Notes On Section Formula - CBSE Class 12 Maths
The position vector for R in both these cases. Case I: R divides AB internally. $\stackrel{\to }{\text{OA}}$  =            = $\stackrel{\to }{\text{b}}$       = $\stackrel{\to }{\text{r}}$    Point R divide line segment AB in the ratio m:n AR:RB = m:n In ∆OAR: $\stackrel{\to }{\text{OR}}$   = $\stackrel{\to }{\text{OA}}$   +  $\stackrel{\to }{\text{AR}}$         ⇒ $\stackrel{\to }{\text{AR}}$ = $\stackrel{\to }{\text{OR}}$ - $\stackrel{\to }{\text{OA}}$ ⇒ $\stackrel{\to }{\text{AR}}$ = $\stackrel{\to }{\text{r}}$ - $\stackrel{\to }{\text{a}}$ In ∆ORB: $\stackrel{\to }{\text{OB}}$ = $\stackrel{\to }{\text{OR}}$ + $\stackrel{\to }{\text{RB}}$ ⇒ $\stackrel{\to }{\text{RB}}$ = $\stackrel{\to }{\text{OB}}$ - $\stackrel{\to }{\text{OR}}$ ⇒ $\stackrel{\to }{\text{RB}}$ = $\stackrel{\to }{\text{b}}$ - $\stackrel{\to }{\text{r}}$ $\stackrel{\to }{\text{AR}}$ :$\stackrel{\to }{\text{RB}}$ = m:n ⇒ m $\stackrel{\to }{\text{RB}}$ = n $\stackrel{\to }{\text{AR}}$ ⇒ m ($\stackrel{\to }{\text{b}}$ - $\stackrel{\to }{\text{r}}$) = n ( $\stackrel{\to }{\text{r}}$   - $\stackrel{\to }{\text{a}}$  ) ⇒ m $\stackrel{\to }{\text{b}}$ - m $\stackrel{\to }{\text{r}}$ = n $\stackrel{\to }{\text{r}}$ - n $\stackrel{\to }{\text{a}}$ ⇒ m $\stackrel{\to }{\text{r}}$ + n $\stackrel{\to }{\text{r}}$= m $\stackrel{\to }{\text{b}}$ + n $\stackrel{\to }{\text{a}}$ ⇒ (m + n )$\stackrel{\to }{\text{r}}$= m $\stackrel{\to }{\text{b}}$ + n $\stackrel{\to }{\text{a}}$ ⇒ $\stackrel{\to }{\text{r}}$=  $\frac{\text{m}\stackrel{\to }{\text{b}}\text{+ n}\stackrel{\to }{\text{a}}}{\text{m+n}}$  ⇒ $\stackrel{\to }{\text{OR}}$ = $\frac{\text{m}\stackrel{\to }{\text{b}}\text{+ n}\stackrel{\to }{\text{a}}}{\text{m+n}}$  The point R lies at the midpoint of line segment AB. When m : n = 1:1 ⇒ m = n ⇒ $\stackrel{\to }{\text{OR}}$ =   $\frac{\text{m}\stackrel{\to }{\text{b}}\text{+ n}\stackrel{\to }{\text{a}}}{\text{m+n}}$  =      ⇒ $\stackrel{\to }{\text{OR}}$ =      $\frac{\text{m (}\stackrel{\to }{\text{a}}\text{+}\stackrel{\to }{\text{b}}\text{)}}{\text{2m}}$          ⇒ $\stackrel{\to }{\text{OR}}$ =    =        Case II: R divides AB externally $\stackrel{\to }{\text{OA}}$  =            = $\stackrel{\to }{\text{b}}$       = $\stackrel{\to }{\text{r}}$    Let R divide AB externally such that AR:BR = m:n In ∆OAR: $\stackrel{\to }{\text{OR}}$  = $\stackrel{\to }{\text{OA}}$  + $\stackrel{\to }{\text{AR}}$  ⇒ $\stackrel{\to }{\text{AR}}$  = $\stackrel{\to }{\text{OR}}$  - $\stackrel{\to }{\text{OA}}$  ⇒ $\stackrel{\to }{\text{AR}}$  = $\stackrel{\to }{\text{r}}$  - $\stackrel{\to }{\text{a}}$  In ∆OBR: $\stackrel{\to }{\text{OR}}$  = $\stackrel{\to }{\text{OB}}$  + $\stackrel{\to }{\text{BR}}$  ⇒ $\stackrel{\to }{\text{BR}}$  = $\stackrel{\to }{\text{OR}}$  - $\stackrel{\to }{\text{OB}}$  ⇒ $\stackrel{\to }{\text{BR}}$  = $\stackrel{\to }{\text{r}}$  - $\stackrel{\to }{\text{b}}$  $\stackrel{\to }{\text{AR}}$  :$\stackrel{\to }{\text{BR}}$  = m:n ⇒ m ($\stackrel{\to }{\text{BR}}$ )= n ($\stackrel{\to }{\text{AR}}$ ) ⇒ m ($\stackrel{\to }{\text{r}}$  - $\stackrel{\to }{\text{b}}$ ) = n ($\stackrel{\to }{\text{r}}$  - $\stackrel{\to }{\text{a}}$  ) ⇒ m $\stackrel{\to }{\text{r}}$ - m $\stackrel{\to }{\text{b}}$ = n $\stackrel{\to }{\text{r}}$ - n $\stackrel{\to }{\text{a}}$ ⇒ m $\stackrel{\to }{\text{r}}$ - n $\stackrel{\to }{\text{r}}$= m $\stackrel{\to }{\text{b}}$ - n $\stackrel{\to }{\text{a}}$ ⇒ (m - n )$\stackrel{\to }{\text{r}}$ = m $\stackrel{\to }{\text{b}}$ - n $\stackrel{\to }{\text{a}}$ ⇒ $\stackrel{\to }{\text{r}}$ = $\frac{\text{m}\stackrel{\to }{\text{b}}\text{- n}\stackrel{\to }{\text{a}}}{\text{m-n}}$  ⇒ $\stackrel{\to }{\text{OR}}$ = $\frac{\text{m}\stackrel{\to }{\text{b}}\text{- n}\stackrel{\to }{\text{a}}}{\text{m-n}}$ .

#### Summary

The position vector for R in both these cases. Case I: R divides AB internally. $\stackrel{\to }{\text{OA}}$  =            = $\stackrel{\to }{\text{b}}$       = $\stackrel{\to }{\text{r}}$    Point R divide line segment AB in the ratio m:n AR:RB = m:n In ∆OAR: $\stackrel{\to }{\text{OR}}$   = $\stackrel{\to }{\text{OA}}$   +  $\stackrel{\to }{\text{AR}}$         ⇒ $\stackrel{\to }{\text{AR}}$ = $\stackrel{\to }{\text{OR}}$ - $\stackrel{\to }{\text{OA}}$ ⇒ $\stackrel{\to }{\text{AR}}$ = $\stackrel{\to }{\text{r}}$ - $\stackrel{\to }{\text{a}}$ In ∆ORB: $\stackrel{\to }{\text{OB}}$ = $\stackrel{\to }{\text{OR}}$ + $\stackrel{\to }{\text{RB}}$ ⇒ $\stackrel{\to }{\text{RB}}$ = $\stackrel{\to }{\text{OB}}$ - $\stackrel{\to }{\text{OR}}$ ⇒ $\stackrel{\to }{\text{RB}}$ = $\stackrel{\to }{\text{b}}$ - $\stackrel{\to }{\text{r}}$ $\stackrel{\to }{\text{AR}}$ :$\stackrel{\to }{\text{RB}}$ = m:n ⇒ m $\stackrel{\to }{\text{RB}}$ = n $\stackrel{\to }{\text{AR}}$ ⇒ m ($\stackrel{\to }{\text{b}}$ - $\stackrel{\to }{\text{r}}$) = n ( $\stackrel{\to }{\text{r}}$   - $\stackrel{\to }{\text{a}}$  ) ⇒ m $\stackrel{\to }{\text{b}}$ - m $\stackrel{\to }{\text{r}}$ = n $\stackrel{\to }{\text{r}}$ - n $\stackrel{\to }{\text{a}}$ ⇒ m $\stackrel{\to }{\text{r}}$ + n $\stackrel{\to }{\text{r}}$= m $\stackrel{\to }{\text{b}}$ + n $\stackrel{\to }{\text{a}}$ ⇒ (m + n )$\stackrel{\to }{\text{r}}$= m $\stackrel{\to }{\text{b}}$ + n $\stackrel{\to }{\text{a}}$ ⇒ $\stackrel{\to }{\text{r}}$=  $\frac{\text{m}\stackrel{\to }{\text{b}}\text{+ n}\stackrel{\to }{\text{a}}}{\text{m+n}}$  ⇒ $\stackrel{\to }{\text{OR}}$ = $\frac{\text{m}\stackrel{\to }{\text{b}}\text{+ n}\stackrel{\to }{\text{a}}}{\text{m+n}}$  The point R lies at the midpoint of line segment AB. When m : n = 1:1 ⇒ m = n ⇒ $\stackrel{\to }{\text{OR}}$ =   $\frac{\text{m}\stackrel{\to }{\text{b}}\text{+ n}\stackrel{\to }{\text{a}}}{\text{m+n}}$  =      ⇒ $\stackrel{\to }{\text{OR}}$ =      $\frac{\text{m (}\stackrel{\to }{\text{a}}\text{+}\stackrel{\to }{\text{b}}\text{)}}{\text{2m}}$          ⇒ $\stackrel{\to }{\text{OR}}$ =    =        Case II: R divides AB externally $\stackrel{\to }{\text{OA}}$  =            = $\stackrel{\to }{\text{b}}$       = $\stackrel{\to }{\text{r}}$    Let R divide AB externally such that AR:BR = m:n In ∆OAR: $\stackrel{\to }{\text{OR}}$  = $\stackrel{\to }{\text{OA}}$  + $\stackrel{\to }{\text{AR}}$  ⇒ $\stackrel{\to }{\text{AR}}$  = $\stackrel{\to }{\text{OR}}$  - $\stackrel{\to }{\text{OA}}$  ⇒ $\stackrel{\to }{\text{AR}}$  = $\stackrel{\to }{\text{r}}$  - $\stackrel{\to }{\text{a}}$  In ∆OBR: $\stackrel{\to }{\text{OR}}$  = $\stackrel{\to }{\text{OB}}$  + $\stackrel{\to }{\text{BR}}$  ⇒ $\stackrel{\to }{\text{BR}}$  = $\stackrel{\to }{\text{OR}}$  - $\stackrel{\to }{\text{OB}}$  ⇒ $\stackrel{\to }{\text{BR}}$  = $\stackrel{\to }{\text{r}}$  - $\stackrel{\to }{\text{b}}$  $\stackrel{\to }{\text{AR}}$  :$\stackrel{\to }{\text{BR}}$  = m:n ⇒ m ($\stackrel{\to }{\text{BR}}$ )= n ($\stackrel{\to }{\text{AR}}$ ) ⇒ m ($\stackrel{\to }{\text{r}}$  - $\stackrel{\to }{\text{b}}$ ) = n ($\stackrel{\to }{\text{r}}$  - $\stackrel{\to }{\text{a}}$  ) ⇒ m $\stackrel{\to }{\text{r}}$ - m $\stackrel{\to }{\text{b}}$ = n $\stackrel{\to }{\text{r}}$ - n $\stackrel{\to }{\text{a}}$ ⇒ m $\stackrel{\to }{\text{r}}$ - n $\stackrel{\to }{\text{r}}$= m $\stackrel{\to }{\text{b}}$ - n $\stackrel{\to }{\text{a}}$ ⇒ (m - n )$\stackrel{\to }{\text{r}}$ = m $\stackrel{\to }{\text{b}}$ - n $\stackrel{\to }{\text{a}}$ ⇒ $\stackrel{\to }{\text{r}}$ = $\frac{\text{m}\stackrel{\to }{\text{b}}\text{- n}\stackrel{\to }{\text{a}}}{\text{m-n}}$  ⇒ $\stackrel{\to }{\text{OR}}$ = $\frac{\text{m}\stackrel{\to }{\text{b}}\text{- n}\stackrel{\to }{\text{a}}}{\text{m-n}}$ .

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