Question

If a line segment joining the points $P$ and $Q$ whose position vectors are $a$ and $b$ respectively, then the position vector of a point $R$, which divides the line segment in the ratio $m$ : $n$.
I. Internally, is given by $\frac{m b +n a }{m+n}$
II. Externally, is given by $\frac{m b -n a }{m-n}$
Choose the correct option.

• A. Only I is true
• B. Only II is true
• C. Both I and II are true
• D. Both I and II are false

Answer 1

Answer: (C)

Let $P$ and $Q$ be two points represented by the position vectors $OP$ and $OQ$, respectively with respect to the origion $O$. Then the line segment joining the points $P$ and $Q$ may be divided by a third point, say R in two ways - internally [Fig (i)] and externally [Fig (ii)]. Here, we intend to find the position vector OR for the point $R$ with respect to the origin $O$. We take the two cases one by one.

Case I When $R$ divides $P Q$ internally [Fig. (i)]. If $R$ divides $P Q$ such that $m RQ =n PR$,
where $m$ and $n$ are positive scalars, we say that the point $R$ divides $P Q$ internally in the ratio of $m: n$. Now, from triangles $O R Q$ and $O P R$, we have
$RQ = OQ -OR = b -r$
and $ P R=O R-O P=r-a$,
Therefore, we have $m(b-r)=n(r-a)$
or $r=\frac{m b+n a}{m+n} $ (on simplification)
Hence, the position vector of the point $R$ which divides $P$ and $Q$ internally in the ratio of $m: n$ is given by
$OR =\frac{m b+n a}{m+n}$
Case II When $R$ divides the line segment $P Q$ externally in the ratio $m: n$

i.e., $\frac{P R}{Q R}=\frac{m}{n} $
$\frac{O R-O P}{O R-O Q}=\frac{m}{n}$
$\Rightarrow \frac{r-a}{r-b} =\frac{m}{n} $
$\Rightarrow n r-n a =m r-m b $
$m r-n r =m b-n a$
$r(m-n) =m b-n a $
$r =\frac{m b-n a}{m-n} $
$OR =\frac{m b-n a}{m-n}$
So, option (c) is correct.