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 $\frac{mb+na}{m+n}$
II. Externally, is given by $\frac{mb-na}{m-n}$
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 $PQ$ internally [Fig. (i)]. If $R$ divides $PQ$ such that $mRQ=nPR$,
where $m$ and $n$ are positive scalars, we say that the point $R$ divides $PQ$ internally in the ratio of $m:n$. Now, from triangles $ORQ$ and $OPR$, we have
$RQ=OQ-OR=b-r$
and $PR=OR-OP=r-a$,
Therefore, we have $m(b-r)=n(r-a)$
or $r=\frac{mb+na}{m+n}$ (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
$OR=\frac{mb+na}{m+n}$
Case II When $R$ divides the line segment $PQ$ externally in the ratio $m:n$

i.e., $\frac{PR}{QR}=\frac{m}{n}$
$\frac{OR-OP}{OR-OQ}=\frac{m}{n}$
$\Rightarrow \frac{r-a}{r-b}=\frac{m}{n}$
$\Rightarrow nr-na=mr-mb$
$mr-nr=mb-na$
$r(m-n)=mb-na$
$r=\frac{mb-na}{m-n}$
$OR=\frac{mb-na}{m-n}$
So, option (c) is correct.