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 $\displaystyle \vec{OA}=a$ and $\displaystyle \vec{OB}=b$

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

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 $\displaystyle \vec{OA}=a$ and $\displaystyle \vec{OB}=b$

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

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

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

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