1. Suppose that we want to smoothly move from point {$A$} (in 3D, or {$E^3$}) to point {$B$}.
2. Our goal is to move along a great circle from {$A$} to {$B$} by varying {$\theta$}.
3. That will be a 2D rotation in the plane {$\cal P$} determined by {$OAB$}.
4. The normal to the plane {$\cal P$} is {$ N = A \times B $}.
5. However {$|N|\neq 1 $} most of the time, so we'll have to normalize its length later.
6. In {$\cal P$}, {$A$} serves as the X-axis.
7. {$T=N \times A$} serves as the Y-axis.
8. {$T= (A\times B) \times A = (A\cdot A)B - (A\cdot B) A = B - (A\cdot B) A $}
9. If the angle between {$A$} and {$B$} is {$\alpha$}, then {$A\cdot B = \cos\alpha $}.
10. We have to normalize {$T$} to {$\widehat{T} $}. To do that, divide by {$|T| = \sqrt{1-(A\cdot B)^2} $}.
11. {$\widehat{T} = \frac{B}{\sqrt{1-(A\cdot B)^2} } - \frac{ (A\cdot B)} { \sqrt{1-(A\cdot B)^2}} A $}.
12. Note that, if {$A\perp B$} then {$\widehat{T} = B $}.
13. Now, to interpolate, {$ C = A \cos\theta + \widehat{T} \sin\theta $}
14. {$ C = A\cos\theta + \left(\frac{B}{\sqrt{1-(A\cdot B)^2} } - \frac{ (A\cdot B)} { \sqrt{1-(A\cdot B)^2}} A \right) \sin\theta $}
15. {$ C = A\left(\cos\theta - \frac{(A\cdot B)\sin\theta}{\sqrt{1-(A\cdot B)^2}}\right) + B\frac{\sin\theta}{\sqrt{1-(A\cdot B)^2}} $}
16. We could rewrite it thus: {$$ C(\theta) = \frac{A\left(\sin\alpha\cos\theta - \cos\alpha\sin\theta\right) + B\sin\theta}{\sin\alpha} $$}
17. ... and then thus: {$$ C(\theta) = A \cos\theta + \left(\frac{B-A\cos\alpha }{\sin\alpha}\right)\sin\theta $$}
18. {$\theta=0$} gives {$A$} and {$\theta=\alpha$} gives {$B$}.

Summarizing:

Example:

1. Let {$A=(1,0,0), B = \left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}, 0\right) $}
2. Note that {$ |A| = |B| = 1 $}.
3. Then {$A\cdot B=\frac{1}{\sqrt{2}} $}, {$C = (\cos\theta, \sin\theta, 0) $}.