Owen Ramirez
01/05/2024 · Senior High School

40. Обчислити площу паралелограму побудованого на векторах \( \overrightarrow{\mathrm{a}}=\overrightarrow{\mathrm{p}}-3 \overrightarrow{\mathrm{q}}, \overrightarrow{\mathrm{b}}=\overrightarrow{\mathrm{p}}+2 \overrightarrow{\mathrm{q}} \), якщо \( |\overrightarrow{\mathrm{p}}|=1,|\overrightarrow{\mathrm{q}}|=5,(\overrightarrow{\mathrm{p}}, \overrightarrow{\mathrm{q}})=90^{\circ} \). 41. Дано вершини піраміди \( \mathrm{A}(4,-1,3) ; \mathrm{B}(-2,1,0) ; \mathrm{C}(0,-5,1), \mathrm{D}(3,2,-6) \). Знайти висоту,

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40. Обчислити площу паралелограму побудованого на векторах \( \overrightarrow{\mathrm{a}}=\overrightarrow{\mathrm{p}}-3 \overrightarrow{\mathrm{q}}, \overrightarrow{\mathrm{b}}=\overrightarrow{\mathrm{p}}+2 \overrightarrow{\mathrm{q}} \), якщо \( |\overrightarrow{\mathrm{p}}|=1,|\overrightarrow{\mathrm{q}}|=5,(\overrightarrow{\mathrm{p}}, \overrightarrow{\mathrm{q}})=90^{\circ} \). Рішення: Площа паралелограму, побудованого на векторах \( \overrightarrow{\mathrm{a}} \) і \( \overrightarrow{\mathrm{b}} \), обчислюється за формулою: \[ S = |\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}| \] Спочатку знайдемо векторний продукт \( \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}} \): \[ \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}} = (\overrightarrow{\mathrm{p}} - 3 \overrightarrow{\mathrm{q}}) \times (\overrightarrow{\mathrm{p}} + 2 \overrightarrow{\mathrm{q}}) \] Використовуючи правила векторного продукту: \[ \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}} = \overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{p}} + \overrightarrow{\mathrm{p}} \times 2 \overrightarrow{\mathrm{q}} - 3 \overrightarrow{\mathrm{q}} \times \overrightarrow{\mathrm{p}} - 3 \overrightarrow{\mathrm{q}} \times 2 \overrightarrow{\mathrm{q}} \] Оскільки \( \overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{p}} = \overrightarrow{0} \) і \( \overrightarrow{\mathrm{q}} \times \overrightarrow{\mathrm{q}} = \overrightarrow{0} \), то: \[ \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}} = \overrightarrow{\mathrm{p}} \times 2 \overrightarrow{\mathrm{q}} - 3 \overrightarrow{\mathrm{q}} \times \overrightarrow{\mathrm{p}} \] Оскільки \( \overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}} = -\overrightarrow{\mathrm{q}} \times \overrightarrow{\mathrm{p}} \), то: \[ \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}} = 2 \overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}} + 3 \overrightarrow{\mathrm{q}} \times \overrightarrow{\mathrm{p}} \] \[ \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}} = 5 \overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}} \] Тепер знайдемо модуль векторного продукту: \[ |\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}| = |5 \overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}| = 5 |\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}| \] Модуль векторного продукту \( \overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}} \) обчислюється за формулою: \[ |\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}| = |\overrightarrow{\mathrm{p}}| |\overrightarrow{\mathrm{q}}| \sin(\angle(\overrightarrow{\mathrm{p}}, \overrightarrow{\mathrm{q}})) \] Знаючи, що \( |\overrightarrow{\mathrm{p}}| = 1 \), \( |\overrightarrow{\mathrm{q}}| = 5 \), і \( \angle(\overrightarrow{\mathrm{p}}, \overrightarrow{\mathrm{q}}) = 90^{\circ} \), ми маємо: \[ |\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}| = 1 \cdot 5 \cdot \sin(90^{\circ}) = 5 \] Отже: \[ |\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}| = 5 \cdot 5 = 25 \] П

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Площа паралелограму дорівнює 25.
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