Date | May 2008 | Marks available | 6 | Reference code | 08M.2.hl.TZ1.5 |
Level | HL only | Paper | 2 | Time zone | TZ1 |
Command term | Find | Question number | 5 | Adapted from | N/A |
Question
Find the vector equation of the line of intersection of the three planes represented by the following system of equations.
\[2x - 7y + 5z = 1\]
\[6x + 3y - z = - 1\]
\[ - 14x - 23y + 13z = 5\]
Markscheme
METHOD 1
(from GDC)
\(\left( {\begin{array}{*{20}{ccc|c}}
1&0&{\frac{1}{6}}&{ - \frac{1}{{12}}} \\
0&1&{ - \frac{2}{3}}&{ - \frac{1}{6}} \\
0&0&0&0
\end{array}} \right)\) (M1)
\(x + \frac{1}{6}\lambda = - \frac{1}{{12}}\) A1
\(y - \frac{2}{3}\lambda = - \frac{1}{6}\) A1
\(\boldsymbol{r} = \left( { - \frac{1}{{12}}\boldsymbol{i} - \frac{1}{6}\boldsymbol{j}} \right) + \lambda \left( { - \frac{1}{6}\boldsymbol{i} + \frac{2}{3}\boldsymbol{j} + \boldsymbol{k}} \right)\) A1A1A1 N3
[6 marks]
METHOD 2
(Elimination method either for equations or row reduction of matrix)
Eliminating one of the variables M1A1
Finding a point on the line (M1)A1
Finding the direction of the line M1
The vector equation of the line A1 N3
[6 marks]
Examiners report
A large number of candidates did not use their GDC in this question. Some candidates who attempted analytical solutions looked for a point solution although the question specifically states that the planes intersect in a line. Other candidates eliminated one variable and then had no clear strategy for proceeding with the solution.
Some candidates failed to write ‘r =’, and others did not give the equation in vector form.