1. ## vector

The diagram shows non-collinear points O , A , B with P on the line OA such that OP:PA=2:1 and Q on the line AB such that AQ:QB=2:3 . The lines PQ and OB produced meet at the point R . If $\vec{OA}=a$ and $\vec{OB}=b$

(1) Show that $\vec {PQ}=-\frac{1}{15}a+\frac{2}{5}b$

I managed to prove this .

(2) Find the position vector of R , relative to O , in terms of b

need help on this .

2. Hello hooke
Originally Posted by hooke
The diagram shows non-collinear points O , A , B with P on the line OA such that OP:PA=2:1 and Q on the line AB such that AQ:QB=2:3 . The lines PQ and OB produced meet at the point R . If $\vec{OA}=a$ and $\vec{OB}=b$

(1) Show that $\vec {PQ}=-\frac{1}{15}a+\frac{2}{5}b$

I managed to prove this .

(2) Find the position vector of R , relative to O , in terms of b

need help on this .
The key to this is to realise that (i) $\vec{OR}$ is a multiple of $\vec{OB}$ and (ii) $\vec{PR}$ is a multiple of $\vec{PQ}$, since the points $O,B,R$ are collinear and $P, Q, R$ are collinear. So, suppose:
$\vec{OR} =m \vec{OB}$
and
$\vec{PR}=n \vec{PQ}$
Therefore:
$\vec{OR}=m\vec{b}$
and
$\vec{PR} = n\left(-\frac{1}{15}\vec a + \frac25\vec b\right)$
But:
$\vec{OR} = \vec{OP}+\vec{PR}$

$\Rightarrow m\vec b = \frac23\vec a+n\left(-\frac{1}{15}\vec a + \frac25\vec b\right)$
$= \left(\frac23-\frac{1}{15}n\right)\vec a + \frac25n\vec b$
Now the coefficient of $\vec a$ on the LHS is zero. Therefore:
$0 = \frac23-\frac{1}{15}n$

$\Rightarrow n = 10$
And if we now compare coefficients of $\vec b$, we get:
$m = \frac25n = 4$
Hence:
$\vec{OR} = m\vec b = 4\vec b$