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Date	November 2016	Marks available	3	Reference code	16N.1.SL.TZ0.S_4
Level	Standard Level	Paper	Paper 1	Time zone	Time zone 0
Command term	Find	Question number	S_4	Adapted from	N/A

Question

The position vectors of points P and Q are i $+$ 2 j $-$ k and 7i $+$ 3j $-$ 4k respectively.

Find a vector equation of the line that passes through P and Q.

[4]

a.

The line through P and Q is perpendicular to the vector 2i $+$ nk. Find the value of $n$ .

[3]

b.

Markscheme

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

valid attempt to find direction vector (M1)

eg $\,\,\,\,\,$ $\overrightarrow {{\text{PQ}}} ,{\text{ }}\overrightarrow {{\text{QP}}}$

correct direction vector (or multiple of) (A1)

eg $\,\,\,\,\,$ 6i $+$ j $-$ 3k

any correct equation in the form r $=$ a $+$ tb (any parameter for $t$ ) A2 N3

where a is i $+$ 2j $-$ k or 7i $+$ 3j $-$ 4k , and b is a scalar multiple of 6i $+$ j $-$ 3k

eg $\,\,\,\,\,$ r $=$ 7i $+$ 3j $-$ 4k $+$ t(6i $+$ j $-$ 3k), r $= \left( {\begin{array}{*{20}{c}} {1 + 6s} \\ {2 + 1s} \\ { - 1 - 3s} \end{array}} \right),{\text{ }}r = \left( {\begin{array}{*{20}{c}} 1 \\ 2 \\ { - 1} \end{array}} \right) + t\left( {\begin{array}{*{20}{c}} { - 6} \\ { - 1} \\ 3 \end{array}} \right)$

Notes: Award A1 for the form a $+$ tb, A1 for the form L $=$ a $+$ tb, A0 for the form r $=$ b $+$ ta.

[4 marks]

a.

correct expression for scalar product (A1)

eg $\,\,\,\,\,$ $6 \times 2 + 1 \times 0 + ( - 3) \times n,{\text{ }} - 3n + 12$

setting scalar product equal to zero (seen anywhere) (M1)

eg $\,\,\,\,\,$ u $\bullet$ v $= 0,{\text{ }} - 3n + 12 = 0$

$n = 4$ A1 N2

[3 marks]

b.

Examiners report

[N/A]

a.

[N/A]

b.

Syllabus sections

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

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