find intermediate points on small circle of a sphere What I know/have:


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*A sphere S with radius $r_0=1$ centered in Cartesian space at $c_0=(0,0,0)$

*Three points on the sphere $S$: $p_1$, $p_2$, and $p_3$

*A plane $P$ through the points $p_1$, $p_2$, and $p_3$

*A circle $C$, which is produced by intersecting plane $P$ with sphere $S$. All points $p_1$, $p_2$, and $p_3$ are on the circle $C$

*The arc $A$ when going from $p_1$ over circle $C$ to $p_3$ has the interesting property that at fraction $0.5$ of its total distance, we find $p_2$

*Hence, at fraction $0$ of the total distance of arc $A$ we find $p_1$, and at fraction $1$ of the total distance of arc $A$ we find $p_3$


What I want:
A point $p_4$ in Cartesian space ($x$,$y$,$z$ with respect to origin $c_0$), which lies on the arc $A$ at an arbitrary fraction $X$ of the total distance that arc $A$ spans from $p_1$ to $p_3$
From a computational perspective this would mean:


*

*input coordinates of 3 points and a fraction

*output coordinates of 1 point


Unfortunately all the steps in between are a black box to me in terms of the mathematical/computational steps involved and I would appreciate a solution ... or help with finding the solution myself.
Visualization of the Problem
The visualization was drawn by hand using geogebra so please just assume the assumptions above hold true (although it might look different in the visualization)

Previous research:


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*The Great Circle is the circle that comes from intersecting a sphere and a plane that contains the origin of the sphere as a point

*There are computations for finding intermediate points along the Great Circle Distance in Ed William's Aviation Formulary

*In my case, the plane does NOT contain the origin of the sphere but three other points on the sphere

*What I am looking for are intermediate points on a "Small Circle Distance"


Appendix (not required for answering the question) - Real-Life application:
When measuring EEG data, there is a conventional set of rules how to place electrodes (white buttons in image below) on the scalp of a human. Usually this is measured with a measuring tape but of course we can also model the human head by a geometrical sphere and compute the relative electrode positions on that sphere (based only on rules and the initial, arbitrary placement of one electrode ... no measurements involved then).

Looking at figure B of the image above, and relating to this question, let's assume I know the positions $F7$, $F8$, and $Fz$. Knowing the rules, $F3$ and $F4$ will lie at fractions on the contour connecting $F7$ and $F8$ through $Fz$.
 A: In principle, you can just establish a parametric equation of the arc, where the parameter is the arc length. Especially in the case of circles, their parametric equation is fairly simple.
Say you have a point $O$, and two orthogonal unit vectors $\mathbf u$ and $\mathbf v$. Then the parametric equation of the circle of center $O$, radius $r>0$, and in the plane through $O$ with directions $\mathbf u$ and $\mathbf v$, is given by
$$M(\theta) = O+r\left(\cos\theta\mathbf u+\sin\theta\mathbf v\right)$$
for $\theta\in\mathbb R$. Restricting the value of $\theta$ to a range of length less than $2\pi$ will yield circular arcs supported by the circle.

To apply this to your problem, you just have to figure out what choice of $\mathbf u$ and $\mathbf v$ will make your life simple.
One possibility is to compute the center $O$ and radius $r$ of your circle,
then set $\hat{\mathbf u} = \vec{Op_1}$ and
$\hat{\mathbf v} = \vec{Op_2}$. 
Let $\mathbf u$ and $\mathbf v$ be the Gram–Schmidt orthonormalisation of $\hat{\mathbf u}$ and $\hat{\mathbf v}$. Let also $\varphi$ be the unique real number in $(O,2\pi)$ such that $p_3=O+r(\cos\varphi\mathbf u+\sin\varphi\mathbf v)$.
Then, your arc $A$ is the collection of points $M(\theta)$ with $0\le\theta\le\varphi$. If you want to use a fraction of the total arc length, you can use $\theta = f\times\varphi$ with $0\le f\le 1$ your fraction.
A: Forget about small circles on a sphere. Consider the generalization that we have $3$ points $p_1, p_2, p_3$ lying on some circle centered at $c$ with radius $r$ and $p_2$ is the mid-point of the circular arc $p_1p_3$. Let $\theta_0$ be the half-angle of the arc $p_1p_3$ subtended at $c$.
We can parametrize the circular arc by an angle $\theta \in [-\theta_0,\theta_0]$ using following formula:
$$p(\theta) = c + (p_2 - c)\cos\theta + \frac{p_3-p_1}{2}\frac{\sin\theta}{\sin\theta_0} \tag{*1}$$
Independent of value of $c$, we have $p(0) = p_2$. In order for $p(-\theta_0) = p_1$ and $p(\theta_0) = p_3$, we need
$$c + (p_2 - c)\cos\theta_0 = \frac{p_1+p_3}{2}
\quad\implies\quad
c = \frac{1}{1-\cos\theta_0} \left(\frac{p_1+p_3}{2} - p_2 \cos\theta_0\right)$$
Plug this back in $(*1)$, we obtain
$$\bbox[padding: 1em;border: 1px solid blue]{
p(\theta) = \frac{1-\cos\theta}{1-\cos\theta_0}\frac{p_3+p_2}{2} + \frac{\sin\theta}{\sin\theta_0} \frac{p_3-p_1}{2} + \frac{\cos\theta-\cos\theta_0}{1-\cos\theta_0} p_2}\tag{*2}$$
What remains is to compute $\theta_0$. 
Notice $|p_3 - p_1| = 2r\sin\theta_0$ and $|p_2 - p_1| = 2r\sin\frac{\theta_0}{2}$,
we have 
$$\cos\frac{\theta_0}{2} = \frac{\sin\theta_0}{2\sin\frac{\theta_0}{2}}
= \frac{|p_3-p_1|}{2|p_2-p_1|}
\quad\implies\quad \theta_0 = 2\cos^{-1}\left(\frac{|p_3-p_1|}{2|p_2-p_1|}\right)
$$
To parametrize the arc using a fraction $t$, one just need to associate
$t \in [0,1]$ with the point $p((2t-1)\theta_0)$ given in formula $(*2)$.
