Предмет: Математика, автор: vbash911

знайти кут між прямими x-1/-3=y+2/2 та 2x-y+3=0​

Ответы

Автор ответа: pushpull
1

Ответ:

82.9°

Пошаговое объяснение:

Приведем уравнение прямых к одному виду, например, к уравнению с угловым коэффициентом

\displaystyle \frac{x-1}{-3} =\frac{y+2}{2}    умножим на 6, затем у влево, всё остальное вправо.

\displaystyle -2x+2=3y+6\\\\3y = -2x -4\\\\y=-\frac{2}{3} x-\frac{4}{3}

Второе уравнение просто перенесем всё, что при х и свободные члены вправо и умножим на -1

\displaystyle y=2x+3

Из уравнений прямых найдем их направляющие векторы

\displaystyle \vec a= \bigg\{ -2/3;1} \bigg\}\\ \\\\  \vec b =\bigg \{2;1\bigg \}

Формула для вычисления угла

\displaystyle cos \phi = \frac{|\vec a*\vec b|}{|a|*|b|}  =\frac{| a_x * b_x + a_y* b_y |}{\sqrt{a_x^2 + a_y^2} *\sqrt{b_x^2 + b_y^2}   }

подставим наши значения

\displaystyle cos \phi=\frac{| (-2/3)*2 + 1*1 |}{\sqrt{(-2/3)^2 + 1^2 }*\sqrt{2^2+1^2}  } =\frac{1/3}{\sqrt{13/9}*\sqrt{5}  } =\frac{\sqrt{65} }{65} \approx 0.124

\phi \approx 82.9^o

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