+180 A and B are two points on a unit sphere. We know the space distance between A and B is sqrt3 What is distance from A to B along the (minor) arc of a great circle?
+1199 The distance between two points on a sphere along the minor arc of a great circle is equal to the central angle between the two points in radians. The central angle between two points on a sphere is equal to the arccosine of the dot product of the unit vectors pointing to the two points.
Therefore, to find the distance from A to B along the minor arc of a great circle, we first need to find the unit vectors pointing to A and B. We can do this by dividing the coordinates of A and B by the radius of the sphere, which is 1.
Let a and b be the unit vectors pointing to A and B, respectively. Then the distance from A to B along the minor arc of a great circle is equal to the central angle between a and b, which is given by:
theta = arccos(\mathbf{a} \cdot \mathbf{b})
We know that the space distance between A and B is 3. We can use this information to find the dot product of a and b.
\mathbf{a} \cdot \mathbf{b} = \cos(theta) = \frac{\sqrt{3}}{2}
Therefore, the distance from A to B along the minor arc of a great circle is given by:
theta = arccos(\mathbf{a} \cdot \mathbf{b}) = arccos(\frac{\sqrt{3}}{2})
theta = \frac{\pi}{3}
Therefore, the distance from A to B along the minor arc of a great circle is pi/3.
+180 thanks for trying, but the correct answer I found out on my own. I guess ill post an explanation here
We take a cross-section along the great circle. Let O be the center of the sphere, and let M be the midpoint of line AB
Since OA = OB = 1, angle OMA = 90, Am is sqrt3/2 so triangle AOM is a 30 60 90 triangle
So AOM is 60 degrees, BOM is also 60 degrees, therefore AOB is 120 degrees, therefore arc AB has length
120/360 *2pi=2/3pi
