17N.1.AHL.TZ0.H_9

pestleMathematicsAIHLPaper 117N· ahl-3-10-vector-definitionssource ↗

In the following diagram, $\overrightarrow {{\text{OA}}}$ = a, $\overrightarrow {{\text{OB}}}$ = b. C is the midpoint of [OA] and $\overrightarrow {{\text{OF}}} = \frac{1}{6}\overrightarrow {{\text{FB}}}$ .

N17/5/MATHL/HP1/ENG/TZ0/09

It is given also that $\overrightarrow {{\text{AD}}} = \lambda \overrightarrow {{\text{AF}}}$ and $\overrightarrow {{\text{CD}}} = \mu \overrightarrow {{\text{CB}}}$ , where $\lambda ,{\text{ }}\mu \in \mathbb{R}$ .

Find, in terms of a and b $\overrightarrow {{\text{OF}}}$ .

[1]

a.i.

Find, in terms of a and b $\overrightarrow {{\text{AF}}}$ .

[2]

a.ii.

Find an expression for $\overrightarrow {{\text{OD}}}$ in terms of a, b and $\lambda$ ;

[2]

b.i.

Find an expression for $\overrightarrow {{\text{OD}}}$ in terms of a, b and $\mu$ .

[2]

b.ii.

Show that $\mu = \frac{1}{{13}}$ , and find the value of $\lambda$ .

[4]

Deduce an expression for $\overrightarrow {{\text{CD}}}$ in terms of a and b only.

[2]

Given that area $\Delta {\text{OAB}} = k({\text{area }}\Delta {\text{CAD}})$ , find the value of $k$ .

[5]

Markscheme / solution

$\overrightarrow {{\text{OF}}} = \frac{1}{7}$ b A1

[1 mark]

a.i.

$\overrightarrow {{\text{AF}}} = \overrightarrow {{\text{OF}}} - \overrightarrow {{\text{OA}}}$ (M1)

$= \frac{1}{7}$ b – a A1

[2 marks]

a.ii.

$\overrightarrow {{\text{OD}}} =$ a $+ \lambda \left( {\frac{1}{7}b -a} \right){\text{ }}\left( { = (1 - \lambda )a + \frac{\lambda }{7}b} \right)$ M1A1

[2 marks]

b.i.

$\overrightarrow {{\text{OD}}} = \frac{1}{2}$ a $+ \mu \left( { - \frac{1}{2}a + b} \right){\text{ }}\left( { = \left( {\frac{1}{2} - \frac{\mu }{2}} \right)a + \mu b} \right)$ M1A1

[2 marks]

b.ii.

equating coefficients: M1

$\frac{\lambda }{7} = \mu ,{\text{ }}1 - \lambda = \frac{{1 - \mu }}{2}$ A1

solving simultaneously: M1

$\lambda = \frac{7}{{13}},{\text{ }}\mu = \frac{1}{{13}}$ A1AG

[4 marks]

$\overrightarrow {{\text{CD}}} = \frac{1}{{13}}\overrightarrow {{\text{CB}}}$

$= \frac{1}{{13}}\left( {b - \frac{1}{2}a} \right){\text{ }}\left( { = - \frac{1}{{26}}a + \frac{1}{{13}}b} \right)$ M1A1

[2 marks]

METHOD 1

${\text{area }}\Delta {\text{ACD}} = \frac{1}{2}{\text{CD}} \times {\text{AC}} \times \sin {\rm{A\hat CB}}$ (M1)

${\text{area }}\Delta {\text{ACB}} = \frac{1}{2}{\text{CB}} \times {\text{AC}} \times \sin {\rm{A\hat CB}}$ (M1)

${\text{ratio }}\frac{{{\text{area }}\Delta {\text{ACD}}}}{{{\text{area }}\Delta {\text{ACB}}}} = \frac{{{\text{CD}}}}{{{\text{CB}}}} = \frac{1}{{13}}$ A1

$k = \frac{{{\text{area }}\Delta {\text{OAB}}}}{{{\text{area }}\Delta {\text{CAD}}}} = \frac{{13}}{{{\text{area }}\Delta {\text{CAB}}}} \times {\text{area }}\Delta {\text{OAB}}$ (M1)

$= 13 \times 2 = 26$ A1

METHOD 2

${\text{area }}\Delta {\text{OAB}} = \frac{1}{2}\left| {a \times b} \right|$ A1

${\text{area }}\Delta {\text{CAD}} = \frac{1}{2}\left| {\overrightarrow {{\text{CA}}} \times \overrightarrow {{\text{CD}}} } \right|$ or $\frac{1}{2}\left| {\overrightarrow {{\text{CA}}} \times \overrightarrow {{\text{AD}}} } \right|$ M1

$= \frac{1}{2}\left| {\frac{1}{2}a \times \left( { - \frac{1}{{26}}a + \frac{1}{{13}}b} \right)} \right|$

$= \frac{1}{2}\left| {\frac{1}{2}a \times \left( { - \frac{1}{{26}}a} \right) + \frac{1}{2}a \times \frac{1}{{13}}b} \right|$ (M1)

$= \frac{1}{2} \times \frac{1}{2} \times \frac{1}{{13}}\left| {a \times b} \right|{\text{ }}\left( { = \frac{1}{{52}}\left| {a \times b} \right|} \right)$ A1

${\text{area }}\Delta {\text{OAB}} = k({\text{area }}\Delta {\text{CAD}})$

$\frac{1}{2}\left| {a \times b} \right| = k\frac{1}{{52}}\left| {a \times b} \right|$

$k = 26$ A1

[5 marks]

Examiners’ report

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a.i.

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a.ii.

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b.i.

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b.ii.

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