21M.1.AHL.TZ1.5

pestleMathematicsAIHLPaper 121M· ahl-3-10-vector-definitionssource ↗

A garden has a triangular sunshade suspended from three points $A (2, 0, 2), B (8, 0, 2)$ and $C (5, 4, 3)$ , relative to an origin in the corner of the garden. All distances are measured in metres.

Find $\vec{CA}$ .

[1]

a.i.

Find $\vec{CB}$ .

[1]

a.ii.

Find $\vec{CA} \times \vec{CB}$ .

[2]

Hence find the area of the triangle $ABC$ .

[2]

Markscheme / solution

$\vec{CA} = (\begin{matrix} - 3 \\ - 4 \\ - 1 \end{matrix})$ A1

[1 mark]

a.i.

$\vec{CB} = (\begin{matrix} 3 \\ - 4 \\ - 1 \end{matrix})$ A1

[1 mark]

a.ii.

$\vec{CA} \times \vec{CB} = (\begin{matrix} 0 \\ - 6 \\ 24 \end{matrix})$ (M1)A1

Note: Do not award (M1) if less than 2 entries are correct.

[2 marks]

area is $\frac{1}{2} \sqrt{6^{2} + 24^{2}} = 12.4 m^{2} (12.3693 \dots, 3 \sqrt{17})$ (M1)A1

[2 marks]

Examiners’ report

[N/A]

a.i.

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a.ii.

[N/A]