Question

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

Hint:

Section formula for vectors: \( \vec{r} = \frac{n \vec{a} + m \vec{b}}{m + n} \) for internal division in ratio \( m:n \).

Answer 1

Let \( R \) divide \( AB \) internally in ratio \( m:n \). By section formula: 
Position vector of \( R \): 
\[ \vec{r} = \frac{n \cdot \vec{a} + m \cdot \vec{b}}{m + n}. \] Proof using vectors: 
\[ \vec{AR} = \vec{r} - \vec{a}, \vec{RB} = \vec{b} - \vec{r}. \] Since \( R \) divides \( AB \) in \( m:n \), \( \vec{AR} : \vec{RB} = m:n \). 
\[ n \vec{AR} = m \vec{RB}. \] \[ n (\vec{r} - \vec{a}) = m (\vec{b} - \vec{r}). \] \[ n \vec{r} - n \vec{a} = m \vec{b} - m \vec{r}. \] \[ n \vec{r} + m \vec{r} = m \vec{b} + n \vec{a} \Rightarrow (m + n) \vec{r} = m \vec{b} + n \vec{a}. \] \[ \vec{r} = \frac{m \vec{b} + n \vec{a}}{m + n}. \] Answer: Proved.