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Date November 2017 Marks available 4 Reference code 17N.2.SL.TZ0.S_3
Level Standard Level Paper Paper 2 Time zone Time zone 0
Command term Find Question number S_3 Adapted from N/A

Question

Let  AB = ( 4 1 2 ) .

Find | AB | .

[2]
a.

Let  AC = ( 3 0 0 ) . Find B A ^ C .

[4]
b.

Markscheme

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

correct substitution     (A1)

eg 4 2 + 1 2 + 2 2

4.58257

| AB | = 21 (exact), 4.58     A1     N2

[2 marks]

a.

finding scalar product and | AC |     (A1)(A1)

scalar product = ( 4 × 3 ) + ( 1 × 0 ) + ( 2 × 0 )   ( = 12 )

| AC | = 3 2 + 0 + 0   ( = 3 )

substituting their values into cosine formula     (M1)

eg cos B A ^ C  =  4 × 3 + 0 + 0 3 2 × 21 ,   4 21 ,   cos θ = 0.873

0.509739 (29.2059°)

B A ^ C = 0.510 (29.2°)     A1     N2

[4 marks]

b.

Examiners report

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

Syllabus sections

Topic 3— Geometry and trigonometry » AHL 3.12—Vector definitions
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