In 3D computer graphics, one needs to represent 3-dimensional objects on a plane screen, by drawing a given point P at the place where the line from P to the eye meets the screen. Suppose that the screen is the yz-plane, and the eye is at E : (2, 0, 0).
At what point Q : $(y, z)$ in the yz-plane should one represent the point P : $(x_0, y_0, z_0)$? (Express y and z in terms of the coordinates of P. Assume that $x_0 < 2$. Why is this assumption legitimate?
I was trying to:
we know that a normal vector to the plane is $N = \langle 1,0,0\rangle$ hence:
$$\begin{align*} \mathrm{proj}P_N &= \langle x_0,0,0\rangle\\ Q + \mathrm{proj}P_N &= P\\ Q &= P -\mathrm{proj}P_N \end{align*}$$
hence $$Q = \langle 0,y_0,z_0\rangle$$
is this correct?
No, what you have described is an orthogonal projection onto the screen. What you want is a projection through the point $E$. So you want to solve $$Q=\lambda E+(1-\lambda)P\tag{1}$$ (This is the parametrization for a line throught two points) $$Q\cdot \langle 1,0,0\rangle=0$$ (This says $Q$ must lie in plane perpendicular to $x$-axis)
So you get
\begin{align*} 0 = (\lambda E+(1-\lambda)P)\cdot \langle 1,0,0\rangle = 2\lambda +(1-\lambda)P_x \end{align*} So $$\lambda=\frac{P_x}{P_x-2}\text{,}$$ and your point is obtained by substituting this value of $\lambda$ into (1) (observe $1-\lambda = \frac{P_x-2}{P_x-2}-\frac{P_x}{P_x-2}=-\frac{2}{P_x-2}$) :
\begin{align*} Q &=\frac{P_x}{P_x-2}E-\frac{2}{P_x-2}P\\ & = \langle \frac{P_x}{P_x-2}\cdot 2,0,0\rangle-\langle \frac{2}{P_x-2}\cdot P_x,\frac{2}{P_x-2}\cdot P_y, \frac{2}{P_x-2}\cdot P_z\rangle\\ &=\langle 0,\frac{2P_y}{2-P_x},\frac{2P_z}{2-P_x}\rangle \end{align*}
I will note you could also determine this pretty easily from similar triangles, but perhaps the method I used might be more mechanical and easier to generalize if, for example, you have a plane determined by an arbitrary normal vector.