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Date November 2008 Marks available 6 Reference code 08N.1.hl.TZ0.10
Level HL only Paper 1 Time zone TZ0
Command term Show that Question number 10 Adapted from N/A

Question

Three distinct non-zero vectors are given by \(\overrightarrow {{\text{OA}}} \) = a, \(\overrightarrow {{\text{OB}}} \) = b, and \(\overrightarrow {{\text{OC}}} \) = c .

If \(\overrightarrow {{\text{OA}}} \) is perpendicular to \(\overrightarrow {{\text{BC}}} \) and \(\overrightarrow {{\text{OB}}} \) is perpendicular to \(\overrightarrow {{\text{CA}}} \) , show that \(\overrightarrow {{\text{OC}}} \) is perpendicular to \(\overrightarrow {{\text{AB}}} \).

Markscheme

\(\overrightarrow {{\text{BC}}} \) = cb

\(\overrightarrow {{\text{CA}}} \) = ac

\( \Rightarrow \) a\( \cdot \)(cb) = 0     M1

and b\( \cdot \)(ac) = 0     M1

\( \Rightarrow \) a\( \cdot \)c = a\( \cdot \)b     A1

and a\( \cdot \)b = b\( \cdot \)c     A1

\( \Rightarrow \) a\( \cdot \)c = b\( \cdot \)c     M1

\( \Rightarrow \) b\( \cdot \)ca\( \cdot \)c = 0

c\( \cdot \)(ba) = 0     A1

\( \Rightarrow \) \(\overrightarrow {{\text{OC}}} \) is perpendicular to \(\overrightarrow {{\text{AB}}} \) , as b \( \ne \) a .     AG

[6 marks]

Examiners report

Only the better candidates were able to make significant progress with this question. Many candidates understood how to begin the question, but did not see how to progress to the last stage. On the whole the candidates' use of notation in this question was poor.

Syllabus sections

Topic 4 - Core: Vectors » 4.2 » Properties of the scalar product: \({\boldsymbol{v}} \cdot {\boldsymbol{w}} = {\boldsymbol{w}} \cdot {\boldsymbol{v}}\) ; \({\boldsymbol{u}} \cdot \left( {{\mathbf{v}} + {\boldsymbol{w}}} \right) = {\boldsymbol{u}} \cdot {\boldsymbol{v}} + {\boldsymbol{u}} \cdot {\boldsymbol{w}}\) ; \(\left( {k{\boldsymbol{v}}} \right) \cdot {\boldsymbol{w}} = k\left( {{\boldsymbol{v}} \cdot {\boldsymbol{w}}} \right)\) ; \({\boldsymbol{v}} \cdot {\boldsymbol{v}} = {\left| {\boldsymbol{v}} \right|^2}\) .

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