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Date	May 2017	Marks available	4	Reference code	17M.2.AHL.TZ2.H_7
Level	Additional Higher Level	Paper	Paper 2	Time zone	Time zone 2
Command term		Question number	H_7	Adapted from	N/A

Question

Given that a $\times$ b $=$ b $\times$ c $\ne$ 0 prove that a $+$ c $=$ sb where s is a scalar.

Markscheme

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

METHOD 1

a $\times$ b = b $\times$ c

(a $\times$ b) $-$ (b $\times$ c) = 0

(a $\times$ b) + (c $\times$ b) = 0 M1A1

(a + c) $\times$ b = 0 A1

(a + c) is parallel to b $\Rightarrow$ 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 $\times$ b = b $\times$ c $\Rightarrow \left( {\begin{array}{*{20}{c}} {{a_2}{b_3} - {a_3}{b_2}} \\ {{a_3}{b_1} - {a_1}{b_3}} \\ {{a_1}{b_2} - {a_2}{b_1}} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {{b_2}{c_3} - {b_3}{c_2}} \\ {{b_3}{c_1} - {b_1}{c_3}} \\ {{b_1}{c_2} - {b_2}{c_1}} \end{array}} \right)$ M1A1

${a_2}{b_3} - {a_3}{b_2} = {b_2}{c_3} - {b_3}{c_2} \Rightarrow {b_3}({a_2} + {c_2}) = {b_2}({a_3} + {c_3})$

${a_3}{b_1} - {a_1}{b_3} = {b_3}{c_1} - {b_1}{c_3} \Rightarrow {b_1}({a_3} + {c_3}) = {b_3}({a_1} + {c_1})$

${a_1}{b_2} - {a_2}{b_1} = {b_1}{c_2} - {b_2}{c_1} \Rightarrow {b_2}({a_1} + {c_1}) = {b_1}({a_2} + {c_2})$

$\frac{{({a_1} + {c_1})}}{{{b_1}}} = \frac{{({a_2} + {c_2})}}{{{b_2}}} = \frac{{({a_3} + {c_3})}}{{{b_3}}} = s$ A1

$\Rightarrow {a_1} + {c_1} = s{b_1}$

$\Rightarrow {a_2} + {c_2} = s{b_2}$

$\Rightarrow {a_3} + {c_3} = s{b_3}$

$\Rightarrow \left( {\begin{array}{*{20}{c}} {{a_1}} \\ {{a_2}} \\ {{a_3}} \end{array}} \right) + \left( {\begin{array}{*{20}{c}} {{c_1}} \\ {{c_2}} \\ {{c_3}} \end{array}} \right) = s\left( {\begin{array}{*{20}{c}} {{b_1}} \\ {{b_2}} \\ {{b_3}} \end{array}} \right)$ A1

$\Rightarrow$ a + c = sb AG

[4 marks]

Examiners report

[N/A]

Syllabus sections

Topic 3— Geometry and trigonometry » AHL 3.16—Vector product

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