Mapping of the midpoint on a tilted plane

Question

There is a sphere and a light source that projects the sphere onto a screen. Considering the sphere and the light source to be in a straight line and the screen to be tilted at unknown angles w.r.t the vertical and horizontal plane about point O, how to find the coordinates of the point on the screen that is collinear to the midpoint of the sphere.

Answer 1

I hope I have understood your set up correctly. First I state the results. There are two of them. Simply the second one is a slight generalization of the first one because I wasn't absolutely sure about how exactly your coordinate system is positioned so I decided to cover my bases. You can simply read the results, but I advise you to look up the derivation and double check the calculations.

Statement 1. The coordinates of the point $M'$ in the coordinate system centered at $O$, with $x$-axis aligned with $OL$, with $z$-axis passing through $O$ and parallel to $NM$, and with $y$-axis chosen uniquely as the axis orthogonal to the $x,z$-coordinate plane and forming right-hand oriented coordinate system with the $x$-axis and $z$-axis, are
$$M' = \left(\,\frac{|OL||LM| - |LM'||LN|}{|LM|} \, , \, \, 0 \, , \,\, \frac{|LM'||MN|}{|LM|} \right)$$ where the lengths $|OL|, \, |LN|$ and $|LM| = m$ are given and $$|LM'| = \sqrt{a_1 a_2 - \frac{a_1 a_2\, a^2}{(a_1+a_2)^2}}$$ with $a_1$ and $a_2$ the two roots of the polynomial $$f(x) = x^2 \, - \, 2\,\left(\frac{\sqrt{\big( m^2 -r^2\big)\, b^2 + r^2 a^2}}{r}\right) \, x \, + \, \frac{m^2 b^2}{ r^2}$$

Statement 2. If the coordinate system is chosen so that $OL$ is the $x$-axis but the $y$-axis and $z$-axis are chosen arbitrarily, so that the line $MN$ is orthogonal to the $x$-axis $OL$ but the angle between $MN$ and the $x, y$-coordinate plane is $\theta$, then the coordinates of the point $M'$ are $$M' = \left(\,\frac{|OL||LM| - |LM'||LN|}{|LM|} \, , \, \, \frac{|LM'||MN|}{|LM|} \cos(\theta) \, , \,\, \frac{|LM'||MN|}{|LM|} \sin(\theta) \right)$$ where the lengths $|OL|, \, |LN|$ and $|LM| = m$ are given and $$|LM'| = \sqrt{a_1 a_2 - \frac{a_1 a_2\, a^2}{(a_1+a_2)^2}}$$ with $a_1$ and $a_2$ the two roots of the polynomial $$f(x) = x^2 \, - \, 2\,\left(\frac{\sqrt{\big( m^2 -r^2\big)\, b^2 + r^2 a^2}}{r}\right) \, x \, + \, \frac{m^2 b^2}{ r^2}$$

Derivation of the statements. First, I will use some coordinate-free geometry, which will allow us to find the distance $|LM'|$ and basically, it will determine the geometry of the tilted circular cone formed by the vertex $L$ and the elliptical shadow on the screen.

Let $L$ be the vertex of the cone of light, $M$ the center of the sphere, $r$ the radius of the sphere, $|LM| = m$, the length of the major semi-axis of the projected ellipse is $a$, while the length of the minor semi-axis is $b$.

Now, let $A_1$ and $A_2$ be the points of intersection of the major axis of the projected ellipse with the contour of that ellipse. Assume that the order of $A_1$ and $A_2$ is chosen so that $|LA_1| < |LA_2|$. Let $B'_1$ and $B'_2$ be the points of intersection of the minor axis of the projected ellipse with the contour of that ellipse. Let us look more carefully at the plane $\omega$ formed by $L, A_1, A_2$. Then, the line $LM$ lies in $\omega$ and $M'$ is the intersection of $LM$ with $A_1A_2$. The sphere intersected by $\omega$ forms a circle tangent to the lines $LA_1$ and $LA_2$. Therefore the line $LM$ is the angle bisector of the triangle $LA_1A_2$ and $M'$ is the intersection of that angle bisector with $A_1A_2$. Thus by a famous property of the angle bisector in a triangle $$\frac{|LA_1|}{|LA_2|} = \frac{|A_1M'|}{|A_2M'|}$$ Let $K$ be the midpoint of $A_1A_2$. Then $|A_1K| = |A_2K| = a$, so $K$ is the center of the ellipse and hence the segment (minor axis) $B'_1B'_2$ passes through $K$ and $K$ is its midpoint too. Now, draw a line in $\omega$ through the point $K$ and perpendicular to the angle bisector $LM \equiv LM'$ and denote the intersection points of that line with $LA_1$ and $LA_2$ by $B_1$ and $B_2$ respectively. Let for simplicity introduce the notations $$|LA_1| = a_1, \,\,\,\, |LA_2| = a_2,\,\,\,\, |B_1K| = b_1, \,\,\,\, |B_2K| = b_2$$ If you define the unique plane $\sigma$ through midpoint $K$ orthogonal to the line $LM$, then the intersection of $\sigma$ and $\omega$ is the line $B_1B_2$ and the intersection of $\sigma$ with the extended (infinite) shadow cone through $L$ is a circle (the shadow of the sphere on the plane $\sigma$) with center the intersection point of $LM$ and $B_1B_2$ of diameter $b_1 + b_2$. Tha latter circle passes through the four points $B_1, B_1', B_2, B'_2$ and $K$ is the intersection point of the two chords in that circle $B_1B_2$ and $B_1'B_2'$. Then by a standard geometric theorem, $$|B_1K|\cdot|B_2K| = |B_1'K|\cdot|B_2'K|$$ Recall $|B_1'K|=|B_2'K| = b$ so the identity above becomes $$b_1 b_2 = b^2$$

Next, let angles $\angle \,A_1LM = \angle \, A_2LM = \alpha$ (recall $LM$ is the angle bisector of $\angle \, A_1LA_2$). Since the circle obtained by the intersection of the sphere with plane $\omega$ is tangent to $LA_1$ and $LA_2$ I can find the angle $\alpha$ as follows $$\sin(\alpha) = \frac{\text{radius of the sphere}}{|LM|} = \frac{r}{m}.$$ Form trigonometry, $\sin(2\alpha) = 2\sin(\alpha)\cos(\alpha)$.

Lemma. $\, b_1 = a_2 \sin(\alpha)$ and $\, b_2 = a_1 \sin(\alpha)$.

Proof: Draw through point $A_1$ the unique line $s_1$ orthogonal to $LM$ and denote by $\hat{A}_1$ the intersection point of $s_1$ with line $LA_2$ and by $M'_1$ the intersection point of line $s_1 \equiv A_1\hat{A}_1$ with line $LM$. Since both lines $A_1\hat{A}_1$ and $B_1B_2$ are orthogonal to $LM$, they are parallel to each other and since $LM$ is orthogonal to $A_1\hat{A}_1$ and also $LM$ is the angle bisector of $\angle \, A_1LA_2 = \angle \, A_1L\hat{A}_1 = 2\alpha$, we can see that $LM$ is the orthogonal bisector of edge $A_1\hat{A}_1$ with midpoint $M'_1$ which means that $|A_1M'_1| = |\hat{A}_1M'_1|$. Focus on triangle $A_1\hat{A}_1A_2$. Point $K$ is the midpoint of $A_1A_2$ and $KB_2$ is parallel to $A_1\hat{A}_1$ (recall $K \in B_1B_2$) and so $$b_2 = |B_2K| = \frac{1}{2} |A_1\hat{A}_1| = |A_1M'_1|$$ Consequently, as triangle $LA_1M'_1$ is right angled with right angle at vertex $M'_1$, we conclude that $$\sin(\alpha) = \frac{|A_1M'_1|}{|LA_1|} = \frac{|B_2K|}{|LA_1|} = \frac{b_2}{a_1} \,\,\, \text{ that is } \,\,\,\, b_2 = a_1 \sin(\alpha).$$ Absolutely analogous arguments lead to the other equality $b_1 = a_2 \sin(\alpha). \,\,\,\,\,\,\, \,\,\,\,\,\, \square$

Corollary. $\, b^2 = b_1 b_2 = a_1 a_2 \sin^2(\alpha)$.

Next step is to apply the cosine theorem to the triangle $A_1A_2L$. It states that $$|LA_1|^2 + |LA_2|^2 - 2 |LA_1| |LA_2| \cos{2 \alpha} = |A_1A_2|^2$$ which in terms of our shorter notation is $$a_1^2 + a_2^2 - 2 a_1 a_2 \cos{2\alpha} = 4a^2$$ Combining the latter equations with the one from the corollary, we get the system of equations \begin{align} & a_1^2 + a_2^2 - 2 \, a_1 a_2 \cos{2\alpha} = 4a^2\\ & a_1 a_2 = \frac{b^2}{ \sin^2(\alpha)} \end{align} By trigonometry $\cos(2\alpha) = \cos^2(\alpha) - \sin^2(\alpha) = 1 - 2\, \sin^2(\alpha)$. Let us reorganize this system a bit \begin{align} & (a_1 + a_2)^2 - 2 ( 1 + \cos{2\alpha} )\, a_1 a_2 = 4a^2\\ & a_1 a_2 = \frac{b^2}{ \sin^2(\alpha)} \end{align} and get to \begin{align} & (a_1 + a_2)^2 = 2 ( 1 + \cos{2\alpha} )\, \frac{b^2}{ \sin^2(\alpha)} + 4a^2 = 4\, \frac{( 1 - \sin^2{\alpha} )\, b^2}{ \sin^2(\alpha)} + 4a^2 \\ & a_1 a_2 = \frac{b^2}{ \sin^2(\alpha)} \end{align} which leads to \begin{align} & a_1 + a_2 = 2\,\frac{\sqrt{\big( m^2 -r^2\big)\, b^2 + r^2 a^2}}{r} \\ & a_1 a_2 = \frac{m^2 b^2}{ r^2} \end{align} If you define the quadratic polynomial $$f(x) = x^2 \, - \, 2\,\left(\frac{\sqrt{\big( m^2 -r^2\big)\, b^2 + r^2 a^2}}{r}\right) \, x \, + \, \frac{m^2 b^2}{ r^2}$$ then $a_1$ and $a_2$ are the two roots of $f(x)$. Now we know how to calculate the ratio, $$\frac{|A_1M'|}{|A_2M'|} = \frac{|LA_1|}{|LA_2|} = \frac{a_1}{a_2}.$$ Consequently, $$|A_1M'| = \frac{a_1\, a}{a_1+a_2} \,\,\, \text{ and } \,\,\, |A_2M'| = \frac{a_2\, a}{a_1+a_2}$$ Now, by another theorem from geometry about a bisector in a triangle we have $|LM'|^2 = |LA_1| |LA_2| - |A_1 M'| |A_2 M'|$ which translates to $$|LM'| = \sqrt{a_1 a_2 - \frac{a_1 a_2\, a^2}{(a_1+a_2)^2}}$$

Now, I am not sure how your reference frame (i.e. orthonormal coordinate system) is exactly positioned, but according to your post $O$ is the origin, and the unit vector $\overrightarrow{e_1}$ is aligned with line $OL$ directed from $O$ to $L$. I guess the vertical unit vector $\overrightarrow{e_3}$ is parallel to line $NM$ which, according to you, is orthogonal to $OL$ with $N \in OL$. Thus $\overrightarrow{e_1}$ and $\overrightarrow{e_3}$ are unit and orthogonal and if we define $\overrightarrow{e_2} = \overrightarrow{e_3} \times \overrightarrow{e_1}$, we end up with the (positively oriented) orthonormal frame $O\overrightarrow{e_1}\overrightarrow{e_2}\overrightarrow{e_3}$. Then the point $M$ lies in the vertical plane spanned by the vectors $O\overrightarrow{e_1}\overrightarrow{e_3}$ and so $$\overrightarrow{LM} = - |LN| \overrightarrow{e_1}+ |MN| \overrightarrow{e_3}$$ A unit vector aligned with the latter vector is then $$\frac{1}{|LM|}\overrightarrow{LM} = - \frac{|LN|}{|LM|} \overrightarrow{e_1} + \frac{|MN|}{|LM|} \overrightarrow{e_3}$$ Points $L, M$ and $M'$ are collinear so
$$\overrightarrow{LM'}=\frac{|LM'|}{|LM|}\overrightarrow{LM} = - \frac{|LM'||LN|}{|LM|} \overrightarrow{e_1} + \frac{|LM'||MN|}{|LM|} \overrightarrow{e_3}$$ By vector addition $$\overrightarrow{OM} = \overrightarrow{OL} + \overrightarrow{LM'} = |OL|\overrightarrow{e_1} - \frac{|LM'||LN|}{|LM|} \overrightarrow{e_1} + \frac{|LM'||MN|}{|LM|} \overrightarrow{e_3}$$ which is $$\overrightarrow{OM} = \left(|OL| - \frac{|LM'||LN|}{|LM|}\right) \overrightarrow{e_1} + \left(\frac{|LM'||MN|}{|LM|}\right) \overrightarrow{e_3}$$ In the case of $MN$ being inclined at angle $\theta$ with respect to the $x, y$-plane the formula becomes

$$\overrightarrow{OM} =$$ $$=\left(\frac{|OL||LM|-|LM'||LN|}{|LM|}\right) \overrightarrow{e_1} + \left(\frac{|LM'||MN|}{|LM|}\right) \cos({\theta}) \, \overrightarrow{e_2} + \left(\frac{|LM'||MN|}{|LM|}\right) \, \sin(\theta) \, \overrightarrow{e_3}$$