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Date	May 2019	Marks available	2	Reference code	19M.1.SL.TZ1.S_2
Level	Standard Level	Paper	Paper 1	Time zone	Time zone 1
Command term	Write down	Question number	S_2	Adapted from	N/A

Question

A line, ${L_1}$ , has equation $r = \left( {\begin{array}{*{20}{c}} { - 3} \\ 9 \\ {10} \end{array}} \right) + s\left( {\begin{array}{*{20}{c}} 6 \\ 0 \\ 2 \end{array}} \right)$ . Point ${\text{P}}\left( {15{\text{,}}\,\,9{\text{,}}\,\,c} \right)$ lies on ${L_1}$ .

Find $c$ .

[4]

A second line, ${L_2}$ , is parallel to ${L_1}$ and passes through (1, 2, 3).

Write down a vector equation for ${L_2}$ .

[2]

Markscheme

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

correct equation (A1)

eg $- 3 + 6s = 15$ , $6s = 18$

$s = 3$ (A1)

substitute their $s$ value into $z$ component (M1)

eg $10 + 3\left( 2 \right)$ , $10 + 6$

$c = 16$ A1 N3

[4 marks]

$r = \left( {\begin{array}{*{20}{c}} 1 \\ 2 \\ 3 \end{array}} \right) + t\left( {\begin{array}{*{20}{c}} 6 \\ 0 \\ 2 \end{array}} \right)$ (=(i + 2j + 3k) + $t$ (6i + 2k)) A2 N2

Note: Accept any scalar multiple of $\left( {\begin{array}{*{20}{c}} 6 \\ 0 \\ 2 \end{array}} \right)$ for the direction vector.

Award A1 for $\left( {\begin{array}{*{20}{c}} 1 \\ 2 \\ 3 \end{array}} \right) + t\left( {\begin{array}{*{20}{c}} 6 \\ 0 \\ 2 \end{array}} \right)$ , A1 for ${L_2} = \left( {\begin{array}{*{20}{c}} 1 \\ 2 \\ 3 \end{array}} \right) + t\left( {\begin{array}{*{20}{c}} 6 \\ 0 \\ 2 \end{array}} \right)$ , A0 for $r = \left( {\begin{array}{*{20}{c}} 6 \\ 0 \\ 2 \end{array}} \right) + t\left( {\begin{array}{*{20}{c}} 1 \\ 2 \\ 3 \end{array}} \right)$ .

[2 marks]

Examiners report

[N/A]

Syllabus sections

Topic 3—Geometry and trigonometry » AHL 3.11—Vector equation of a line in 2d and 3d

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