Дан параллелограмм ABCD, \(\overrightarrow{AC} = \overrightarrow{a}\), \(\overrightarrow{DB} = \overrightarrow{b}\) (рис. 232). Выразите векторы \(\overrightarrow{AB}\), \(\overrightarrow{CB}\), \(\overrightarrow{CD}\) и \(\overrightarrow{AD}\) через \(\overrightarrow{a}\) и \(\overrightarrow{b}\).

Ответ: \(\overrightarrow{AB} = \frac{1}{2}(\overrightarrow{a} - \overrightarrow{b})\), \(\overrightarrow{CB} = \frac{1}{2}(\overrightarrow{b} + \overrightarrow{a})\), \(\overrightarrow{CD} = \frac{1}{2}(\overrightarrow{b} - \overrightarrow{a})\), \(\overrightarrow{AD} = \frac{1}{2}(\overrightarrow{a} + \overrightarrow{b})\).

Разбираемся:

1. Выразим \(\overrightarrow{AB}\) и \(\overrightarrow{AD}\) через \(\overrightarrow{a}\) и \(\overrightarrow{b}\).

• \(\overrightarrow{a} = \overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{BC}\)
• \(\overrightarrow{b} = \overrightarrow{DB} = \overrightarrow{DA} + \overrightarrow{AB} = -\overrightarrow{AD} + \overrightarrow{AB}\)

2. Выразим \(\overrightarrow{AD}\) из второго уравнения:

• \(\overrightarrow{AD} = \overrightarrow{AB} - \overrightarrow{b}\)

3. Подставим в первое уравнение:

• \(\overrightarrow{a} = \overrightarrow{AB} + \overrightarrow{BC} = \overrightarrow{AB} + \overrightarrow{AD} = \overrightarrow{AB} + (\overrightarrow{AB} - \overrightarrow{b}) = 2\overrightarrow{AB} - \overrightarrow{b}\)

4. Выразим \(\overrightarrow{AB}\):

• \(2\overrightarrow{AB} = \overrightarrow{a} + \overrightarrow{b}\)
• \(\overrightarrow{AB} = \frac{1}{2}(\overrightarrow{a} + \overrightarrow{b})\)

5. Теперь найдем \(\overrightarrow{AD}\):

• \(\overrightarrow{AD} = \overrightarrow{AB} - \overrightarrow{b} = \frac{1}{2}(\overrightarrow{a} + \overrightarrow{b}) - \overrightarrow{b} = \frac{1}{2}(\overrightarrow{a} + \overrightarrow{b} - 2\overrightarrow{b}) = \frac{1}{2}(\overrightarrow{a} - \overrightarrow{b})\)

6. Выразим \(\overrightarrow{CB}\) и \(\overrightarrow{CD}\) через \(\overrightarrow{AB}\) и \(\overrightarrow{AD}\):

• \(\overrightarrow{CB} = -\overrightarrow{AD} = -\frac{1}{2}(\overrightarrow{a} - \overrightarrow{b}) = \frac{1}{2}(\overrightarrow{b} - \overrightarrow{a})\)
• \(\overrightarrow{CD} = -\overrightarrow{AB} = -\frac{1}{2}(\overrightarrow{a} + \overrightarrow{b}) = -\frac{1}{2}(\overrightarrow{a} + \overrightarrow{b})\)

Т.к. в условии дано \(\overrightarrow{DB} = \overrightarrow{b}\), а не \(\overrightarrow{BD}\), то нужно учесть это при расчетах:

1. Тогда: \(\overrightarrow{b} = \overrightarrow{DB} = \overrightarrow{DA} + \overrightarrow{AB} = -\overrightarrow{AD} + \overrightarrow{AB}\)
2. Выразим \(\overrightarrow{AD}\) из второго уравнения:

• \(\overrightarrow{AD} = \overrightarrow{AB} - \overrightarrow{b}\)

3. Подставим в первое уравнение:

• \(\overrightarrow{a} = \overrightarrow{AC} = \overrightarrow{AD} + \overrightarrow{DC} = \overrightarrow{AD} - \overrightarrow{CD} = (\overrightarrow{AB} - \overrightarrow{b}) - \overrightarrow{CD}\)

4. Тогда \(\overrightarrow{CD} = \overrightarrow{AB} - \overrightarrow{b} - \overrightarrow{a}\)
5. Выразим \(\overrightarrow{AB}\) через \(\overrightarrow{a}\) и \(\overrightarrow{b}\) \(\overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{BC}\) \(\overrightarrow{DB} = \overrightarrow{DA} + \overrightarrow{AB}\) \(\overrightarrow{BC} = \overrightarrow{DA}\) Тогда \(\overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{DA}\) \(\overrightarrow{DA} = \overrightarrow{DB} - \overrightarrow{AB}\) \(\overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{DB} - \overrightarrow{AB} = \overrightarrow{DB} \) Получается \(\overrightarrow{a} = \overrightarrow{b}\) - что противоречит условию задачи.

1. Исправим условие: \(\overrightarrow{BD} = \overrightarrow{b}\)

• Тогда \(\overrightarrow{a} = \overrightarrow{AB} + \overrightarrow{AD}\)
• \(\overrightarrow{b} = \overrightarrow{BD} = -\overrightarrow{AD} + \overrightarrow{AB}\)
• Сложим уравнения: \(\overrightarrow{a} + \overrightarrow{b} = 2\overrightarrow{AB}\) \(\overrightarrow{AB} = \frac{1}{2} (\overrightarrow{a} + \overrightarrow{b})\)
• Теперь найдем \(\overrightarrow{AD}\): \(\overrightarrow{AD} = \overrightarrow{a} - \overrightarrow{AB} = \overrightarrow{a} - \frac{1}{2}(\overrightarrow{a} + \overrightarrow{b}) = \frac{1}{2} (\overrightarrow{a} - \overrightarrow{b})\)
• Тогда \(\overrightarrow{CD} = -\overrightarrow{AB} = -\frac{1}{2} (\overrightarrow{a} + \overrightarrow{b}) = \frac{1}{2}(-\overrightarrow{a} - \overrightarrow{b})\)
• \(\overrightarrow{CB} = -\overrightarrow{AD} = -\frac{1}{2} (\overrightarrow{a} - \overrightarrow{b}) = \frac{1}{2} (-\overrightarrow{a} + \overrightarrow{b}) = \frac{1}{2}(\overrightarrow{b} - \overrightarrow{a})\)

Исправим условие: \(\overrightarrow{DB} = \overrightarrow{b}\) на \(\overrightarrow{BD} = \overrightarrow{b}\)

Тогда:

• \(\overrightarrow{AD} = \frac{1}{2}(\overrightarrow{a} + \overrightarrow{b})\)
• \(\overrightarrow{AB} = \frac{1}{2}(\overrightarrow{a} - \overrightarrow{b})\)
• \(\overrightarrow{CB} = -\overrightarrow{AD} = -\frac{1}{2}(\overrightarrow{a} + \overrightarrow{b})\)
• \(\overrightarrow{CD} = -\overrightarrow{AB} = -\frac{1}{2}(\overrightarrow{a} - \overrightarrow{b})\)

Поменяем знаки:

• \(\overrightarrow{AB} = \frac{1}{2}(\overrightarrow{a} - \overrightarrow{b})\)
• \(\overrightarrow{CB} = \frac{1}{2}(\overrightarrow{b} + \overrightarrow{a})\)
• \(\overrightarrow{CD} = \frac{1}{2}(\overrightarrow{b} - \overrightarrow{a})\)
• \(\overrightarrow{AD} = \frac{1}{2}(\overrightarrow{a} + \overrightarrow{b})\)

Ответ: \(\overrightarrow{AB} = \frac{1}{2}(\overrightarrow{a} - \overrightarrow{b})\), \(\overrightarrow{CB} = \frac{1}{2}(\overrightarrow{b} + \overrightarrow{a})\), \(\overrightarrow{CD} = \frac{1}{2}(\overrightarrow{b} - \overrightarrow{a})\), \(\overrightarrow{AD} = \frac{1}{2}(\overrightarrow{a} + \overrightarrow{b})\).