| Date | November 2008 | Marks available | 5 | Reference code | 08N.2.hl.TZ0.5 |
| Level | HL only | Paper | 2 | Time zone | TZ0 |
| Command term | Describe and Find | Question number | 5 | Adapted from | N/A |
Question
(a) Find the set of values of k for which the following system of equations has
no solution.
x + 2y − 3z = k
3x + y + 2z = 4
5x + 7z = 5
(b) Describe the geometrical relationship of the three planes represented by this system of equations.
Markscheme
(a)
\(\left( {\begin{array}{*{20}{c}}
1&2&{ - 3}&k \\
3&1&2&4 \\
5&0&7&5
\end{array}} \right)\) M1
\({R_1} - 2{R_2}\)
\(\left( {\begin{array}{*{20}{c}}
{ - 5}&0&{ - 7}&{k - 8} \\
3&1&2&4 \\
5&0&7&5
\end{array}} \right)\) (A1)
\({R_1} + {R_3}\)
\(\left( {\begin{array}{*{20}{c}}
0&0&0&{k - 3} \\
3&1&2&4 \\
5&0&7&5
\end{array}} \right)\) (A1)
Hence no solutions if \(k \in \mathbb{R}\), \(k \ne 3\) A1
(b) Two planes meet in a line and the third plane is parallel to that line.
[5 marks]
Examiners report
Most candidates realised that some form of row operations was appropriate here but arithmetic errors were fairly common. Many candidates whose arithmetic was correct gave their answer as k = 3 instead of \(k \ne 3\) . Very few candidates gave a correct answer to (b) with most failing to realise that stating that there was no common point was not enough to answer the question.
