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Question Number 186701 by Mingma last updated on 08/Feb/23

Commented by aba last updated on 08/Feb/23

$$\frac{4}{5}$$

Commented by Mingma last updated on 09/Feb/23

can you show workings?

Answered by som(math1967) last updated on 09/Feb/23

$$(a^2+b^2+c^2)(x^2+y^2+z^2)-{(ax+by+cz)}^2$$$$=25\times 16-{20}^2$$$$a^2{x}^2+a^2{y}^2+a^2{z}^2+b^2{x}^2+b^2{y}^2+b^2{z}^2$$$$c^2{x}^2+c^2{y}^2+c^2{z}^2-a^2{x}^2-b^2{y}^2-c^2{z}^2$$$$-2abxy-2bcyz-2cazx=0$$$${(ay-bx)}^2+{(bz-cy)}^2+{(az-cx)}^2=0$$$$\Rightarrow (ay-bx)=0\Rightarrow \frac{x}{y}=\frac{a}{b}\Rightarrow \frac{x}{a}=\frac{y}{b}$$$$\Rightarrow bz-cy=0\Rightarrow \frac{y}{z}=\frac{b}{c}\Rightarrow \frac{y}{b}=\frac{z}{c}$$$$\therefore \frac{x}{a}=\frac{y}{b}=\frac{z}{c}=k\left(let\right)$$$$x=ak,y=bk,z=ck$$$$x^2+y^2+z^2=25$$$$k^2(a^2+b^2+c^2)=25$$$$k=\frac{5}{4}$$$$\frac{a+b+c}{x+y+z}=\frac{(a+b+c)}{k(a+b+c)}=\frac{1}{k}=\frac{4}{5}$$

Answered by mr W last updated on 09/Feb/23

$$vectormethod$$$$\mathit{p}=ai+bj+ck$$$$\mathit{q}=xi+yj+zk$$$$\mid \mathit{p}\mid =\sqrt{a^2+b^2+c^2}=\sqrt{16}=4$$$$\mid \mathit{q}\mid =\sqrt{x^2+y^2+z^2}=\sqrt{25}=5$$$$\mathit{p}\cdot \mathit{q}=ax+by+cz=20$$$$\mathit{p}\cdot \mathit{q}=\mid \mathit{p}\mid \mid \mathit{q}\mid \cos \theta =4\times 5\cos \theta =20\cos \theta$$$$20\cos \theta =20\Rightarrow \cos \theta =1\Rightarrow \theta =0°\Rightarrow \mathit{p}//\mathit{q}$$$$\Rightarrow \frac{a}{x}=\frac{b}{y}=\frac{c}{z}=\lambda ,say$$$$\lambda =\frac{\mid \mathit{p}\mid }{\mid \mathit{q}\mid }=\frac{4}{5}$$$$\frac{a+b+c}{x+y+z}=\frac{\lambda x+\lambda y+\lambda z}{x+y+z}=\lambda =\frac{4}{5}✓$$

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