Consider a line segment AB.
Let R be any point on it such that point R divides AB internally in the ratio m : n.
bar(OA) = bara, bar(OR) = barr and bar(OB) = barb are the position vectors of points A, R, B respectively. Since point R divides AB internally in the ratio m : n,
$\begin{aligned} & \frac{A R}{R B}=\frac{m}{n} \\ & n(A R)=m(R B)\end{aligned}$
$\overline{A R}$ and $\overline{R B}$ are in the same direction.
$\therefore n(\overline{A R})=m(\overline{R B})$
$n(\bar{r}-\bar{a})=m(\bar{b}-\bar{r})$
$n \bar{r}-n \bar{a}=m \bar{b}-m \bar{r}$
$n \bar{r}+m \bar{r}=m \bar{b}+m \bar{a}$
$\bar{r}(m+n)=m \bar{b}+n \bar{a}$
$\bar{r}=\frac{m \bar{b}+n \bar{a}}{m+n}$......(1)
This is the section formula for internal division
Let P. V. of point $A \bar{a}=\hat{i}-2 \hat{j}+\widehat{k}$
P. V. of point $B \bar{b}=\hat{i}+4 \hat{j}-2 \widehat{k}$
Given, $\frac{m}{n}=\frac{2}{1}$
Now, $\bar{r}=\frac{2(\hat{i}+4 \hat{j}-2 \widehat{k})+1(\hat{i}-2 \hat{j}+\widehat{k})}{2+1}$......from (1)
$\bar{r}=\frac{3 \hat{i}+6 \hat{j}-3 \widehat{k}}{3}$
P. V. of $R$ is $\bar{r}=\hat{i}+2 \hat{j}-\widehat{k}$
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