Mean Q of all points q on a cone whose origin vectors 0q are perpendicular to a given point P.

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.

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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.

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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:-

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

expanding...

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

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 $$q$$.

• 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. – Aretino Mar 11 at 22:40
• @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 $Qy=0$. – steveOw Mar 11 at 23:02
• You are right: I was assuming $P_y=0$. – Aretino Mar 11 at 23:04
• I checked that your formula for $\mathbf{Q}$ is right. – Aretino Mar 12 at 12:12
• @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 $q$. – steveOw Mar 12 at 14:29

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 $$\mathbf{q}-\mathbf{p}$$: $$\mathbf{a}\cdot(\mathbf{q}-\mathbf{p})=0, \quad\text{that is:}\quad \mathbf{a}\cdot(k\mathbf{a}-\mathbf{p})=0, \quad\text{whence:}\quad k={\mathbf{p}\cdot\mathbf{a}\over|\mathbf{a}|^2}.$$