Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2011 / Paper 2 / 12

Using the physics convention for spherical coordinates, the polar angle $\theta$ is measured from positive $z$ and the azimuth $\varphi$ from positive $x$ toward positive $y$. The Cartesian coordinates are

Hold $r$ fixed and differentiate the position vector $\mathbf{X}(\theta ,\varphi )$. Its coordinate tangent vectors areTheir dot product is zero, and their squared Euclidean norms are $r^2$ and $r^2{\sin }^2\theta$. To first order, the displacement is $\delta \mathbf{X}={\mathbf{X}}_{\theta }\delta \theta +{\mathbf{X}}_{\varphi }\delta \varphi +O((|\delta \theta |+|\delta \varphi |{)}^2)$. Squaring givesThis is the line element on a sphere. The chord distance and the short surface arc length agree to the order retained here.

A plane through the origin has equation $lx+my+nz=0$ for some nonzero normal $(l,m,n)$. Dividing that normal by its Euclidean norm makes $l^2+m^2+n^2=1$ without changing the plane. Substitute the spherical coordinates from (a) and cancel the nonzero radius:Its intersection with the sphere is the desired great circle; changing the sign of all three coefficients describes the same circle.

At the first endpoint, the Cartesian coordinates are $(r,0,0)$, so its great circle must have $l=0$. At the second endpoint, the plane condition gives $m\sin {\theta }_0\sin {\varphi }_0+n\cos {\theta }_0=0$. Since $\cos {\theta }_0>0$, an equivalent regular plane equation isWhere $\cos \theta \ne 0$, divide by $\cos {\theta }_0\cos \theta$ to obtainThe regular implicit equation includes the equatorial first endpoint; the tangent form is undefined there and must be read by a limit. The shorter connecting arc of this great circle is the shortest path. The endpoints are not antipodal, so this shorter arc is unique. For example, the central angle is $\delta =\arccos (\sin {\theta }_0\cos {\varphi }_0)$, giving spherical distance $r\delta$.