Даны координаты вершин пирамиды А1А2А3А4:..

Задача: Даны координаты вершин пирамиды A₁A₂A₃A₄: A₁A₂A₃A₄: A₁(7, 2, 4), A₂(7, −1, −2), A₃(3, 3, 1), A₄(−4, 2, 1). Найти угол между ребрами A₁A₂ и A₁A₄.

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$A_1A_2 = \sqrt{(7-7)^2+(-1-2)^2 + (-2-4)^2} = \sqrt{9+36} =\sqrt{45}\\A_2A_4 = \sqrt{(-4-7)^2+(2-(-1))^2 + (1-(-2))^2} = \sqrt{121+9+9} =\sqrt{139}\\A_1A_4 = \sqrt{(-4-7)^2+(2-2)^2 + (1-4)^2} = \sqrt{121+9} =\sqrt{130}\\\\$

$cos \angle A_2A_1A_4 = \frac{(A_1A_4)^2+(A_1A_2)^2-(A_2A_4)^2}{2\cdot (A_1A_4)\cdot (A_1A_2) }\\\\cos \angle A_2A_1A_4 = \frac{(\sqrt{130} )^2+(\sqrt{45})^2-(\sqrt{139})^2}{2\cdot \sqrt{130}\cdot \sqrt{45} } =\\\\= \frac{130+45-139}{2\sqrt{5\cdot 26} \cdot \sqrt{5\cdot 9}} =\frac{36}{30\sqrt{26}} =\frac{6}{5\sqrt{26}} \approx 0,235 \approx 76 \°$

Ответ: Угол между ребрами A₁A₂ и A₁A₄ равен $\approx$ 76°.