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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]

Examiners report

[N/A]

Syllabus sections

Topic 5 - Core: Statistics and probability » 5.5 » Concept of discrete and continuous random variables and their probability distributions.
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