $\textbf{P}$ in X-Y plane is vertically (along Z axis) backprojected onto the light blue plane, can I know the position of its backprojection $\textbf{P'}$, namely $(x'_0, y'_0, z'_0)$? Note an arbitrary X-Y coordiante system on the blue plane is acceptable as long as it is Cartesian.
Any suggestion is highly appreciated.
We find three equations and three unknowns: Since $P'$ is above or below $P$ we must have $x'=x$ and $y'=y.$ The third equations follows from the fact that $P$ lies on the plane, hence $n\cdot P' = n_1x'+n_2y'+n_3z' = 0$ where $n=(n_1,n_2,n_3)$. Since we are interested in $z'$ we solve the third equation provided $n_3\neq 0$ for $z'$: $$z'= \frac{n_1x'+n_2y'}{n_3}.$$ If $n_3=0$ there is either no or infinitely many solutions depending on $n_1x'+n_2y' \overset ?= 0$. This coincides with the geometric interpretation: $n_3=0$ means the plane contains the entire $z$-axis. If now the plane contains $P$, which is the case if $n_1x'+n_2y' = 0$, every point above and below also lies on the plane. Else if the plane does not contain $P$, which is the case if $n_1x'+n_2y' \neq 0$, there is no point on the plane above or below $P$.