if-a-2b-3c-4d-5e-6f-0-find-the-maximum-of-a-b-c-d-e-f-a-2-b-2-c-2-d-2-e-2-f-2-

Question Number 182000 by mr W last updated on 03/Dec/22

i f a - 2 b + 3 c - 4 d + 5 e - 6 f = 0, f i n d

t h e m a x i m u m o f

\frac{∣ a + b + c + d + e + f ∣}{\sqrt{a^{2} + b^{2} + c^{2} + d^{2} + e^{2} + f^{2}}} .

Commented by mr W last updated on 04/Dec/22

A (0, 0, 0, 0, 0, 0)

B (1, - 2, 3, - 4, 5, - 6)

P (1, 1, 1, 1, 1, 1)

∣ A P ∣= \sqrt{1^{2} + 1^{2} + 1^{2} + 1^{2} + 1^{2} + 1^{2}} = \sqrt{6}

∣ A B ∣= \sqrt{1^{2} + {(- 2)}^{2} + 3^{2} + {(- 4)}^{2} + 5^{2} + {(- 6)}^{2}} = \sqrt{91}

\cos θ = \frac{1 - 2 + 3 - 4 + 5 - 6}{\sqrt{6} \times \sqrt{91}} = - \frac{3}{\sqrt{6} \times \sqrt{91}}

\sin θ = \frac{\sqrt{6 \times 91 - 3^{2}}}{\sqrt{6} \times \sqrt{91}} = \frac{\sqrt{537}}{\sqrt{6} \times \sqrt{91}}

∣ A P ∣ \sin θ = \sqrt{6} \times \frac{\sqrt{537}}{\sqrt{6} \times \sqrt{91}} = \sqrt{\frac{537}{91}}

\Rightarrow {(\frac{∣ a + b + c + d + e + f ∣}{\sqrt{a^{2} + b^{2} + c^{2} + d^{2} + e^{2} + f^{2}}})}_{m a x} = \sqrt{\frac{537}{91}}

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