Question

calcular o ângulo interno referente ao vértice A do triangulo de vértices A=(2,1,3), B(1,0-1), e C(-1,2,1)

Answer 1

O ângulo interno no vértice A será o ângulo entre os vetores AB e AC.

Vamos então começar determinando estes vetores.

$\vec{AB}~=~B~-~A\\\\\\\vec{AB}~=~(1,0,-1)~-~(2,1,3)\\\\\\\vec{AB}~=~(1-2~,~0-1~,\,-1-3)\\\\\\\boxed{\vec{AB}~=~(-1~,\,-1~,\,-4)}\\\\\\\\\vec{AC}~=~C~-~A\\\\\\\vec{AC}~=~(-1,2,1)~-~(2,1,3)\\\\\\\vec{AC}~=~(-1-2~,~2-1~,\,1-3)\\\\\\\boxed{\vec{AC}~=~(-3~,~1~,\,-2)}$

O ângulo entre dois vetores é dado por:

$cos(\theta)~=~\frac{\vec{AB}\,.\,\vec{AC}}{\left|\vec{AB}\right|\,.\,\left|\vec{AC}\right|}\\\\\\cos(\theta)~=~\frac{(-1~,\,-1~,\,-4)~.~(-3~,~1~,\,-2)}{\sqrt{(-1)^2+(-1)^2+(-4)^2}~.~\sqrt{(-3)^2+(1)^2+(-2)^2}}\\\\\\cos(\theta)~=~\frac{-1~.~(-3)~+\,(-1)~.~1~+\,(-4)~.~(-2)}{\sqrt{1+1+16}~.~\sqrt{9+1+4}}\\\\\\cos(\theta)~=~\frac{3~-~1~+~8}{\sqrt{18}~.~\sqrt{14}}\\\\\\cos(\theta)~=~\frac{10}{\sqrt{18~.~14}}\\\\\\cos(\theta)~=~\frac{10}{\sqrt{252}}$

$cos(\theta)~=~\frac{10}{\sqrt{2^2~.~3^2~.~7}}\\\\\\cos(\theta)~=~\frac{10}{2~.~3\sqrt{7}}\\\\\\cos(\theta)~=~\frac{5}{3\sqrt{7}}~.~\frac{\sqrt{7}}{\sqrt{7}}\\\\\\cos(\theta)~=~\frac{5\sqrt{7}}{21}\\\\\\\theta~=~arcos\left(\frac{5\sqrt{7}}{21}\right)\\\\\\\boxed{\theta~\approx~50,95^\circ}$