A circle of radius $10$ units is centered at $(0, 0, 50)$. The plane it lies in, has a unit normal vector $(1, 0, 0)$. A uniform direction ray of light falls on the circle, generating a shadow on the floor, which is the $xy$ plane. The direction vector of the ray is $(4, -2, -3)$. Find the equation of the shadow (which is an ellipse).
My Attempt:
The parametric equation of the circle is
$ c(t) = (0, 0, 50) + 10 \cos t \ (0, 1, 0) + 10 \sin t \ (0, 0, 1) $
Projecting this on the plane that passes through the origin and whose normal is $(4, -2, -3)$, can be performed as follows:
First the projection matrix is
$P = I - \dfrac{ {n n}^T }{n^T n} $
This evaluates to
$P = I - \dfrac{1}{29} \begin{bmatrix} 16 && -8 && -12 \\ -8 && 4 && 6 \\ -12 && 6 && 9 \end{bmatrix} = \dfrac{1}{29} \begin{bmatrix} 13 && 8 && 12 \\ 8 && 25 && -6 \\ 12 && -6 && 20 \end{bmatrix} $
Applying $P$ to $c(t)$ gives
$ c'(t) = \dfrac{1}{29} \bigg( (600, -300, 1000) + 10 \cos t (8, 25, -6) + 10 \sin t (12, -6, 20) \bigg) $
From this parametric equation, we can get the implicit algebraic equation of the elliptic cylinder that the projection lies on.
Note that
$ c'(t) = V_0 + \cos t \ V_1 + \sin t \ V_2 $
So that points on the cylinder are given by
$ r = V_0 + \cos t \ V_1 + \sin t \ V_2 + s V_3 $
where $V_3 = (4, -2, -3) $ and $s \in \mathbb{R} $.
I now have,
$ r - V_0 = V u + s V_3 $
where $ V = [V_1, V_2] $ and $u = [\cos t , \sin t ]^T $
Pre-multiplying by $V^T$, gives, (because $V_1$, and $V_2$ are orthogonal to $V_3$)
$ V^T (r - V_0) = (V^T V) u $
So that
$ u = (V^T V)^{-1} V^T (r - V_0) $
Now, $u^T u = 1 $, so, the equation of the cylinder will be
$ (r - V_0)^T V (V^T V)^{-2} V^T (r - V_0) = 1 $
The matrix $G = V (V^T V)^{-2} V^T $ equals
$ G = \begin{bmatrix} 0.008125 && 0.005 && 0.0075 \\ 0.005 && 0.01 && 0 \\ 0.0075 && 0 && 0.01 \end{bmatrix} $
Now putting $z = 0 $ in the equation of the cylinder, gives the equation of the shadow as
$ 0.008125 x^2 + 0.01 xy + 0.01 y^2 - 0.75 x + 24 = 0 $
I look forward to your comments, hints, verification of my solution, and alternative solutions. Thank you all.
The corresponding points on the $xy$ plane are given by:
$$\vec {p_2}=\vec {p_1}+\frac{\lambda}{|\vec r|}\vec r$$
Where $$0=p_{1,z}+\frac{\lambda}{|\vec r|}r_z\Rightarrow\lambda=-p_{1,z}\frac{|\vec r|}{r_z}$$
So, $$\vec {p_2}=\vec {p_1}-\frac{p_{1,z}}{r_z}\vec r$$
Using your parametrisation:
$$x_2=0-\frac{50+10\sin t}{-3}\times4\\ y_2=10\cos t-\frac{50+10\sin t}{-3}\times(-2)\\ z_2=0=50+10\sin t-\frac{50+10\sin t}{-3}\times(-3)$$
$$\therefore x_2=\frac{200+40\sin t}{3}, y_2=\frac{30\cos t -20\sin t-100}{3}\\ $$
We eliminate $t$: $$\sin t=\frac{3x_2-200}{40}\\ \cos t = \pm\sqrt{1-\left(\frac{3x_2-200}{40}\right)^2}=\frac{3y_2+20\sin t + 100}{30}\\ \left(\frac{3y_2+20\frac{3x_2-200}{40}+ 100}{30}\right)^2=1-\left(\frac{3x_2-200}{40}\right)^2$$
The final equation is the following: $$\begin{align} \frac{13 x^2}{1600} + \frac{x y}{100} - \frac{3 x}{4} + \frac{y^2}{100} + 24 = 0\end{align}$$
Our coefficents of $x$ differ but the other coefficients are the same