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Question Number 193296 by MATHEMATICSAM last updated on 09/Jun/23

$$\mathrm{If}{a}^2+b^2+c^2=16,x^2+y^2+z^2=25$$$$\mathrm{and}ax+by+cz=20\mathrm{then}\mathrm{what}\mathrm{is}\mathrm{the}$$$$\mathrm{value}\mathrm{of}\frac{a+b+c}{x+y+z}?$$

Commented by kapoorshah last updated on 11/Jun/23

$$let\overrightarrow{u}=\left(\begin{array}{c}a\\ b\\ c\end{array}\right)and\overrightarrow{v}=\left(\begin{array}{c}x\\ y\\ z\end{array}\right)$$$$\mid \overrightarrow{u}\mid =4and\mid \overrightarrow{v}\mid =5$$$$\cos \theta =\frac{\overrightarrow{u}.\overrightarrow{v}}{\mid \overrightarrow{u}\mid \mid \overrightarrow{v}\mid }=1$$$$\overrightarrow{u}parallel\overrightarrow{v}\Rightarrow \frac{a+b+c}{x+y+z}=\mp \frac{\mid \overrightarrow{u}\mid }{\mid \overrightarrow{v}\mid }$$$$=\mp \frac{4}{5}$$

Answered by York12 last updated on 09/Jun/23

$$trivialyyoucanobtaina=0,b=0\Rightarrow c=\mp 4$$$$x=0,y=0\Rightarrow z=\mp 5$$$$whichsatifiesthethirdequation$$$$\Rightarrow$$$$\frac{a+b+c}{x+y+z}=\frac{\mp 4}{5}$$$$$$

Answered by York12 last updated on 09/Jun/23

$$$$$$(a^2+b^2+c^2)(x^2+y^2+z^2)={(ax+by+cz)}^2=400$$$$\Rightarrow (a^2+b^2+c^2)(x^2+y^2+z^2)-{(ax+by+cz)}^2=0$$$$\Rightarrow {(ay-bx)}^2+{(ax-cx)}^2+{(bz-cy)}^2=0$$$$\Rightarrow \frac{a}{x}=\frac{b}{y}=\frac{c}{z}=\frac{a+b+c}{x+y+z}=\lambda$$$$\Rightarrow a=x\lambda ,b=y\lambda ,c=z\lambda$$$$\Rightarrow {\lambda }^2\left(\underset{25}{\underbrace{x^2+y^2+z^2}}\right)=16\Rightarrow \lambda =\frac{\mp 4}{5}★$$$$$$$$$$

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