№ 11.44 Геометрія = № 22.44 Математика
$|\overrightarrow{a}|=2$, $|\overrightarrow{b}|=1$, $\angle(\overrightarrow{a},\overrightarrow{b})=45^{\circ}$. Знайдіть:
1) $\overrightarrow{a}\overrightarrow{b}$;
2) $\overrightarrow{a}(\overrightarrow{a}+\overrightarrow{b})$;
3) $\overrightarrow{b}(\overrightarrow{a}-\overrightarrow{b})$;
4) $\overrightarrow{a}(2\overrightarrow{a}+5\overrightarrow{b})$.
Розв’язок:
1) $\overrightarrow{a}\overrightarrow{b}=|\overrightarrow{a}|\cdot|\overrightarrow{b}|cos\angle(\overrightarrow{a},\overrightarrow{b})=$
$=2\cdot1\cdot cos45^{\circ}=2\cdot\frac{\sqrt{2}}{2}=\sqrt{2}$
2) $\overrightarrow{a}\left( \overrightarrow{a}+\overrightarrow{b} \right)={\overrightarrow{a}}^{2}+\overrightarrow{a}\overrightarrow{b}=$
$=|\overrightarrow{a}|^{2}+\overrightarrow{a}\overrightarrow{b}=$
$=2^{2}+\sqrt{2}=4+\sqrt{2}$
3) $\overrightarrow{b}\left( \overrightarrow{a}-\overrightarrow{b} \right)=\overrightarrow{b}\overrightarrow{a}-{\overrightarrow{b}}^{2}=$
$=|\overrightarrow{b}|\cdot|\overrightarrow{a}|cos\angle(\overrightarrow{a},\overrightarrow{b})-|\overrightarrow{b}|^{2}=$
$=2\cdot1\cdot\frac{\sqrt{2}}{2}-1^{2}=\sqrt{2}-1$
4) $\overrightarrow{a}\left( 2\overrightarrow{a}+5\overrightarrow{b} \right)=2{\overrightarrow{a}}^{2}+5\overrightarrow{a}\overrightarrow{b}=$
$=2|\overrightarrow{a}|^{2}+5\overrightarrow{a}\overrightarrow{b}=$
$=2\cdot2^{2}+5\cdot\sqrt{2}=8+5\sqrt{2}$