Advanced Vectors - Part 1

Geometry Level 3

Vector O A → , O B → , O C → \overrightarrow{OA},\ \overrightarrow{OB},\ \overrightarrow{OC} are on the same plane, ∣ O A → ∣ = 1 , ∣ O B → ∣ = 1 , ∣ O C → ∣ = 2 |\overrightarrow{OA}|=1,\ |\overrightarrow{OB}|=1,\ |\overrightarrow{OC}|=\sqrt{2} , the relative position is shown above in the picture.

t a n ( < O A → , O C → > ) = 7 , < O B → , O C → > = π 4 tan(<\overrightarrow{OA},\overrightarrow{OC}>)=7,\ <\overrightarrow{OB},\overrightarrow{OC}>\ =\dfrac{\pi}{4} .

If O C → = m O A → + n O B → , ( m , n ∈ R ) \overrightarrow{OC}=m\overrightarrow{OA}+n\overrightarrow{OB},\ (m,n \in \mathbb R) , Find m + n m+n .


The answer is 3.

Alice Smith by Alice Smith , 20, China

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1 solution

tan ⁡ α = 7 ⟹ sin ⁡ α = 7 5 2 , cos ⁡ α = 1 5 2 \tan α=7\implies \sin α=\dfrac{7}{5\sqrt 2}, \cos α=\dfrac{1}{5\sqrt 2}

The given conditions imply that

2 sin ⁡ ( 3 π 4 − α ) = m sin ⁡ π 4 = n sin ⁡ α ⟹ \dfrac{\sqrt 2 }{\sin (\frac{3π}{4}-α)}=\dfrac{m}{\sin \frac{π}{4}}=\dfrac{n}{\sin α}\implies

m = 5 4 , n = 7 4 ⟹ m + n = 3 m=\dfrac{5}{4}, n=\dfrac{7}{4}\implies m+n=\boxed 3 .

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