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Date	May 2017	Marks available	3	Reference code	17M.1.AHL.TZ1.H_5
Level	Additional Higher Level	Paper	Paper 1 (without calculator)	Time zone	Time zone 1
Command term	Find	Question number	H_5	Adapted from	N/A

Question

ABCD is a parallelogram, where $\overrightarrow {{\text{AB}}}$ = –i + 2j + 3k and $\overrightarrow {{\text{AD}}}$ = 4i – j – 2k.

Find the area of the parallelogram ABCD.

[3]

By using a suitable scalar product of two vectors, determine whether ${\rm{A\hat BC}}$ is acute or obtuse.

[4]

Markscheme

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

$\overrightarrow {{\text{AB}}} \times \overrightarrow {{\text{AD}}} = -$ i $+ 10$ j – 7k M1A1

${\text{area}} = \left| {\overrightarrow {{\text{AB}}} \times \overrightarrow {{\text{AD}}} } \right|{\text{ = }}\sqrt {{1^2} + {{10}^2} + {7^2}}$

$= 5\sqrt 6 \left( {\sqrt {150} } \right)$ A1

[3 marks]

METHOD 1

$\overrightarrow {{\text{AB}}} \bullet \overrightarrow {{\text{AD}}} = - 4 - 2 - 6$ M1A1

$= - 12$

considering the sign of the answer

$\overrightarrow {{\text{AB}}} \bullet \overrightarrow {{\text{AD}}} < 0$ , therefore angle ${\rm{D\hat AB}}$ is obtuse M1

(as it is a parallelogram), ${\rm{A\hat BC}}$ is acute A1

[4 marks]

METHOD 2

$\overrightarrow {{\text{BA}}} \bullet \overrightarrow {{\text{BC}}} = + 4 + 2 + 6$ M1A1

$= 12$ considering the sign of the answer M1

$\overrightarrow {{\text{BA}}} \bullet \overrightarrow {{\text{BC}}} > 0 \Rightarrow {\rm{A\hat BC}}$ is acute A1

[4 marks]

Examiners report

[N/A]

Syllabus sections

Topic 3— Geometry and trigonometry » AHL 3.13—Scalar (dot) product

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