Question
If $A (\bar{a})$ and $B (\bar{b})$ be any two points in the space $R (\bar{r})$ be a point on the line segment $AB$ dividing it internally in the ratio $m: n$, then prove that $\bar{r}=\frac{m \bar{b}+n \bar{a}}{m+n}$.
✓
Answer
$R$ is a point on the line segment $A B(A-R-B)$ and $\overline{A R}$ and $\overline{R B}$ are in the same direction.
Point $R$ divides $A B$ internally in the ratio $m: n$
$\therefore \frac{ AR }{ RB }=\frac{m}{n}$
$\therefore n ( AR )= m ( RB )$
As $n(\overline{ AR })$ and $m(\overline{ RB })$ have same direction and magnitude,
$n(\overline{ AR })=m(\overline{ RB })$
$\therefore n(\overline{ OR }-\overline{ OA })=m(\overline{ OB }-\overline{ OR })$
$\therefore n(\vec{r}-\vec{a})=m(\vec{b}-\vec{r})$
$\therefore n \vec{r}-n \vec{a}=m \vec{b}-m \vec{r}$
$\therefore m \vec{r}+n \vec{r}=m \vec{b}+n \vec{a}$
$\therefore(m+n) \vec{r}=m \vec{b}+n \vec{a}$
$\therefore \vec{r}=\frac{m \vec{b}+n \vec{a}}{m+n}$
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