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Date	May 2019	Marks available	1	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	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 $- - \to AB$ $\overrightarrow {{\text{AB}}}$ .

[1]

a.i.

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

[1]

a.ii.

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

[4]

Markscheme

$- - \to AB = ⎛ ⎜ ⎝ \begin{matrix} 02 - 2 \end{matrix} ⎞ ⎟ ⎠$ $\overrightarrow {{\text{AB}}} = \left( {\begin{array}{*{20}{c}} 0 \\ 2 \\ { - 2} \end{array}} \right)$ A1

Note: Accept row vectors or equivalent.

[1 mark]

a.i.

$- - \to AC = ⎛ ⎜ ⎝ \begin{matrix} 31 - 2 \end{matrix} ⎞ ⎟ ⎠$ $\overrightarrow {{\text{AC}}} = \left( {\begin{array}{*{20}{c}} 3 \\ 1 \\ { - 2} \end{array}} \right)$ A1

Note: Accept row vectors or equivalent.

[1 mark]

a.ii.

METHOD 1

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

±(2i + 6j +6k) A1

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

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

METHOD 2

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

$⎛ ⎜ ⎝ \begin{matrix} 02 - 2 \end{matrix} ⎞ ⎟ ⎠ \cdot ⎛ ⎜ ⎝ \begin{matrix} 31 - 2 \end{matrix} ⎞ ⎟ ⎠ = \sqrt{0^{2} + 2^{2} + {(- 2)}^{2}} \sqrt{3^{2} + 1^{2} + {(- 2)}^{2}} cos θ$ $\left( {\begin{array}{*{20}{c}} 0 \\ 2 \\ { - 2} \end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}} 3 \\ 1 \\ { - 2} \end{array}} \right) = \sqrt {{0^2} + {2^2} + {{\left( { - 2} \right)}^2}} \sqrt {{3^2} + {1^2} + {{\left( { - 2} \right)}^2}} {\text{cos}}\,\theta$

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

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

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

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