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Vector method to determine the distance from Betelgeuse to the great
circle that contains Capella and Canopus.

Coordinates from M. Couëtte:
Cap 280,8283333° 45,9983333°
Bet 271,206667° 7,406667°
Can 264,0116667° -52,6950000°

Assume the vectors have magnitude 1. Then the vector cross product Cap ×
Can = (.98559, 3.419°, 97.099°). Change the magnitude from .98559 to 1.

The dot product of the modified vector and Bet = -.05392. That's the
sine of the separation between Betelgeuse and the great circle that
contains Cap and Can. The angle is 3.091° (disregard the sign).

Explanation: the cross product Cap × Can is a third vector which is
perpendicular to Cap and Can. Therefore the great circle that contains
Cap and Can is fully defined by their cross product: a vector directed
to the pole of the circle.

The magnitude of the pole vector is the sine of the separation angle
between Cap and Can. We don't need that angle. To make next step work
properly, change the pole vector magnitude to 1.

To compute the separation angle between the pole vector and Bet, take
the arc cosine of their dot product. But we really want the complement
of that angle, i.e., the separation angle from the great circle, so take
the arc sine.

In this case we know the angle is small, so a simple arc sine of the dot
product is sufficient. To generalize the algorithm you should also
compute the cross product of the same two vectors. The magnitude of the
cross product is the cosine of the desired angle. Now with sine and
cosine the angle can be calculated accurately even if the third point is
very close to the pole.

I ignored the sign of the answer because in this case it doesn't matter
if Bet is "north" or "south" of the great circle. If that's important,
then the order of the cross product operands is important, because the
operation is not commutative. If you reverse the order, the cross
product points to the other pole of the great circle.

   

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