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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 AB = –i + 2j + 3k and AD = 4ij – 2k.

Find the area of the parallelogram ABCD.

[3]
a.

By using a suitable scalar product of two vectors, determine whether A B ^ C is acute or obtuse.

[4]
b.

Markscheme

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

AB × AD = i + 10 j – 7k     M1A1

area = | AB × AD |  =  1 2 + 10 2 + 7 2

  = 5 6 ( 150 )     A1

[3 marks]

a.

METHOD 1

AB AD = 4 2 6      M1A1

= 12

considering the sign of the answer

AB AD < 0 , therefore angle D A ^ B is obtuse     M1

(as it is a parallelogram), A B ^ C is acute     A1

[4 marks]

METHOD 2

BA BC = + 4 + 2 + 6      M1A1

= 12 considering the sign of the answer     M1

BA BC > 0 A B ^ C is acute     A1

[4 marks]

b.

Examiners report

[N/A]
a.
[N/A]
b.

Syllabus sections

Topic 3— Geometry and trigonometry » AHL 3.13—Scalar (dot) product
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Topic 3— Geometry and trigonometry » AHL 3.16—Vector product
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