Past exam of the mathematics course of the University of Cambridge / 2007 / ia / Paper 1 / 5B (Algebra and Geometry)

Decompose the position vector into its components parallel and perpendicular to the unit vector $\mathbf{u}$:This is an orthogonal projection, and the Pythagorean theorem gives $|\mathbf{r}{|}^2=s^2+|\mathbf{v}{|}^2$. The two surfaces are distinguished by fixing an angle in part (a), and fixing distance from an axis in part (b).

For $\mathbf{r}\ne 0$, the dot product equation says that the angle $\theta$ between $\mathbf{r}$ and the oriented unit vector $\mathbf{u}$ satisfies $\cos \theta =\alpha$. Thus $\theta =\arccos \alpha$ is fixed. More explicitly, using the orthogonal projection from the introductory solution,Consequently the surface is one infinite nappe of a right circular cone, with apex at the origin, including the apex itself. The cone's axis points along $\mathrm{sgn}(\alpha )\mathbf{u}$, and its acute semi-angle is $\arccos |\alpha |$. Equivalently, $\alpha$ is the cosine of the angle from the oriented $\mathbf{u}$ axis. Squaring the original equation without retaining the sign of $s$ would incorrectly add the opposite nappe of a double circular cone. No base disk is included in this infinite lateral surface.

The vector $\mathbf{r}-(\mathbf{r}\cdot \mathbf{u})\mathbf{u}$ is the perpendicular component of $\mathbf{r}$ relative to the axis $\mathbb{R}\mathbf{u}$, so its length is the distance from that axis. Hence the surface is an infinite circular cylinder of radius $\beta$ about the line through the origin parallel to $\mathbf{u}$. Here $\beta$ is the radius and $\mathbf{u}$ fixes the axis direction; reversing $\mathbf{u}$ leaves the same circular cylinder.

If ${\mathbf{v}}_1,{\mathbf{v}}_2$ are an orthonormal basis of the plane perpendicular to $\mathbf{u}$, a parameterization isThis verifies that all heights and all angular positions, but no interior points, belong to the surface.

The sphere-plane intersection is governed by the orthogonal projection of the sphere's centre onto the plane. Since $\mathbf{n}$ has unit length, let $h=p-\mathbf{n}\cdot \mathbf{c}$, the signed distance from $\mathbf{c}$ to the plane along $\mathbf{n}$. The following parts determine the centre and radius and retain this sign; the radii alone cannot distinguish the two possible sides of the oriented plane.

Set ${\mathbf{b}}_0=\mathbf{c}+h\mathbf{n}$, where $h=p-\mathbf{n}\cdot \mathbf{c}$. Then $\mathbf{n}\cdot {\mathbf{b}}_0=p$, so ${\mathbf{b}}_0$ lies in the plane. For any position $\mathbf{r}$ in that plane, $\mathbf{r}-{\mathbf{b}}_0$ is perpendicular to ${\mathbf{b}}_0-\mathbf{c}=h\mathbf{n}$. The Pythagorean theorem therefore givesOn the sphere the left side is $a^2$. Thus the intersection is the circle in the plane centred at ${\mathbf{b}}_0$, with squared radius $a^2-h^2$. Its centre must therefore be $\mathbf{b}={\mathbf{b}}_0$, andIn terms of the radii, $\lambda =\pm \sqrt{a^2-{\rho }^2}$, with the sign fixed by $\lambda =p-\mathbf{n}\cdot \mathbf{c}$. In the central-plane case $\rho =a$, both signs give zero. This is the center and radius of a sphere-plane intersection formula with its orientation made explicit.

Substitute the signed displacement $\lambda =p-\mathbf{n}\cdot \mathbf{c}$ into the Pythagorean theorem relation from part (a):A genuine circle requires $|p-\mathbf{n}\cdot \mathbf{c}|<a$. Equality gives a tangent plane and a single point, interpreted as radius zero; a larger distance gives no intersection. Under the stipulated circle intersection, the radicand is positive. Its centre is $\mathbf{c}+(p-\mathbf{n}\cdot \mathbf{c})\mathbf{n}$.