Предмет: Математика, автор: KD637

Найти угол между прямыми 3х-2у+7=0 и 2х+3у-3=0. Срочно, с объяснением, пожалуйста​

Ответы

Автор ответа: zveryakovanata
2

Ответ: π/2

Пошаговое объяснение:

Угол φ между двумя прямыми, заданными общими уравнениями                                A₁x + B₁y + C₁ = 0  и A₂x + B₂y + C₂ = 0, вычисляется по формуле:

Cosφ =(A₁A₂+B₁B₂)/(√(A₁²+B₁²) · √(A₂²+B₂²)) ⇒

угол между прямыми 3x - 2y + 7 = 0 и 2x + 3y - 3 = 0

Cos φ =(3·2+(-2)·3) /(√(3²+(-2)²) ·√(2²+3²))= 0

Cos φ =0 ⇒  φ =π/2

⇒ острый угол между данными прямыми равен   φ =π/2

KD637: Спасибо, а что это за формула ? Cosφ =(A₁A₂+B₁B₂)/(√(A₁²+B₁²) · √(A₂²+B₂²))
zveryakovanata: Из учебника Аналитическая геометрия
KD637: Как она называется ? Точнее конкретно эта форма, ведь в учебнике она записана по другому
zveryakovanata: Из определения скалярного произведения и из выражения в координатах длин векторов. https://function-x.ru/line7.html
zveryakovanata: Пусть две прямые и заданы общими уравнениями (1)
A₁х+B₁у+С₁=0 и (2) A₂х+B₂у+С₂=0
Так как нормальным вектором (1) прямой является вектор а(A₁;B₁), а нормальным вектором прямой (2) является вектор b( A₂;B₂), то задача об определении угла между прямыми и сводится к определению угла между векторами a и b. Cosφ =(A₁A₂+B₁B₂)/(√(A₁²+B₁²) · √(A₂²+B₂²))
zveryakovanata: Понятно теперь?
KD637: Да, спасибо большое
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