Year: 2007
Paper: 1
Question Number: 7
Course: UFM Pure
Section: Vectors
No solution available for this problem.
There were significantly more candidates attempting this paper this year (an increase of nearly 50%), but many found it to be very difficult and only achieved low scores. In particular, the level of algebraic skill required by the questions was often lacking. The examiners' express their concern that this was the case despite a conscious effort to make the paper more accessible than last year's. At this level, the fluent, confident and correct handling of mathematical symbols (and numbers) is necessary and is expected; many good starts to questions soon became unstuck after a simple slip. Graph sketching was usually poor: if future candidates wanted to improve one particular skill, they would be well advised to develop this. There were of course some excellent scripts, full of logical clarity and perceptive insight. It was pleasing to note that the applied questions were more popular this year, and many candidates scored well on at least one of these. It was however surprising how rarely answers to questions such as 5, 9, 10, 11 and 12 began with a diagram. However, the examiners were left with the overall feeling that some candidates had not prepared themselves well for the examination. The use of past papers to ensure adequate preparation is strongly recommended. A student's first exposure to STEP questions can be a daunting, demanding experience; it is a shame if that takes place during a public examination on which so much rides. Further, and fuller, discussion of the solutions to these questions can be found in the Hints and Answers document.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
\begin{questionparts}
\item The line $L_1$ has vector equation
$\displaystyle
{\bf r} =
\begin{pmatrix}
1 \\
0 \\
2
\end{pmatrix}
+
\lambda
\begin{pmatrix}
\hphantom{-} 2 \\
\hphantom{-} 2 \\
-3
\end{pmatrix}
$.
The line $L_2$ has vector equation
$\displaystyle
{\bf r} =
\begin{pmatrix}
\hphantom{-} 4 \\
-2 \\
\hphantom{-} 9
\end{pmatrix}
+
\mu
\begin{pmatrix}
\hphantom{-} 1 \\
\hphantom{-} 2 \\
-2
\end{pmatrix}
.
$
Show that the distance $D$
between a point on $L_1$ and a point on $L_2$
can be expressed in the form
\[
D^2 = \left(3\mu -4 \lambda-5 \right)^2 + \left( \lambda -1 \right)^2 + 36\,.
\]
Hence determine the minimum distance
between these two lines and find the coordinates
of the points on the two lines that are the minimum distance apart.
\item
The line $L_3$ has vector equation
${\bf r} =
\begin{pmatrix}
2 \\
3 \\
5
\end{pmatrix}
+
\alpha
\begin{pmatrix}
0 \\
1 \\
0
\end{pmatrix}
.
$
The line $L_4$ has vector equation
$
{\bf r} =
\begin{pmatrix}
\hphantom{-} 3 \\
\hphantom{-} 3 \\
-2
\end{pmatrix}
+
\beta
\begin{pmatrix}
\, 4k\\
1-k \\
\!\!\! -3k
\end{pmatrix}
.
$
Determine the minimum distance between these two lines,
explaining geometrically the two different cases that arise
according to the value of $k$.
\end{questionparts}
Part (i) was well done by most of those who attempted this question, but many then found it difficult to develop the strategy in part (ii). A certain amount of trial and error is needed to complete the squares in an expression in terms of both α and β, but the coefficients (in particular, 1α2, 1β2 and 26β2k2) do not permit many possibilities. This question demanded some stamina, as Mathematics at university level also does.