Year: 2007
Paper: 2
Question Number: 8
Course: UFM Pure
Section: Vectors
Although the paper was by no means an easy one, it was generally found a more accessible paper than last year's, with most questions clearly offering candidates an attackable starting-point. The candidature represented the usual range of mathematical talents, with a pleasingly high number of truly outstanding students; many more who were able to demonstrate a thorough grasp of the material in at least three questions; and the few whose three-hour long experience was unlikely to have been a particularly pleasant one. However, even for these candidates, many were able to make some progress on at least two of the questions chosen. Really able candidates generally produced solid attempts at five or six questions, and quite a few produced outstanding efforts at up to eight questions. In general, it would be best if centres persuaded candidates not to spend valuable time needlessly in this way – it is a practice that is not to be encouraged, as it uses valuable examination time to little or no avail. Weaker brethren were often to be found scratching around at bits and pieces of several questions, with little of substance being produced on more than a couple. It is an important examination skill – now more so than ever, with most candidates now not having to employ such a skill on the modular papers which constitute the bulk of their examination experience – for candidates to spend a few minutes at some stage of the examination deciding upon their optimal selection of questions to attempt. As a rule, question 1 is intended to be accessible to all takers, with question 2 usually similarly constructed. In the event, at least one – and usually both – of these two questions were among candidates' chosen questions. These, along with questions 3 and 6, were by far the most popularly chosen questions to attempt. The majority of candidates only attempted questions in Section A (Pure Maths), and there were relatively few attempts at the Applied Maths questions in Sections B & C, with Mechanics proving the more popular of the two options. It struck me that, generally, the working produced on the scripts this year was rather better set-out, with a greater logical coherence to it, and this certainly helps the markers identify what each candidate thinks they are doing. Sadly, this general remark doesn't apply to the working produced on the Mechanics questions, such as they were. As last year, the presentation was usually appalling, with poorly labelled diagrams, often with forces missing from them altogether, and little or no attempt to state the principles that the candidates were attempting to apply.
Difficulty Rating: 1600.0
Difficulty Comparisons: 0
Banger Rating: 1529.3
Banger Comparisons: 4
The points $B$ and $C$ have position vectors $\mathbf{b}$ and $\mathbf{c}$, respectively, relative to the origin $A$, and $A$, $B$ and $C$ are not collinear.
\begin{questionparts}
\item The point $X$ has position vector
$s \mathbf{b}+t\mathbf{c}$. Describe the locus of $X$ when $s+t=1$.
\item The point $P$ has position vector $\beta \mathbf{b}+\gamma\mathbf{c}$, where $\beta$ and $\gamma$ are non-zero, and $\beta+\gamma\ne1$.
The line $AP$ cuts the line $BC$ at $D$. Show that $BD:DC=\gamma:\beta$.
\item The line $BP$ cuts the line $CA$ at $E$, and the line $CP$ cuts the line $AB$ at $F$. Show that
\[
\frac{AF}{FB} \times \frac{BD}{DC} \times \frac{CE}{EA}=1\,.
\]
\end{questionparts}\begin{questionparts}
\item $X$ lies on the line including $B$ and $C$.
\item points on the line $AP$ have the form $\lambda(\beta \mathbf{b}+\gamma\mathbf{c})$, and the point $D$ will be the point where $\lambda\beta + \lambda \gamma = 1$.
\begin{align*}
&& \frac{|BD|}{|DC|} &= \frac{|\mathbf{b} -\lambda(\beta \mathbf{b}+\gamma\mathbf{c})| }{|\lambda(\beta \mathbf{b}+\gamma\mathbf{c})- \mathbf{c}|} \\
&&&= \frac{|(1-\lambda \beta)\mathbf{b} - \lambda \gamma \mathbf{c}|}{|\lambda \beta \mathbf{b}+(\lambda \gamma -1)\mathbf{c}|}\\
&&&= \frac{|\lambda \gamma\mathbf{b} - \lambda \gamma \mathbf{c}|}{|\lambda \beta \mathbf{b}-(\lambda \beta)\mathbf{c}|} \\
&&&= \frac{\gamma}{\beta}
\end{align*}
\item The line $BP$ is $\mathbf{b} + \mu(\beta \mathbf{b}+\gamma\mathbf{c})$ and will meet $CA$ when $1+\mu\beta = 0$, ie $\mu = -\frac{1}{\beta}$, therefore $E$ is $-\frac{\gamma}{\beta}\mathbf{c}$, and so $\frac{|CE|}{|EA|} = \frac{1+\gamma/\beta}{\gamma/\beta} = \frac{\beta+\gamma}{\gamma}$.
Similarly, $F$ is $-\frac{\beta}{\gamma}\mathbf{b}$ and $\frac{|AF|}{|FB|} = \frac{\beta/\gamma}{1+\frac{\beta}{\gamma}} = \frac{\beta}{\gamma+\beta}$, and so
\[\frac{AF}{FB} \times \frac{BD}{DC} \times \frac{CE}{EA} = \frac{\beta}{\gamma+\beta} \frac{\gamma}{\beta} \frac{\beta+\gamma}{\gamma} = 1 \]
\end{questionparts}
This is really just half of the (⇔) proof of Ceva's Theorem. Several candidates even recognised it as such. Of the remarkably small number of attempts submitted, most fell down at some stage (again) by failing to be sufficiently careful with signs/arithmetic/the modest amounts of algebra involved. It often didn't help those candidates who chose completely different symbols each time they did a stage of the working.