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 →AB−−→AB.
Find the vector →AC−−→AC.
Hence or otherwise, find the area of the triangle ABC.
Markscheme
→AB=(02−2)−−→AB=⎛⎜⎝02−2⎞⎟⎠ A1
Note: Accept row vectors or equivalent.
[1 mark]
→AC=(31−2)−−→AC=⎛⎜⎝31−2⎞⎟⎠ A1
Note: Accept row vectors or equivalent.
[1 mark]
METHOD 1
attempt at vector product using →AB−−→AB and →AC−−→AC. (M1)
±(2i + 6j +6k) A1
attempt to use area =12|→AB×→AC|=12∣∣∣−−→AB×−−→AC∣∣∣ M1
=√762(=√19)=√762(=√19) A1
METHOD 2
attempt to use →AB∙→AC=|→AB||→AC|cosθ−−→AB∙−−→AC=∣∣∣−−→AB∣∣∣∣∣∣−−→AC∣∣∣cosθ M1
(02−2)⋅(31−2)=√02+22+(−2)2√32+12+(−2)2cosθ⎛⎜⎝02−2⎞⎟⎠⋅⎛⎜⎝31−2⎞⎟⎠=√02+22+(−2)2√32+12+(−2)2cosθ
6=√8√14cosθ6=√8√14cosθ A1
cosθ=6√8√14=6√112cosθ=6√8√14=6√112
attempt to use area =12|→AB×→AC|sinθ=12∣∣∣−−→AB×−−→AC∣∣∣sinθ M1
=12√8√14√1−36112(=12√8√14√76112)=12√8√14√1−36112(=12√8√14√76112)
=√762(=√19)=√762(=√19) A1
[4 marks]