In the diagram, $\overrightarrow{OA} = 6a$, $\overrightarrow{OB} = 6b$ and $3\overrightarrow{AN} = \overrightarrow{AB}$. M is the mid-point of OB.

1. Express $\overrightarrow{AB}$ in terms of $a$ and $b$.

$\overrightarrow{AB}=\overrightarrow{AO}+\overrightarrow{OB}$
Since $\overrightarrow{OA}=6a$, then $\overrightarrow{AO}=-6a$.

$\overrightarrow{AB}=-6a+6b=6b-6a$

2. Use $3\overrightarrow{AN}=\overrightarrow{AB}$ to find $\overrightarrow{AN}$.

$3\overrightarrow{AN}=\overrightarrow{AB}$
$3\overrightarrow{AN}=6b-6a$
$\overrightarrow{AN}=\frac{1}{3}(6b-6a)=2b-2a$

1. Express $\overrightarrow{ON}$ using a suitable vector path and simplify.

$\overrightarrow{ON}=\overrightarrow{OA}+\overrightarrow{AN}$
$=6a+(2b-2a)$
$=4a+2b$

1. Find the position vector of $M$ using that $M$ is the midpoint of $OB$.

Since $M$ is the mid-point of $OB$,
$\overrightarrow{OM}=\frac{1}{2}\overrightarrow{OB}=\frac{1}{2}(6b)=3b.$

2. Express $\overrightarrow{NM}$ using a vector path through $O$ and simplify.

From part (b), $\overrightarrow{ON}=4a+2b$.

$\overrightarrow{NM}=\overrightarrow{NO}+\overrightarrow{OM}$
$=-\overrightarrow{ON}+\overrightarrow{OM}$
$=-(4a+2b)+3b$
$=b-4a.$

1. Find $\overrightarrow{MP}$ using the given relation and the earlier result for $\overrightarrow{NM}$.

From part (c),
$\overrightarrow{NM}=b-4a$.

Hence
$\overrightarrow{MN}=-(b-4a)=4a-b$.

Given $\overrightarrow{MP}=3\overrightarrow{MN}$,
$\overrightarrow{MP}=3(4a-b)=12a-3b.$

2. Use the path from $O$ to $P$ through $M$ to find the position vector of $P$.

Since $M$ is the mid-point of $OB$ and $\overrightarrow{OB}=6b$,
$\overrightarrow{OM}=3b.$

Therefore
$\overrightarrow{OP}=\overrightarrow{OM}+\overrightarrow{MP}$
$=3b+(12a-3b)$
$=12a.$

1. Compare the position vectors of A and P.

Given $\overrightarrow{OA}=6a$ and from part (d)(i), $\overrightarrow{OP}=12a$.

So
$\overrightarrow{OP}=2\overrightarrow{OA}$.

2. State the two geometrical facts about O, A and P.

Since $\overrightarrow{OP}$ is a positive scalar multiple of $\overrightarrow{OA}$, the points $O$, $A$ and $P$ lie on the same straight line.

Also, $\overrightarrow{OP}=2\overrightarrow{OA}$, so $OA=AP$. Hence $A$ is the midpoint of $OP$.