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

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

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

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

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

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

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

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

[1 mark]

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

$= \frac{1}{7}$b – a     A1

[2 marks]

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

[2 marks]

$\overrightarrow {{\text{OD}}}  = \frac{1}{2}$ 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]

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]