In the triangle OAB, OA = a and OB = b
The point P is the point on AB such that AP = 2PB and Q is a point such that OP = 3QP

Okay, this is a long question, I just couldn't do the last part.

The point R is on OA so that RQ is parallel to AB. Show that OR = 2/3 a (Hint: Let QR = cBA, where c is a scalar quantity)

I figured that OR = OQ + QR, and
OQ = 2/9 a + 4/9 b

now I am stuck.... any help will be appreciated. Thanks

2. Hello, anonymous_maths!

In $\Delta OAB,\;\overrightarrow{OA} = \vec a,\;\overrightarrow{OB} = \vec b$

$P$ is on $\overrightarrow{AB}$ such that:. $\overrightarrow{AP} \:=\: 2\!\cdot\!\overrightarrow{PB}$

. . and $Q$ is a point such that: . $\overrightarrow{OP} \:=\: 3\!\cdot\!\overrightarrow{QP}$

Point $R$ is on $\overrightarrow{OA}$ so that $\overrightarrow{RQ} \parallel \overrightarrow{AB}$

Show that: . $\overrightarrow{OR} \:=\:\tfrac{2}{3}\vec a$

As you said, we have: . $\overrightarrow{OR} \;=\;\overrightarrow{OQ} + \overrightarrow{QR}$ .[1]

Since $\Delta OQR \sim \Delta OPA\!:\;\;\overrightarrow{OQ} \:=\:\tfrac{2}{3}\,\overrightarrow{OP}\:\text{ and }\:\overrightarrow{QR} \:=\:\tfrac{2}{3}\,\overrightarrow{PA}$

Substitute into [1]: . $\overrightarrow{OR} \;=\;\tfrac{2}{3}\,\overrightarrow{OP} + \tfrac{2}{3}\,\overrightarrow{PA} \;=\;\tfrac{2}{3}\left(\overrightarrow{OP} + \overrightarrow{PA}\right) \;=\;\tfrac{2}{3}\,\overrightarrow{OA}$

. . Therefore: . $\overrightarrow{OR} \;=\;\tfrac{2}{3}\vec a$