Vector Fundamentals

Geometry Level pending

a → , b → \overrightarrow{a},\overrightarrow{b} are two vectors such that ∣ 2 a → − b → ∣ ≤ 3 |2\overrightarrow{a}-\overrightarrow{b}| \leq 3 , find the minimum value of ⌊ 1000 a → ⋅ b → ⌋ \lfloor 1000\overrightarrow{a} \cdot \overrightarrow{b} \rfloor .

Note: ∣ n → ∣ |\overrightarrow{n}| notes the length of the vector on the Euclidean plane. i.e. The Euclidean norm.

Extra: How many degrees of freedom does this system have?


The answer is -1125.

Alice Smith by Alice Smith , 20, China

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1 solution

From the given inequality we get a → ⋅ b → ≥ 4 a 2 + b 2 − 9 4 \overrightarrow a \cdot \overrightarrow b\geq \dfrac{4a^2+b^2-9}{4} . Also − a b ≤ a → ⋅ b → ≤ a b ⟹ 4 a 2 + b 2 − 9 4 ≤ a b ⟹ 4 a 2 − 4 a b + b 2 ≤ 9 ⟹ 2 a − b ≤ 3 ⟹ 2 a ≤ b + 3 ⟹ 4 a 2 + b 2 − 9 ≥ 2 b 2 − 6 b -ab\leq \overrightarrow a \cdot \overrightarrow b\leq ab\implies \dfrac{4a^2+b^2-9}{4}\leq ab\implies 4a^2-4ab+b^2\leq 9\implies 2a-b\leq 3\implies 2a\leq b+3\implies 4a^2+b^2-9\geq 2b^2-6b . Therefore 1000 a → ⋅ b → ≥ 250 ( 4 a 2 + b 2 − 9 ) ≥ 500 ( b 2 − 3 b ) = 500 [ ( b − 3 2 ) 2 − 9 4 ] ≥ − 1125 1000\overrightarrow a\cdot \overrightarrow b\geq 250(4a^2+b^2-9)\geq 500(b^2-3b)=500[(b-\dfrac{3}{2})^2-\dfrac{9}{4}]\geq \boxed {-1125}

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