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Question Number 194648 by universe last updated on 12/Jul/23

Commented by Frix last updated on 12/Jul/23

$$\frac{7}{8}$$

Commented by universe last updated on 12/Jul/23

$$solution?$$

Commented by Frix last updated on 12/Jul/23

$$a=b=c\wedge d=e=f$$

Answered by witcher3 last updated on 12/Jul/23

$$(\mathrm{ad}+\mathrm{be}+\mathrm{cf})⩽\sqrt{({\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2)({\mathrm{d}}^2+{\mathrm{e}}^2+{\mathrm{f}}^2)}$$$$\mathrm{cachy}\mathrm{shwartz}$$$$\mathrm{use}\mathrm{when}\mathrm{we}\mathrm{have}\mathrm{equality}$$

Answered by deleteduser1 last updated on 12/Jul/23

$$ad+be+cf⩽\sqrt{a^2+b^2+c^2}\sqrt{d^2+e^2+f^2}=56$$$$Equalityholdswhena=kd,b=ke,c=kf$$$$\Rightarrow \frac{a+b+c}{d+e+f}=k$$$$a^2+b^2+c^2=64=k^2(d^2+e^2+f^2)=49k^2$$$$\Rightarrow k=\frac{8}{7}=\frac{a+b+c}{d+e+f}\Rightarrow \frac{d+ef}{a+b+c}=\frac{7}{8}$$

Answered by mr W last updated on 14/Jul/23

$$\mathit{p}=(a,b,c)$$$$\mathit{q}=(d,e,f)$$$$\mid \mathit{p}\mid =\sqrt{a^2+b^2+c^2}=\sqrt{64}=8$$$$\mid \mathit{q}\mid =\sqrt{d^2+e^2+f^2}=\sqrt{49}=7$$$$ad+be+cf=\mathit{p}\cdot \mathit{q}=\mid \mathit{p}\mid \mid \mathit{q}\mid \cos \theta$$$$56=8\times 7\cos \theta$$$$\Rightarrow \cos \theta =1\Rightarrow \theta =0°\Rightarrow \mathit{p}//\mathit{q}$$$$\Rightarrow \frac{d}{a}=\frac{e}{b}=\frac{f}{c}=k=\pm \frac{\mid \mathit{q}\mid }{\mid \mathit{p}\mid }=\pm \frac{7}{8}$$$$\Rightarrow \frac{d+e+f}{a+b+c}=k=\pm \frac{7}{8}$$

Commented by deleteduser1 last updated on 15/Jul/23

$$Observethat{d}^2+e^2+f^2=k(ad+be+cf)=56k⩾0$$$$\Rightarrow kispositive$$

Commented by mr W last updated on 17/Jul/23

$$yes,youareright!thanks!$$

Commented by kapoorshah last updated on 17/Jul/23

$$nice$$

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