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Date	November 2016	Marks available	5	Reference code	16N.1.AHL.TZ0.H_1
Level	Additional Higher Level	Paper	Paper 1	Time zone	Time zone 0
Command term	Find	Question number	H_1	Adapted from	N/A

Question

Find the coordinates of the point of intersection of the planes defined by the equations $x + y + z = 3, x - y + z = 5$ $x + y + z = 3,{\text{ }}x - y + z = 5$ and $x + y + 2 z = 6$ $x + y + 2z = 6$ .

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

METHOD 1

for eliminating one variable from two equations (M1)

eg, $⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} (x + y + z = 3) 2 x + 2 z = 8 2 x + 3 z = 11 \end{matrix}$ $\left\{ {\begin{array}{*{20}{l}} {(x + y + z = 3)} \\ {2x + 2z = 8} \\ {2x + 3z = 11} \end{array}} \right.$ A1A1

for finding correctly one coordinate

eg, $⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} (x + y + z = 3) (2 x + 2 z = 8) z = 3 \end{matrix}$ $\left\{ {\begin{array}{*{20}{l}} {(x + y + z = 3)} \\ {(2x + 2z = 8)} \\ {z = 3} \end{array}} \right.$ A1

for finding correctly the other two coordinates A1

$\Rightarrow ⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} x = 1 y = - 1 z = 3 \end{matrix}$ $\Rightarrow \left\{ {\begin{array}{*{20}{l}} {x = 1} \\ {y = - 1} \\ {z = 3} \end{array}} \right.$

the intersection point has coordinates $(1, - 1, 3)$ $(1,{\text{ }} - 1,{\text{ }}3)$

METHOD 2

for eliminating two variables from two equations or using row reduction (M1)

eg, $⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} (x + y + z = 3) - 2 = 2 z = 3 \end{matrix}$ $\left\{ {\begin{array}{*{20}{l}} {(x + y + z = 3)} \\ { - 2 = 2} \\ {z = 3} \end{array}} \right.$ or $⎛ ⎜ ⎝ \begin{matrix} 1 & 1 & 1 0 & - 2 & 0 0 & 0 & 1 \end{matrix} ∣ ∣ ∣ ∣ \begin{matrix} 323 \end{matrix} ⎞ ⎟ ⎠$ $\left( {\begin{array}{*{20}{c}} 1&1&1 \\ 0&{ - 2}&0 \\ 0&0&1 \end{array}\left| {\begin{array}{*{20}{c}} 3 \\ 2 \\ 3 \end{array}} \right.} \right)$ A1A1

for finding correctly the other coordinates A1A1

$\Rightarrow ⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} x = 1 y = - 1 (z = 3) \end{matrix}$ $\Rightarrow \left\{ {\begin{array}{*{20}{l}} {x = 1} \\ {y = - 1} \\ {(z = 3)} \end{array}} \right.$ or $⎛ ⎜ ⎝ \begin{matrix} 1 & 0 & 0 0 & 1 & 0 0 & 0 & 1 \end{matrix} ∣ ∣ ∣ ∣ \begin{matrix} 1 - 1 3 \end{matrix} ⎞ ⎟ ⎠$ $\left( {\begin{array}{*{20}{c}} 1&0&0 \\ 0&1&0 \\ 0&0&1 \end{array}\left| {\begin{array}{*{20}{c}} 1 \\ { - 1} \\ 3 \end{array}} \right.} \right)$

the intersection point has coordinates $(1, - 1, 3)$ $(1,{\text{ }} - 1,{\text{ }}3)$

METHOD 3

$∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 1 & - 1 & 1 1 & 1 & 2 \end{matrix} ∣ ∣ ∣ ∣ = - 2$ $\left| {\begin{array}{*{20}{c}} 1&1&1 \\ 1&{ - 1}&1 \\ 1&1&2 \end{array}} \right| = - 2$ (A1)

attempt to use Cramer’s rule M1

$x = \frac{∣ ∣ ∣ ∣ \begin{matrix} 3 & 1 & 1 5 & - 1 & 1 6 & 1 & 2 \end{matrix} ∣ ∣ ∣ ∣}{- 2} = \frac{- 2}{- 2} = 1$ $x = \frac{{\left| {\begin{array}{*{20}{c}} 3&1&1 \\ 5&{ - 1}&1 \\ 6&1&2 \end{array}} \right|}}{{ - 2}} = \frac{{ - 2}}{{ - 2}} = 1$ A1

$y = \frac{∣ ∣ ∣ ∣ \begin{matrix} 1 & 3 & 1 1 & 5 & 1 1 & 6 & 2 \end{matrix} ∣ ∣ ∣ ∣}{- 2} = \frac{2}{- 2} = - 1$ $y = \frac{{\left| {\begin{array}{*{20}{c}} 1&3&1 \\ 1&5&1 \\ 1&6&2 \end{array}} \right|}}{{ - 2}} = \frac{2}{{ - 2}} = - 1$ A1

$z = \frac{∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 3 1 & - 1 & 5 1 & 1 & 6 \end{matrix} ∣ ∣ ∣ ∣}{- 2} = \frac{- 6}{- 2} = 3$ $z = \frac{{\left| {\begin{array}{*{20}{c}} 1&1&3 \\ 1&{ - 1}&5 \\ 1&1&6 \end{array}} \right|}}{{ - 2}} = \frac{{ - 6}}{{ - 2}} = 3$ 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 4—Statistics and probability » SL 4.7—Discrete random variables

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