Let P be the point of intersection of the diagonal of a parallelogram ABCD and O is any point. If $\vec{O A} + \vec{O B} + \vec{O C} + \vec{O D} = λ \vec{O P}, Then λ =__________________.$

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Solution

Given:
P is the point of intersection of the diagonal of a parallelogram ABCD
$\vec{O A} + \vec{O B} + \vec{O C} + \vec{O D} = λ \vec{O P}$

$We know, \vec{O A} + \vec{A P} = \vec{O P} \vec{O B} + \vec{B P} = \vec{O P} \vec{O C} + \vec{C P} = \vec{O P} \vec{O D} + \vec{D P} = \vec{O P} Adding all, we get \vec{O A} + \vec{O B} + \vec{O C} + \vec{O D} + \vec{A P} + \vec{B P} + \vec{C P} + \vec{D P} = 4 \vec{O P} \Rightarrow \vec{O A} + \vec{O B} + \vec{O C} + \vec{O D} - \vec{C P} + \vec{B P} + \vec{C P} + \vec{D P} = 4 \vec{O P} (∵ \vec{A P} = \vec{P C} \Rightarrow \vec{A P} = - \vec{C P}) \Rightarrow \vec{O A} + \vec{O B} + \vec{O C} + \vec{O D} - \vec{D P} + \vec{D P} = 4 \vec{O P} (∵ \vec{B P} = \vec{P D} \Rightarrow \vec{B P} = - \vec{D P}) \Rightarrow \vec{O A} + \vec{O B} + \vec{O C} + \vec{O D} = 4 \vec{O P} \Rightarrow λ \vec{O P} = 4 \vec{O P} (given) \Rightarrow λ = 4$

Hence, $λ = 4 .$

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