I am trying to learn spherical geometry, but I have difficulty resolving a simple issue.

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Let's define a sphere's equator and it's poles N, S. if we create a great circle by tilting the equator circle by degree of $\alpha$. In that great circle point P is the closest to N (the point in the sphere where the latitude is highest). Now, let's look at the great circle that is created by the points N, P on the sphere. Those great circles intersect in point P on the sphere.

My question is about the angle between those great circles at the point P. It seems to me that the angle is indeed $90$. but according the definition of this source (point 9):

"By the angle between two great circles is meant the angle of inclination of the planes of the circles."

It seems that the planes of the great circles are not perpendicular - meaning the angle between them is not 90.

I would appreciate if someone would instruct me what I'm missing here, and how to show, formally, that the angle is 90?


You are right. The angle between the two plane is ${90}°$. Here is two ways to see it.

First, the point $P$ is the one closest to the North pole. Since great circle are the straigh line of spherical geometry, the shortest distance between a point and a line is obtain with a perpendicular.

Second, the angle between the plane is given by the angle between their normal vector.

Let consider the sphere centered at the origin and the poles on the $z$ axis. Rotate the sphere so the point $P$ is on the $yz$-plane. Note: the point $Q$ is also on the $yz$-plane since $P$ and $Q$ are the ends of the same diameter.

The normal of the plane passing thru $P$, $Q$ and the two poles is on the $x$ axis. E.g. the normal could be the vector starting at $O$ pointing toward were the equtorial plane meet the oblique plane.

The normal vector of the oblique plane passing thru $P$ and $Q$ is in the $yz$-plane. The normal could be the vector starting at $O$ pointing toward the mid point between $P$ and $Q$ on the great circle passing by the poles.

The two vectors are perpendicular.

Hope it helps

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  • $\begingroup$ Thanks for the first way - that brings good intuition. but with respect to the secod way, I feel bad, but I can't see this. if $P$ and the poles are in the $yz$-plane, so that point $Q$ as $Q$ is simply the continuation of $PO$ $\endgroup$ – d_e Dec 20 '19 at 22:26
  • $\begingroup$ @d_e yes the point $Q$ is diametrically oppose to $P$. As a matter of facts, the great circle is divided in four equal part: $P$ to equator; equator to $Q$; $Q$ to equator and equator to $P$. $\endgroup$ – Alain Remillard Dec 20 '19 at 22:35
  • $\begingroup$ so the plan passing thru $P$ and $Q$ is again the $yz$-plane - and it's normal is in $x$ axis again. so I can;t follow your last sentence in the answer. $\endgroup$ – d_e Dec 20 '19 at 22:38
  • $\begingroup$ I was refering to your oblique great circle. I'll think of a better way to write it but english is not my first language. $\endgroup$ – Alain Remillard Dec 21 '19 at 1:58

think coordinates for a unit sphere centered at the origin. take the plane of the paper (for your diagram) to be the $xz$-plane. this contains the circle PNQS whose equation is: $$ x^2 + z^2 = 1 $$ the normal to this plane lies along the $y$-axis.

this is also the axis about which the horizontal circle has been rotated around to bring it to the position shown. thus a normal to the rotated circle lies in the $xz$-plane, and is thus always perpendicular to the y-axis, for any angle of rotation.

NB two planes are perpendicular if and only if their normals are perpendicular.

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