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Section Formula

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Section Formula - Lesson Summary

The position vector for R in both these cases.

Case I: R divides AB internally.

   OA →   =  a →        

   OB →   =  b →     

   OR →   =  r →    

Point R divide line segment AB in the ratio m:n

AR:RB = m:n

In ∆OAR:

OR →    = OA →   +  AR →        

⇒ AR → = OR → - OA →

⇒ AR → = r → - a →

In ∆ORB:

OB → = OR → + RB →

⇒ RB → = OB → - OR →

⇒ RB → = b → - r →

AR → :RB → = m:n

m RB → = n AR →

m (b → - r → ) = n ( r →    - a →   )

m b → - m r → = n r → - n a →

m r → + n r → = m b → + n a →

⇒ (m + n )r → = m b → + n a →

⇒ r → =  m b → + n a → m+n  

⇒ OR → = m b → + n a → m+n  

The point R lies at the midpoint of line segment AB.

When m : n = 1:1

m = n

⇒ OR → =   m b → + n a → m+n   =    m b → + m a →  m+m  

⇒ OR → =       m ( a → + b → ) 2m          

⇒ OR → =   ( b → + a → ) 2   =  ( a → + b → ) 2       


Case II: R divides AB externally

   OA →   =  a →        

   OB →   =  b →     

   OR →   =  r →    

Let R divide AB externally such that AR:BR = m:n

In ∆OAR:

OR →   = OA →   + AR →  

⇒ AR →   = OR →   - OA →  

⇒ AR →   = r →   - a →  

In ∆OBR:

OR →   = OB →   + BR →  

⇒ BR →   = OR →   - OB →  

⇒ BR →   = r →   - b →  

AR →   :BR →   = m:n

m (BR →  )= n (AR →  )

m (r →   - b →  ) = n (r →   - a →   )

m r → - m b → = n r → - n a →

m r → - n r → = m b → - n a →

⇒ (m - n )r → = m b → - n a →

⇒ r → = m b → - n a → m-n  

⇒ OR → = m b → - n a → m-n  .

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