Date | November 2016 | Marks available | 5 | Reference code | 16N.1.hl.TZ0.1 |
Level | HL only | Paper | 1 | Time zone | TZ0 |
Command term | Find | Question number | 1 | Adapted from | N/A |
Question
Find the coordinates of the point of intersection of the planes defined by the equations \(x + y + z = 3,{\text{ }}x - y + z = 5\) and \(x + y + 2z = 6\).
Markscheme
METHOD 1
for eliminating one variable from two equations (M1)
eg, \(\left\{ {\begin{array}{*{20}{l}} {(x + y + z = 3)} \\ {2x + 2z = 8} \\ {2x + 3z = 11} \end{array}} \right.\) A1A1
for finding correctly one coordinate
eg, \(\left\{ {\begin{array}{*{20}{l}} {(x + y + z = 3)} \\ {(2x + 2z = 8)} \\ {z = 3} \end{array}} \right.\) A1
for finding correctly the other two coordinates A1
\( \Rightarrow \left\{ {\begin{array}{*{20}{l}} {x = 1} \\ {y = - 1} \\ {z = 3} \end{array}} \right.\)
the intersection point has coordinates \((1,{\text{ }} - 1,{\text{ }}3)\)
METHOD 2
for eliminating two variables from two equations or using row reduction (M1)
eg, \(\left\{ {\begin{array}{*{20}{l}} {(x + y + z = 3)} \\ { - 2 = 2} \\ {z = 3} \end{array}} \right.\) or \(\left( {\begin{array}{*{20}{c}} 1&1&1 \\ 0&{ - 2}&0 \\ 0&0&1 \end{array}\left| {\begin{array}{*{20}{c}} 3 \\ 2 \\ 3 \end{array}} \right.} \right)\) A1A1
for finding correctly the other coordinates A1A1
\( \Rightarrow \left\{ {\begin{array}{*{20}{l}} {x = 1} \\ {y = - 1} \\ {(z = 3)} \end{array}} \right.\) or \(\left( {\begin{array}{*{20}{c}} 1&0&0 \\ 0&1&0 \\ 0&0&1 \end{array}\left| {\begin{array}{*{20}{c}} 1 \\ { - 1} \\ 3 \end{array}} \right.} \right)\)
the intersection point has coordinates \((1,{\text{ }} - 1,{\text{ }}3)\)
METHOD 3
\(\left| {\begin{array}{*{20}{c}} 1&1&1 \\ 1&{ - 1}&1 \\ 1&1&2 \end{array}} \right| = - 2\) (A1)
attempt to use Cramer’s rule M1
\(x = \frac{{\left| {\begin{array}{*{20}{c}} 3&1&1 \\ 5&{ - 1}&1 \\ 6&1&2 \end{array}} \right|}}{{ - 2}} = \frac{{ - 2}}{{ - 2}} = 1\) A1
\(y = \frac{{\left| {\begin{array}{*{20}{c}} 1&3&1 \\ 1&5&1 \\ 1&6&2 \end{array}} \right|}}{{ - 2}} = \frac{2}{{ - 2}} = - 1\) A1
\(z = \frac{{\left| {\begin{array}{*{20}{c}} 1&1&3 \\ 1&{ - 1}&5 \\ 1&1&6 \end{array}} \right|}}{{ - 2}} = \frac{{ - 6}}{{ - 2}} = 3\) A1
Note: Award M1 only if candidate attempts to determine at least one of the variables using this method.
[5 marks]