$A B C D$ is a parallelogram and P is the point of intersection of the diagonals. If O is the origin, then $- - \to O A + - - \to O B + - - \to O C + - - \to O D$ is equal to

4 - - \to O P

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2 - - \to O P

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- - \to O P

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N u l l v e c t o r

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Solution

The correct option is A $4 - - \to O P$
$\to O A + \to O B + \to O C + \to O D$

$= (\to O P + \to P A) + (\to O P + \to P B) + (\to O P + \to P C) + (\to O P + \to P D)$

$= 4 \to O P + (\to P A + \to P D + \to P B + \to P C)$

$= 4 \to O P + (\to P A - \to P A + \to P B - \to P B)$

( $∵$ For a parallelogram $P$ bisects the two diagonal.)

$= 4 \to O P$

Hence, A is the correct option.

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Similar questions

\vec{O A} + \vec{O B} + \vec{O C} + \vec{O D} = 4 \vec{O P}

- - \to O A + - - \to O B + - - \to O C + - - \to O D

- - \to O A + - - \to O B + - - \to O C + - - \to O D

\vec{O A} + \vec{O B} + \vec{O C} + \vec{O D} = λ \vec{O P}, Then λ =__________________.

O \vec{A} + O \vec{B} + O \vec{C} + O \vec{D} =

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