Доказателство за вектори в пространството

[Гост]

Докажете, че за всеки два вектора a,b в пространството е изпълнено:
[tex](\vec{a} х \vec{b})^{2}[/tex] + [tex](\vec{a} . \vec{b})^{2}[/tex]=[tex]\vec{a}^{2}[/tex] . [tex]\vec{b}^{2}[/tex]

[Sup3rlum]

$=(\vec{a} \times \vec{b})^2+(\vec{a} \cdot \vec{b})^2$
$=(\vec{a} \times \vec{b})\cdot(\vec{a} \times \vec{b})+(\vec{a} \cdot \vec{b})^2$
$=((\vec{a} \times \vec{b})\times\vec{a}) \cdot \vec{b} +(\vec{a} \cdot \vec{b})^2$
$=(-\vec{a}\times(\vec{a} \times \vec{b})) \cdot \vec{b} +(\vec{a} \cdot \vec{b})^2$
$=-((\vec{a}\cdot \vec{b})\vec{a} - (\vec{a}\cdot \vec{a})\vec{b}) \cdot \vec{b} +(\vec{a} \cdot \vec{b})^2$
$=\vec{a}^2\vec{b}^2-(\vec{a}\cdot\vec{b})^2+(\vec{a}\cdot\vec{b})^2$
$=\vec{a}^2\vec{b}^2$

[KOPMOPAH]

$$\left | \overrightarrow{a}\times \overrightarrow{b}\right |=\left | \overrightarrow{a} \right |\left | \overrightarrow{b} \right |\sin \alpha \Rightarrow \left | \overrightarrow{a}\times \overrightarrow{b}\right |^2=\left | \overrightarrow{a} \right |^2\left | \overrightarrow{b} \right |^2\sin^2 \alpha~~~~~(1) $$
$$\left | \overrightarrow{a}\cdot \overrightarrow{b}\right |=\left | \overrightarrow{a} \right |\left | \overrightarrow{b} \right |\cos \alpha \Rightarrow \left | \overrightarrow{a}\cdot \overrightarrow{b}\right |^2=\left | \overrightarrow{a} \right |^2\left | \overrightarrow{b} \right |^2\cos^2 \alpha~~~~~(2) $$
Събираме изразите $(1)$ и $(2)$ $$ \left | \overrightarrow{a}\times \overrightarrow{b}\right |^2+ \left | \overrightarrow{a}\cdot \overrightarrow{b}\right |^2=\left | \overrightarrow{a} \right |^2\left | \overrightarrow{b} \right |^2\sin^2 \alpha+\left | \overrightarrow{a} \right |^2\left | \overrightarrow{b} \right |^2\cos^2 \alpha= \\ =\left | \overrightarrow{a} \right |^2\left | \overrightarrow{b} \right |^2 \underbrace{\left(\sin^2 \alpha+\cos^2 \alpha\right)}_{=1} =\left | \overrightarrow{a} \right |^2\left | \overrightarrow{b} \right |^2 $$