Year: 2010
Paper: 1
Question Number: 7
Course: UFM Pure
Section: Vectors
No solution available for this problem.
There were significantly more candidates attempting this paper than last year (just over 1000), and the scores were much higher than last year (presumably due to the easier first question): fewer than 2% of candidates scored less than 20 marks overall, and the median mark was 61. The pure questions were the most popular as usual, though there was much more variation than in some previous years: questions 1, 3, 4 and 6 were the most popular, while question 7 (on vectors) was intensely unpopular. About half of all candidates attempted at least one mechanics question, and 15% attempted at least one probability question. The marks were unsurprising: the pure questions generally gained the better marks, while the mechanics and probability questions generally had poorer marks. A sizeable number of candidates ignored the advice on the front cover and attempted more than six questions, with a fifth of candidates trying eight or more questions. A good number of those extra attempts were little more than failed starts, but suggest that some candidates are not very effective at question-picking. This is an important skill to develop during STEP preparation. Nevertheless, the good marks and the paucity of candidates who attempted the questions in numerical order does suggest that the majority are being wise in their choices. Because of the abortive starts, I have often restricted my attention below to those attempts which counted as one of the six highest-scoring answers, and referred to these as "significant attempts". The majority of candidates did begin with question 1 (presumably as it appeared to be the easiest), but some spent far longer on it than was wise. Some attempts ran to over eight pages in length, especially when they had made an algebraic slip early on, and used time which could have been far better spent tackling another question. It is important to balance the desire to finish the question with an appreciation of when to stop and move on. Many candidates realised that for some questions, it was possible to attempt a later part without a complete (or any) solution to an earlier part. An awareness of this could have helped some of the weaker students to gain vital marks when they were stuck; it is generally better to do more of one question than to start another question, in particular if one has already attempted six questions. It is also fine to write "continued later" at the end of a partial attempt and then to continue the answer later in the answer booklet. As usual, though, some candidates ignored explicit instructions to use the previous work, such as "Hence", or "Deduce". They will get no credit if they do not do what they are asked to! (Of course, "Hence, or otherwise, show . . ." gives them the freedom to use any method of their choosing; often the "hence" will be the easiest, but in Question 5 this year, the "otherwise" approach was very popular.) On some questions, some candidates tried to work forwards from the given question and backwards from the answer, hoping that they would meet somewhere in the middle. While this worked on occasion, it often required fudging. It is wise to remember that STEP questions do require a greater facility with mathematics and algebraic manipulation than the A-level examinations, as well as a depth of understanding which goes beyond that expected in a typical sixth-form classroom.
Difficulty Rating: 1484.0
Difficulty Comparisons: 1
Banger Rating: 1500.0
Banger Comparisons: 0
Relative to a fixed origin
$O$, the points
$A$ and $B$ have position vectors
$\bf{a}$ and $\bf{b}$, respectively. (The points $O$, $A$ and $B$ are not
collinear.)
The point $C$ has position vector $\bf c$ given by
\[
{\bf c} =\alpha {\bf a}+ \beta {\bf b}\,,
\]
where $\alpha$ and $\beta$ are
positive constants with $\alpha+\beta<1\,$.
The lines $OA$
and $BC$ meet at the point $P$ with position vector $\bf p$
and the lines $OB$ and $AC$ meet at the point $Q$ with position vector $\bf q$.
Show that
\[
{\bf p} =\frac{\alpha {\bf a} }{1-\beta}\,,
\]
and write down $\bf q$ in terms of
$\alpha,\ \beta$ and $\bf {b}$.
Show further that the point $R$
with position vector $\bf r$ given by
\[
{\bf r} =\frac{\alpha {\bf a}
+ \beta {\bf b}}{\alpha + \beta}\,,
\]
lies on the lines $OC$ and $AB$.
The lines $OB$ and $PR$ intersect at the point $S$.
Prove that
$
\dfrac{OQ}{BQ} = \dfrac{OS}{BS}\,$.
This was a very unpopular question, attempted by only 20% of candidates and being one of the six best questions of only 10%. It was also the worst-scoring: of the significant attempts, the median mark was 1/20. However, those who managed to get beyond the start of the question generally did quite well, resulting in an upper quartile mark (among significant attempts) of 11/20. It generally indicated a very poor understanding of vectors among those students who attempted this question. Only a handful of students successfully completed the question. Of those who attempted the question, many drew a decent diagram, which is very helpful in understanding what is being asked, though few realised that C lies strictly inside triangle OAB. Many also realised that they could simply write down the formula for q directly by symmetry. Most, however, appeared to be incapable of writing down the equation of a straight line in vector form and were therefore unable to proceed any further. Those who did often made their lives more difficult by writing the equation of OA as r = a+λa rather than choosing the origin as their fixed point, giving the simpler equation r = λa. While the former is clearly correct, it is messier to work with and therefore more likely to lead to errors later. A very common confusion was to write down the equations of the lines but to call them OA and BC instead of OA and BC; this sometimes led to candidates trying to equate the vectors as OA = BC, which was fairly nonsensical (and other candidates tried equating the vectors without writing down the equations of the lines). Another frequent piece of nonsense was an attempt to divide one vector by another or to add a vector to a scalar. Few candidates used notation carefully, and many suffered for it. It is strongly recommended that students are taught to always distinguish their vector variables from their scalar variables by underlining, under-squiggling or using arrows above them, and that this is insisted upon. Some candidates attempted to write the formula for p in words and hoped that, somehow, this would be sufficient justification. This was joined by several attempts to work forwards from what was known and backwards from the desired result, which together with some glue in between was meant to provide a convincing argument for the result. Those who successfully completed the first part were generally successful with finding the position vector of r as well. The final part of the question was attempted by relatively few candidates. There was a mixed level of success. Most realised that they needed to find the position vector of s, and this was done fairly well; the algebra was the trickiest part here, as the ideas were the same as earlier. The last step, proving the equality of the ratios, was a little trickier, and there were a few attempts to divide vectors or to ignore problems with signs.