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Date	May 2019	Marks available	4	Reference code	19M.1.AHL.TZ2.H_2
Level	Additional Higher Level	Paper	Paper 1 (without calculator)	Time zone	Time zone 2
Command term	Find and Hence or otherwise	Question number	H_2	Adapted from	N/A

Question

Three points in three-dimensional space have coordinates A(0, 0, 2), B(0, 2, 0) and C(3, 1, 0).

Find the vector $\overrightarrow {{\text{AB}}}$ .

[1]

a.i.

Find the vector $\overrightarrow {{\text{AC}}}$ .

[1]

a.ii.

Hence or otherwise, find the area of the triangle ABC.

[4]

b.

Markscheme

$\overrightarrow {{\text{AB}}} = \left( {\begin{array}{*{20}{c}} 0 \\ 2 \\ { - 2} \end{array}} \right)$ A1

Note: Accept row vectors or equivalent.

[1 mark]

a.i.

$\overrightarrow {{\text{AC}}} = \left( {\begin{array}{*{20}{c}} 3 \\ 1 \\ { - 2} \end{array}} \right)$ A1

Note: Accept row vectors or equivalent.

[1 mark]

a.ii.

METHOD 1

attempt at vector product using $\overrightarrow {{\text{AB}}}$ and $\overrightarrow {{\text{AC}}}$ . (M1)

±(2i + 6j +6k) A1

attempt to use area $= \frac{1}{2}\left| {\overrightarrow {{\text{AB}}} \times \overrightarrow {{\text{AC}}} } \right|$ M1

$= \frac{{\sqrt {76} }}{2}\,\,\,\left( { = \sqrt {19} } \right)$ A1

METHOD 2

attempt to use $\overrightarrow {{\text{AB}}} \bullet \overrightarrow {{\text{AC}}} = \left| {\overrightarrow {{\text{AB}}} } \right|\left| {\overrightarrow {{\text{AC}}} } \right|{\text{cos}}\,\theta$ M1

$\left( {\begin{array}{*{20}{c}} 0 \\ 2 \\ { - 2} \end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}} 3 \\ 1 \\ { - 2} \end{array}} \right) = \sqrt {{0^2} + {2^2} + {{\left( { - 2} \right)}^2}} \sqrt {{3^2} + {1^2} + {{\left( { - 2} \right)}^2}} {\text{cos}}\,\theta$

$6 = \sqrt 8 \sqrt {14} \,{\text{cos}}\,\theta$ A1

${\text{cos}}\,\theta = \frac{6}{{\sqrt 8 \sqrt {14} }} = \frac{6}{{\sqrt {112} }}$

attempt to use area $= \frac{1}{2}\left| {\overrightarrow {{\text{AB}}} \times \overrightarrow {{\text{AC}}} } \right|{\text{sin}}\,\theta$ M1

$= \frac{1}{2}\sqrt 8 \sqrt {14} \sqrt {1 - \frac{{36}}{{112}}} \,\left( { = \frac{1}{2}\sqrt 8 \sqrt {14} \sqrt {\frac{{76}}{{112}}} } \right)$

$= \frac{{\sqrt {76} }}{2}\,\,\,\left( { = \sqrt {19} } \right)$ A1

[4 marks]

b.

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

b.

Syllabus sections

Topic 3— Geometry and trigonometry » AHL 3.12—Vector definitions

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21N.2.AHL.TZ0.11d.ii:
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parallel.
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17M.1.SL.TZ1.S_8b:
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21N.2.AHL.TZ0.11b:
The line $L$ is the intersection of $Π_{1}$ and $Π_{2}$ . Verify that the vector equation of $L$ can be written as $r = (\begin{matrix} 0 \\ - 2 \\ 0 \end{matrix}) + λ (\begin{matrix} 1 \\ 1 \\ - 1 \end{matrix})$ .

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