Problem source

The surface $S$ in 3-dimensional space is described by the equation
\[
\mathbf{a}\cdot\mathbf{r}+ar=a^{2},
\]
where $\mathbf{r}$ is the position vector with respect to the origin $O$, $\mathbf{a}(\neq\mathbf{0})$ is the position vector of a fixed point, $r=\left|\mathbf{r}\right|$ and $a=\left|\mathbf{a}\right|.$
Show, with the aid of a diagram, that $S$ is the locus of points which are equidistant from the origin $O$ and the plane $\mathbf{r}\cdot\mathbf{a}=a^{2}.$
The point $P$, with position vector $\mathbf{p},$ lies in $S$, and the line joining $P$ to $O$ meets $S$ again at $Q$. Find the position vector of $Q$. 
The line through $O$ orthogonal to $\mathbf{p}$ and $\mathbf{a}$ meets $S$ at $T$ and $T'$. Show that the position vectors of $T$ and $T'$ are 
\[
\pm\frac{1}{\sqrt{2ap-a^{2}}}\mathbf{a}\times\mathbf{p},
\]
where $p=\left|\mathbf{p}\right|.$ 
Show that the area of the triangle $PQT$ is 
\[
\frac{ap^{2}}{2p-a}.
\]

Solution source

The plane is the same as the plane $(\mathbf{r} - \mathbf{a}) \cdot \mathbf{a} = 0$, ie the plane through $\mathbf{a}$ whose normal is parallel to $\mathbf{a}$

The distance from $\mathbf{r}$ to the plane therefore is $\lambda$ where $\mathbf{r}+\lambda \frac{1}{a}\mathbf{a}$ must be on the plane, ie $(\mathbf{r}+\frac{\lambda}{a} \mathbf{a} - \mathbf{a})\cdot \mathbf{a} = 0 \Rightarrow \lambda = \frac{a^2-\mathbf{a} \cdot \mathbf{r}}{a}$

But if $\mathbf{a} \cdot \mathbf{r} = a^2 - ar$ then $\lambda = r$, ie the distance to the plane is the same as the distance to the origin.

$\mathbf{q} = k \mathbf{p}$ and so $\mathbf{a} \cdot k \mathbf{p} + a |k|p = a^2$ if $k > 0$ we will find $k = 1$ the position vector we already know about, therefore suppose $k < 0$ so:

\begin{align*}
&& \mathbf{a} \cdot k \mathbf{p} - ka p &= a^2 \\
\Rightarrow && k(a^2-ap)-kap &= a^2 \\
\Rightarrow && k(a^2-2ap) &= a^2  \\
\Rightarrow && k &= \frac{a^2}{a^2-2ap}
\end{align*}

Therefore $\mathbf{q} = \frac{a^2}{a^2-2ap} \mathbf{p}$

The line through $O$ orthogonal to $\mathbf{p}$ and $\mathbf{a}$ will be parallel to $\mathbf{a} \times \mathbf{p}$. Therefore we should consider points of the from $s \mathbf{a} \times \mathbf{p}$ on the surface $S$.

\begin{align*}
&& s\mathbf{a} \cdot ( \mathbf{a} \times \mathbf{p}) + sa^2p |\sin \theta| &= a^2
\end{align*}

The angle between $\cos \theta = \frac{\mathbf{a} \cdot \mathbf{p}}{ap} = \frac{a^2-ap}{ap} \Rightarrow |\sin \theta| = \sqrt{1-\frac{(a-p)^2}{p^2}} = \frac{1}{p} \sqrt{2ap-a^2}$

Therefore $sa^2 \sqrt{2ap-a^2} = a^2 \Rightarrow s = \frac{1}{\sqrt{2ap-a^2}}$ and so the points are as required.

Noting that $|\mathbf{p} \times \mathbf{t}| = |\frac{1}{p \sin \theta}\mathbf{p} \times (\mathbf{p} \times \mathbf{a}) | = |\frac{1}{p \sin \theta}p^2a \sin \theta | = pa$


The area of triangle $PQT$ is :

\begin{align*}
\frac12 | (\mathbf{p} - \mathbf{t}) \times  (\mathbf{q} - \mathbf{t}) | &= \frac12 |\mathbf{p} \times \mathbf{q} - \mathbf{t} \times \mathbf{q} - \mathbf{p} \times \mathbf{t} - \mathbf{t} \times \mathbf{t}| \\
&= \frac12 |\mathbf{t} \times (\mathbf{p} - \mathbf{q})| \\
&= \frac12 \cdot (1 - \frac{a^2}{a^2-2ap})| \mathbf{t} \times \mathbf{p}| \\
&= \frac12 \frac{2ap}{a^2-2ap} \cdot ap \\
&= \frac{ap^2}{a^2-ap}
\end{align*}