Choose one of the vertex of kite as origion Let the other vertices be Aa→, Bb→ and Cc→In a kite adjacent sides are equal so,OA=OC and BC=BAa→=c→ and BC→=BA→BC→=c→-b→ and BA→=a→-b→So, we hadBC→=BA→c→-b→=a→-b→c→-b→2=a→-b→2c→2+b→2-2 b→ c→=a→2+b→2-2 a→ b→We know a→=c→a→2+b→2-2 b→ c→=a→2+b→2-2 a→ b→-2 b→ c→=-2 a→ b→b→ c→= a→ b→Now doagonal vectors,OB→=b→AC→=a→-c→OA→ AC→=a→-c→ b→=a→ b→-b→ c→Since, a→ b→=b→ c→OA→ AC→=0OA→⊥AC→Hence, diaogonal OA and OC are ⊥

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Please solve this 12. "The diagonals of a kite are at right angles". prove the statement by vector method.

C h o o s e o n e o f t h e v e r t e x o f k i t e a s o r i g i o n . L e t t h e o t h e r v e r t i c e s b e A (\vec{a}), B (\vec{b}) a n d C (\vec{c}) I n a k i t e a d j a c e n t s i d e s a r e e q u a l s o, O A = O C a n d B C = B A |\vec{a}| = |\vec{c}| a n d |\vec{B C}| = |\vec{B A}| \vec{B C} = \vec{c} - \vec{b} a n d \vec{B A} = \vec{a} - \vec{b} S o, w e h a d |\vec{B C}| = |\vec{B A}| |\vec{c} - \vec{b}| = |\vec{a} - \vec{b}| {|\vec{c} - \vec{b}|}^{2} = {|\vec{a} - \vec{b}|}^{2} {|\vec{c}|}^{2} + {|\vec{b}|}^{2} - 2 \vec{b} . \vec{c} = {|\vec{a}|}^{2} + {|\vec{b}|}^{2} - 2 \vec{a} . \vec{b} W e k n o w |\vec{a}| = |\vec{c}| {|\vec{a}|}^{2} + {|\vec{b}|}^{2} - 2 \vec{b} . \vec{c} = {|\vec{a}|}^{2} + {|\vec{b}|}^{2} - 2 \vec{a} . \vec{b} - 2 \vec{b} . \vec{c} = - 2 \vec{a} . \vec{b} \vec{b} . \vec{c} = \vec{a} . \vec{b} N o w d o a g o n a l v e c t o r s, \vec{O B} = \vec{b} \vec{A C} = \vec{a} - \vec{c} \vec{O A} . \vec{A C} = (\vec{a} - \vec{c}) . \vec{b} = \vec{a} . \vec{b} - \vec{b} . \vec{c} S i n c e, \vec{a} . \vec{b} = \vec{b} . \vec{c} \vec{O A} . \vec{A C} = 0 \vec{O A} ⊥ \vec{A C} H e n c e, d i a o g o n a l O A a n d O C a r e ⊥

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Please solve this Please solve this 12. "The diagonals of a kite are at right angles". prove the statement by vector method.

Please solve this

Please solve this 12. "The diagonals of a kite are at right angles". prove the statement by vector method.