Let-ABCD-be-a-parallelogram-whose-diagonals-intersect-at-P-and-ley-O-be-the-origin-then-OA-OB-OC-OD-equals-

Question Number 41291 by Mayur last updated on 04/Aug/18

Let A B C D be a parallelogram whose

diagonals intersect at P and ley O be

the origin, then \vec{O A} + \vec{O B} + \vec{O C} + \vec{O D}

equals

Answered by MJS last updated on 04/Aug/18

4 \overset{⇀}{O P}

Answered by tanmay.chaudhury50@gmail.com last updated on 05/Aug/18

O \vec{A} + A \vec{P} = O \vec{P}

O \vec{B} + B \vec{P} = O \vec{P}

O \vec{C} - P \vec{C} = O \vec{P}

O \vec{D} - P \vec{D} = O \vec{P}

n o w a d d t h e m

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

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

1 . (s i n c e O \vec{P} + P \vec{D} = O D)

2 . ∣ A \vec{P} ∣=∣ P \vec{C} ∣

3 . ∣ \vec{B P} ∣=∣ P \vec{D} ∣ d i o a g o n a l b i s e c t e a c h o t h e r

Commented by tanmay.chaudhury50@gmail.com last updated on 05/Aug/18