\[\boxed{\mathbf{763.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[Дано:\ \]
\[\mathrm{\Delta}ABC;\]
\[AB = 6;\]
\[BC = 8;\ \]
\[\angle B = 90{^\circ}.\]
\[Решение.\]
\[\textbf{а)}\ \left| \overrightarrow{\text{BA}} \right| = 6;\ \ \ \left| \overrightarrow{\text{BC}} \right| = 8:\]
\[\left| \overrightarrow{\text{BA}} \right| - \left| \overrightarrow{\text{BC}} \right| = 6 - 8 = - 2;\]
\[\left| \overrightarrow{\text{BA}} - \overrightarrow{\text{BC}} \right| =\]
\[= \left| \overrightarrow{\text{CA}} \right|\ (по\ правилу\ треугольника);\]
\[CA = \sqrt{AB^{2} + BC^{2}} =\]
\[= \sqrt{36 + 64} = \sqrt{100} = 10;\ \]
\[\left| \overrightarrow{\text{BA}} - \overrightarrow{\text{BC}} \right| = 10.\]
\[\textbf{б)}\ \left| \overrightarrow{\text{AB}} \right| = \left| \overrightarrow{\text{BA}} \right| = 6;\ \ \ \left| \overrightarrow{\text{BC}} \right| = 8:\]
\[\left| \overrightarrow{\text{AB}} \right| + \left| \overrightarrow{\text{BC}} \right| = 6 + 8 = 14;\]
\[\left| \overrightarrow{\text{AB}} + \overrightarrow{\text{BC}} \right| =\]
\[= \left| \overrightarrow{\text{AC}} \right|\ (по\ правилу\ треугольника);\]
\[AC = \sqrt{AB^{2} + BC^{2}} =\]
\[= \sqrt{36 + 64} = \sqrt{100} = 10;\ \]
\[\left| \overrightarrow{\text{AB}} + \overrightarrow{\text{BC}} \right| = 10.\]
\[\textbf{в)}\ \left| \overrightarrow{\text{BA}} \right| = 6;\ \ \ \left| \overrightarrow{\text{BC}} \right| = 8:\]
\[\left| \overrightarrow{\text{BA}} \right| + \left| \overrightarrow{\text{BC}} \right| = 6 + 8 = 14;\]
\[BD = \sqrt{AB^{2} + AD^{2}} =\]
\[= \sqrt{36 + 64} = \sqrt{100} = 10;\]
\[\left| \overrightarrow{\text{BA}} + \overrightarrow{\text{BC}} \right| = 10.\]
\[\textbf{г)}\ \left| \overrightarrow{\text{AB}} \right| = \left| \overrightarrow{\text{BA}} \right| = 6;\ \ \ \left| \overrightarrow{\text{BC}} \right| = 8:\]
\[\left| \overrightarrow{\text{AB}} \right| - \left| \overrightarrow{\text{BC}} \right| = 6 - 8 = - 2;\]
\[\left| \overrightarrow{\text{AB}} - \overrightarrow{\text{BC}} \right| = \left| \overrightarrow{\text{DC}} + \overrightarrow{\text{CB}} \right| =\]
\[= |\overrightarrow{\text{DB}}|\ (по\ правилу\ треугольника);\]
\[DB = \sqrt{AB^{2} + AD^{2}} =\]
\[= \sqrt{36 + 64} = \sqrt{100} = 10;\]
\[\left| \overrightarrow{\text{AB}} - \overrightarrow{\text{BC}} \right| = 10.\ \]