If \(A(\bar{a})\) and \(B(\bar{b})\) are any two points in space and \(R(\bar{r})\) be a point on the line segment \(AB\) dividing internally in the ratio \(m : n\) then prove that \(\bar{r} = \dfrac{m\bar{b} + n\bar{a}}{m + n}\).

Answer: Proof of the section formula (internal division).

[Figure: segment \(AB\) with point \(R\) between \(A\) and \(B\), \(AR : RB = m : n\); \(O\) is the origin with \(\overline{OA} = \bar{a}\), \(\overline{OB} = \bar{b}\), \(\overline{OR} = \bar{r}\).]

Since \(R\) divides \(AB\) internally in the ratio \(m : n\), \(\dfrac{AR}{RB} = \dfrac{m}{n}\), i.e. \(n\cdot AR = m\cdot RB\).

\(\overline{AR}\) and \(\overline{RB}\) have the same direction, so \(n\,\overline{AR} = m\,\overline{RB}\).

Now \(\overline{AR} = \bar{r} - \bar{a}\) and \(\overline{RB} = \bar{b} - \bar{r}\).

\(\therefore n(\bar{r} - \bar{a}) = m(\bar{b} - \bar{r})\)

\(n\bar{r} - n\bar{a} = m\bar{b} - m\bar{r}\)

\((m + n)\bar{r} = m\bar{b} + n\bar{a}\)

\(\therefore \bar{r} = \dfrac{m\bar{b} + n\bar{a}}{m + n}\).