Question

Дополните векторы до ортонормированного базиса

Answer 1

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Answer 2

$\vec{x}=(\frac{1}{\sqrt3},\frac{1}{\sqrt3},\frac{1}{\sqrt3})\; ,\; \; \vec{y}=(\frac{1}{\sqrt{14}},\frac{2}{\sqrt{14}},-\frac{3}{\sqrt{14}})\\\\|\vec{x}|=\sqrt{\frac{1}{3}+\frac{1}{3}+\frac{1}{3}}=1\; ,\; \; |\vec{y}|+\sqrt{\frac{1}{14}+\frac{4}{14}+\frac{9}{14}}=1\\\\\vec{x}\cdot \vec{y}=\frac{1}{\sqrt{42}}+\frac{2}{\sqrt{42}}-\frac{3}{\sqrt{42}}=0\; \; \Rightarrow \; \; \vec{x}\perp \vec{y}\\\\\vec{x}\perp \vec{z}\; ,\; \; \vec{y}\perp \vec{z}\; ,\; \; z=(z_1,z_2,z_3)\\\\\vec{x}\cdot \vec{z}=\frac{z_1}{\sqrt3}+\frac{z_2}{\sqrt3}+\frac{z_3}{\sqrt3}=0\; \; \Rightarrow \; \; z_1+z_2+z_3=0$

$\vec{y}\cdot \vec{z}=\frac{z_1}{\sqrt{14}}+\frac{2z_2}{\sqrt{14}}-\frac{3z_3}{\sqrt{14}}=0\; \; \Rightarrow z_1+2z_2+2z_3=0$

$\left \{ {{z_1+z_2+z_3=0} \atop {z_1+2z_2-3z_3=0}} \right.\; \ominus \; \left \{ {{z_1+z_2+z_3=0} \atop {z_2-4z_3=0}} \right. \; \; \left \{ {{z_1=-z_2-z_3} \atop {z_2=4z_3}} \right.\; \left \{ {{z_1=-5z_3} \atop {z_2=4z_3}} \right.\; \Rightarrow \\\\z_3=t\; \; \to \; \; pyst\; \; t=-1\; ,\; \vec{z}=(5,-4,-1)\\\\|\vec{z}|=\sqrt{25+16+1}=42\\\\\vec{z}^\circ =(\frac{5}{\sqrt{42}},-\frac{4}{\sqrt{42}},-\frac{1}{\sqrt{42}})\\\\Otvet:\; \; \frac{4}{\sqrt{42}}.$