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

$$if\mathit{a}-2\mathit{b}+3\mathit{c}-4\mathit{d}+5\mathit{e}-6\mathit{f}=0,find$$$$themaximumof$$$$\frac{\mid \mathit{a}+\mathit{b}+\mathit{c}+\mathit{d}+\mathit{e}+\mathit{f}\mid }{\sqrt{{\mathit{a}}^2+{\mathit{b}}^2+{\mathit{c}}^2+{\mathit{d}}^2+{\mathit{e}}^2+{\mathit{f}}^2}}.$$

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

$$A(0,0,0,0,0,0)$$$$B(1,-2,3,-4,5,-6)$$$$P(1,1,1,1,1,1)$$$$\mid AP\mid =\sqrt{1^2+1^2+1^2+1^2+1^2+1^2}=\sqrt{6}$$$$\mid AB\mid =\sqrt{1^2+{(-2)}^2+3^2+{(-4)}^2+5^2+{(-6)}^2}=\sqrt{91}$$$$\cos \theta =\frac{1-2+3-4+5-6}{\sqrt{6}\times \sqrt{91}}=-\frac{3}{\sqrt{6}\times \sqrt{91}}$$$$\sin \theta =\frac{\sqrt{6\times 91-3^2}}{\sqrt{6}\times \sqrt{91}}=\frac{\sqrt{537}}{\sqrt{6}\times \sqrt{91}}$$$$\mid AP\mid \sin \theta =\sqrt{6}\times \frac{\sqrt{537}}{\sqrt{6}\times \sqrt{91}}=\sqrt{\frac{537}{91}}$$$$\Rightarrow {\left(\frac{\mid \mathit{a}+\mathit{b}+\mathit{c}+\mathit{d}+\mathit{e}+\mathit{f}\mid }{\sqrt{{\mathit{a}}^2+{\mathit{b}}^2+{\mathit{c}}^2+{\mathit{d}}^2+{\mathit{e}}^2+{\mathit{f}}^2}}\right)}_{max}=\sqrt{\frac{537}{91}}$$

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