(i)     Express $\overrightarrow {{\text{AM}}}$ in terms of $a$ and $c$.

(ii)     Hence show that $\overrightarrow {{\text{BM}}}  = \frac{1}{2}$$a$ – $b$$+ \frac{1}{2}c$.

(i)     Express $\overrightarrow {{\text{RA}}}$ in terms of $a$ and $b$.

(ii)     Show that $\overrightarrow {RT}  =  - \frac{2}{9}a - \frac{2}{9}b + \frac{4}{9}c$.

Prove that $T$ lies on [$BM$].

(i)     $\overrightarrow {{\text{AM}}}  = \frac{1}{2}\overrightarrow {{\text{AC}}}$     (M1)

$= \frac{1}{2}$($c$ – $a$)     A1

(ii)     $\overrightarrow {{\text{BM}}}  = \overrightarrow {{\text{BA}}}  + \overrightarrow {{\text{AM}}}$     M1

$= a - b + \frac{1}{2}$$(c - a)$     A1

$\overrightarrow {{\text{BM}}}  = \frac{1}{2}a - b + \frac{1}{2}c$     AG

[4 marks]

(i)     $\overrightarrow {{\text{RA}}}  = \frac{1}{3}\overrightarrow {{\text{BA}}}$

$= \frac{1}{3}$(a – b)     A1

(ii)     $\overrightarrow {{\text{RT}}}  = \frac{2}{3}\overrightarrow {{\text{RS}}}$

$= \frac{2}{3}\left( {\overrightarrow {{\text{RA}}}  + \overrightarrow {{\text{AS}}} } \right)$     (M1)

$= \frac{2}{3}\left( {\frac{1}{3}(a - b) + \frac{2}{3}(c - a)} \right)\;\;\;$or equivalent.     A1A1

$= \frac{2}{9}$$\left( {a - b} \right)$ $+ \frac{4}{9}$$\left( {c - a} \right)$     A1

$\overrightarrow {{\text{RT}}}  =  - \frac{2}{9}$$a$ – $- \frac{2}{9}$$b$ $+ \frac{4}{9}$$c$     AG

[5 marks]

$\overrightarrow {{\text{BT}}}  = \overrightarrow {{\text{BR}}}  + \overrightarrow {{\text{RT}}}$

$= \frac{2}{3}\overrightarrow {{\text{BA}}}  + \overrightarrow {{\text{RT}}}$     (M1)

$= \frac{2}{3}a - \frac{2}{3}b - \frac{2}{9}a - \frac{2}{9}b + \frac{4}{9}c$    A1

$\overrightarrow {{\text{BT}}}  = \frac{8}{9}\left( {\frac{1}{2}a - b + \frac{1}{2}c} \right)$     A1

point $B$ is common to $\overrightarrow {{\text{BT}}}$ and $\overrightarrow {{\text{BM}}}$ and $\overrightarrow {{\text{BT}}}  = \frac{8}{9}\overrightarrow {{\text{BM}}}$     R1R1

so $T$ lies on [$BM$]     AG

[5 marks]

Total [14 marks]

A fairly straightforward question for candidates confident in the use of and correct notation for relative position vectors. Sign errors were the most common, but the majority of candidates did not gain all the reasoning marks for part (c). In particular, it was necessary to observe that not only were two vectors parallel, but that they had a point in common.

A fairly straightforward question for candidates confident in the use of and correct notation for relative position vectors. Sign errors were the most common, but the majority of candidates did not gain all the reasoning marks for part (c). In particular, it was necessary to observe that not only were two vectors parallel, but that they had a point in common.

A fairly straightforward question for candidates confident in the use of and correct notation for relative position vectors. Sign errors were the most common, but the majority of candidates did not gain all the reasoning marks for part (c). In particular, it was necessary to observe that not only were two vectors parallel, but that they had a point in common.