1
$\begingroup$

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

enter image description here

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?

$\endgroup$

2 Answers 2

1
$\begingroup$

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

$\endgroup$
4
  • $\begingroup$ 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
    Commented Dec 20, 2019 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 $P$. $\endgroup$ Commented Dec 20, 2019 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
    Commented Dec 20, 2019 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. $\endgroup$ Commented Dec 21, 2019 at 1:58
1
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .