Consider a point $P~(P_x,P_y,0)$ which lies somewhere in the Cartesian region $x > 0. $

Consider a simple 3D cone surface with half angle $\gamma$, originating at point $\mathcal{O}$ $(0,0,0)$ where the cone axis is the positive x-axis $(x,0,0)$.

The cone surface can be described by a bundle of an infinite number of straight line segments originating at origin point $\mathcal{O}$ and each making an angle $\gamma$ with the x-axis.

Now there is an infinite set of points $q(qx,qy,qz)$ - one point per line segment - with origin vectors $\textbf{q}$ which are perpendicular to the corresponding vectors $\textbf{Pq}$ which run from point $P$ to the various $q$ points.

I wish to find a formula for the coordinates ($Qx$, $Qy$, $0$) of the "Mean Point" $Q$ defined by the average coordinates of all the $q$ points.


Ideally I would like to follow the pattern used in ElThor's answer to my previous question here (i.e. using the parametrized vector $\textbf{a}=(\cos\gamma, \sin\alpha\,\sin\gamma, \cos\alpha\,\sin\gamma)$. In that question ElThor has provided the formula for the reflected point. Clearly this current question requires the general formula for a perpendicular point, but I haven't been able to figure it out (or find it) so far.


Edit: Based on ElThor's answer to the older question and taking his formula for the reflected point, adding $P$ and dividing by 2 (these operations prompted by drawing a schematic of the vector geometry) gives the following formula for the vector from the origin to the perpendicular point $\textbf{q}$:-

$$\textbf{q}=\frac{\textbf{p}\cdot\textbf{a}}{|\textbf{a}|^2}\textbf{a} ,$$ where $\textbf{p}$ and $\textbf{q}$ are just position vectors of points $P$ and $Q$ respectively.

expanding for each xyz coordinate:-

$$\textbf{q}=\begin{bmatrix} (P_x\cos\gamma + P_y\sin\gamma\,\sin\alpha)\cos\gamma\\ (P_x\cos\gamma + P_y\sin\gamma\,\sin\alpha)\sin\gamma\,\sin\alpha\\ (P_x\cos\gamma + P_y\sin\gamma\,\sin\alpha)\sin\gamma\,\cos\alpha\ \end{bmatrix},$$


$$\textbf{q}=\begin{bmatrix} (P_x\cos\gamma.\cos\gamma\ + P_y\sin\gamma.\sin\alpha.\cos\gamma)\\ (P_x\cos\gamma.\sin\gamma.\sin\alpha + P_y\sin\gamma.\sin\alpha.\sin\gamma.\sin\alpha)\\ (P_x\cos\gamma.\sin\gamma.\cos\alpha + P_y\sin\gamma.\sin\alpha.\sin\gamma.\cos\alpha)\ \end{bmatrix},$$

simplifying, and using $\sin^2 \alpha = (1/2) + (1/2)\cos2\alpha$...

$$\textbf{q}=\begin{bmatrix} P_x\cos^2\gamma. + P_y\sin\gamma.\cos\gamma.\sin\alpha\\ P_x\cos\gamma.\sin\gamma.\sin\alpha + P_y\sin^2\gamma.((1/2) + (1/2)\cos2\alpha\\ P_x\cos\gamma.\sin\gamma.\cos\alpha + P_y\sin^2\gamma.\sin\alpha.\cos\alpha\ \end{bmatrix},$$

We obtain the mean point $Q$ by noticing that when we integrate with respect to $\alpha$ any terms which are functions of $\sin\alpha, \cos\alpha, \sin2\alpha$ or $\cos 2\alpha$ integrate to zero

$$\textbf{Q}=\begin{bmatrix} P_x\cos^2\gamma\\ (1/2)P_y\sin^2\gamma\\ 0\ \end{bmatrix}.$$

Note 1: I still lack a formal source or derivation of the formula for $q$.

Note 2: Aretino's answer provides a derivation of the formula for $q$.

  • $\begingroup$ Perhaps I don't understand your $q$ points, but it seems quite obvious that $Q_y=0$, while $Q_x=P_x\cos^2\gamma$ is correct. $\endgroup$ Mar 11, 2019 at 22:40
  • $\begingroup$ @Aretino note that $Py$ may be non-zero, so if $P$ is not on the x-axis I don't see why it should be obvious that $Qy=0$. $\endgroup$
    – steveOw
    Mar 11, 2019 at 23:02
  • $\begingroup$ You are right: I was assuming $P_y=0$. $\endgroup$ Mar 11, 2019 at 23:04
  • 1
    $\begingroup$ I checked that your formula for $\mathbf{Q}$ is right. $\endgroup$ Mar 12, 2019 at 12:12
  • $\begingroup$ @Aretino Thanks. I have expanded the section in my question which derives the formula for $Q$. But I still lack a formal source or derivation of the formula for $q$. $\endgroup$
    – steveOw
    Mar 12, 2019 at 14:29

1 Answer 1


A formula for $\mathbf{q}$ is easy to derive. Notice first of all that $\mathbf{q}=k\mathbf{a}$, for some real number $k$. To find $k$ we can exploit perpendicularity between $\mathbf{a}$ and $\mathbf{q}-\mathbf{p}$: $$ \mathbf{a}\cdot(\mathbf{q}-\mathbf{p})=0, \quad\text{that is:}\quad \mathbf{a}\cdot(k\mathbf{a}-\mathbf{p})=0, \quad\text{whence:}\quad k={\mathbf{p}\cdot\mathbf{a}\over|\mathbf{a}|^2}. $$

  • $\begingroup$ That is exactly what I needed! Thanks :) $\endgroup$
    – steveOw
    Mar 12, 2019 at 18:25

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