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Date May 2017 Marks available 4 Reference code 17M.2.hl.TZ2.7
Level HL only Paper 2 Time zone TZ2
Command term Prove that Question number 7 Adapted from N/A

Question

Given that a × b = b × c 0 prove that a + c = sb where s is a scalar.

Markscheme

METHOD 1

a × b = b × c

(a × b) (b × c) = 0

(a × b) + (c × b) = 0     M1A1

(a + c) × b = 0     A1

(a + c) is parallel to b a + c = sb     R1AG

 

Note:     Condone absence of arrows, underlining, or other otherwise “correct” vector notation throughout this question.

 

Note:     Allow “is in the same direction to”, for the final R mark.

 

METHOD 2

a × b = b × c (a2b3a3b2a3b1a1b3a1b2a2b1)=(b2c3b3c2b3c1b1c3b1c2b2c1)     M1A1

a2b3a3b2=b2c3b3c2b3(a2+c2)=b2(a3+c3)

a3b1a1b3=b3c1b1c3b1(a3+c3)=b3(a1+c1)

a1b2a2b1=b1c2b2c1b2(a1+c1)=b1(a2+c2)

(a1+c1)b1=(a2+c2)b2=(a3+c3)b3=s     A1

a1+c1=sb1

a2+c2=sb2

a3+c3=sb3

(a1a2a3)+(c1c2c3)=s(b1b2b3)     A1

a + c = sb     AG

[4 marks]

Examiners report

[N/A]

Syllabus sections

Topic 4 - Core: Vectors » 4.5

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