Укажите последовательность положений позиции «Выдаваемые доверенности регистрируются в журнале уче та выданных

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scaier 30.05.2020 14:51

Решаю векторами.

Компоненты векторов $\overrightarrow{AB}$ , $\overrightarrow{AC}$ и $\overrightarrow{CB}$ :

$\overrightarrow{AB} = \textbf{r}_B - \textbf{r}_A = \{ -2; -2; 0 \}$

$\overrightarrow{AC} = \textbf{r}_C - \textbf{r}_A = \{ -2; 0; 2 \}$

$\overrightarrow{CB} = \textbf{r}_B - \textbf{r}_C = \{ 0; -2; -2 \}$

Нормы (модули, величины, длины) этих векторов:

$\left|\overrightarrow{AC}\right| = \sqrt{(-2)^2+0^2+2^2} = \sqrt{8}$

$\left|\overrightarrow{AB}\right| = \sqrt{(-2)^2+(-2)^2+0^2} = \sqrt{8}$

$\left|\overrightarrow{CB}\right| = \sqrt{0^2+(-2)^2+(-2)^2} = \sqrt{8}$

Угол между векторами $\overrightarrow{AC}$ и $\overrightarrow{AB}$ :

$cos \angle \left(\overrightarrow{AC}, \overrightarrow{AB}\right) = \frac{\overrightarrow{AC} \cdot \overrightarrow{AB}}{\left|\overrightarrow{AC}\right| \left|\overrightarrow{AB}\right|}$

Скалярное произведение

$\overrightarrow{AC} \cdot \overrightarrow{AB} = (-2) \cdot (-2) + 0 \cdot (-2) + 2 \cdot 0 = 4$

Имеем

$cos \angle \left(\overrightarrow{AC}, \overrightarrow{AB}\right) = \frac{4}{\sqrt{8} \cdot \sqrt{8}} = \frac{4}{8} = \frac{1}{2}$

В градусах:

$arccos \: \frac{1}{2} = 60^{\circ}$

Угол между векторами $\overrightarrow{AC}$ и $\overrightarrow{CB}$ :

$cos \angle \left(\overrightarrow{AC}, \overrightarrow{CB}\right) = \frac{\overrightarrow{AC} \cdot \overrightarrow{CB}}{\left|\overrightarrow{AC}\right| \left|\overrightarrow{CB}\right|}$

Скалярное произведение

$\overrightarrow{AC} \cdot \overrightarrow{CB} = (-2) \cdot 0 + 0 \cdot (-2) + 2 \cdot (-2) = -4$

Имеем

$cos \angle \left(\overrightarrow{AC}, \overrightarrow{CB}\right) = \frac{-4}{\sqrt{8} \cdot \sqrt{8}} = \frac{-4}{8} = -\frac{1}{2}$

В градусах:

$arccos \left(-\frac{1}{2}\right) = 120^{\circ}$

Нас интересует угол

$\angle \left(\overrightarrow{CA}, \overrightarrow{CB}\right) = 180^{\circ} - \angle \left(\overrightarrow{AC}, \overrightarrow{CB}\right) = 180^{\circ} - 120^{\circ} = 60^{\circ}$

Сумма углов будет

$60^{\circ} + 60^{\circ} = 120^{\circ}$

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