In a parallelogram OABC, OA^⇁ =a^(−⇁) , OC^→ =c^→ , D is a point such that AD^→ :DB^→ =1:2 Express the following in terms of a and c (i)CB^→ (ii)BC^→ (iii)AB^→ (iv) AD^→ (v)OD^→ (vi)DC^→


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Question Number 72832 by necxxx last updated on 03/Nov/19

$I n a p a r a l l e l o g r a m O A B C, O \overset{⇁}{A} = \overset{- ⇁}{a},$ $O \vec{C} = \vec{c}, D i s a p o i n t s u c h t h a t A \vec{D} : D \vec{B} = 1 : 2$ $E x p r e s s t h e f o l l o w i n g i n t e r m s o f a a n d c$ $(i) C \vec{B} (i i) B \vec{C} (i i i) A \vec{B} (i v) A \vec{D} (v) O \vec{D}$ $(v i) D \vec{C}$
Commented by necxxx last updated on 03/Nov/19

$p l e a s e h e l p . T h a n k s i n a d v a n c e .$
	Answered by mind is power last updated on 03/Nov/19

	$C \vec{B} = O \vec{A} = \vec{a}$ $B \vec{C} = - C \vec{B} = - \vec{a}$ $A \vec{B} = O \vec{C} = \vec{c}$ $A \vec{D} = \frac{1}{2} . D \vec{B}$ $A \vec{D} + D \vec{B} = A \vec{B}$ $3 A \vec{D} = A \vec{B} \Rightarrow A \vec{D} = \frac{A \vec{B}}{3} = \frac{\vec{c}}{3}$ $O \vec{D} = O \vec{A} + A \vec{D} = \vec{a} + \frac{\vec{c}}{3}$
	Commented by necxxx last updated on 03/Nov/19

	$I n o w g e t i t . T h a n k y o u s o m u c h s i r$
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