Beginner Vector Geometry #5

Geometry Level 2

There are two points A A and B B on a coordinate plane with an origin O O .

The midpoint of A B ‾ \overline{AB} is M M .

Given that O M → = ( 3 , 7 ) \overrightarrow{OM}=(3,~7) and O A → ⋅ O B → = 6 \overrightarrow{OA}\cdot\overrightarrow{OB}=6 , find the value of O A ‾ 2 + O B ‾ 2 \overline{OA}^2+\overline{OB}^2 .

This problem is a part of <Beginner Vector Geometry> series .


The answer is 220.

Boi (보이) by Boi (보이) , 19, South Korea

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2 solutions

Boi (보이)
Jul 2, 2017

Note that O A → + O B → = 2 O M → \overrightarrow{OA}+\overrightarrow{OB}=2\overrightarrow{OM} .

O A ‾ 2 + O B ‾ 2 = O A → 2 + O B → 2 + 2 O A → ⋅ O B → − 2 O A → ⋅ O B → = ( O A → + O B → ) 2 − 2 O A → ⋅ O B → = 4 O M → 2 − 12 = 4 O M ‾ 2 − 12 = 4 ( 3 2 + 7 2 ) − 12 = 220 \begin{aligned} \overline{OA}^2+\overline{OB}^2&=\overrightarrow{OA}^2+\overrightarrow{OB}^2+2\overrightarrow{OA}\cdot\overrightarrow{OB}-2\overrightarrow{OA}\cdot\overrightarrow{OB} \\ &=(\overrightarrow{OA}+\overrightarrow{OB})^2-2\overrightarrow{OA}\cdot\overrightarrow{OB} \\ &=4\overrightarrow{OM}^2-12 \\ &=4\overline{OM}^2-12 \\ &=4(3^2+7^2)-12 \\ &=\boxed{220} \end{aligned}

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Rab Gani
Jul 4, 2017

Let A(a,b), then B(6 – a,14 – b)
(OA) ⃗.(OB) ⃗=6 = 6a + 14b – (a2 + b2), (OA) ⃗^2+(OB) ⃗^2 = (a2 + b2) + [(6 – a)2+(14 – b)2] = 232 – 2[6a + 14b – (a2 + b2)] = 232 – 2[6] = 220,

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