Determine os ângulos internos ao triângulo A B C, sendo A(1,-2,3), B(1,0,3) \text { e } C(2,3,-1) . Resp.: \alpha=\operatorname{arccos} \left(\frac{5}{\sqrt{42}}\right), \beta=\operatorname{arccos} \left(\frac{3}{\sqrt{26}}\right), \gamma=\operatorname{arccos} \left(\frac{16}{\sqrt{273}}\right). Passo 1Temos o triângulo com vértices A(1,-2,3),\quad B(1,0,3),\quad C(2,3,-1). Queremos os ângulos internos \alpha=\angle A, \beta=\angle B e \gamma=\angle C. Em \mathbb{R}^3, se dois vetores u e v partem de um mesmo ponto, o ângulo \theta entre eles satisfaz \cos\theta = \frac{u\cdot v}{\|u\|\;\|v\|}. Passo 2Para \alpha=\angle A, usamos \overrightarrow{AB}=B-A=(0,2,0), \quad \overrightarrow{AC}=C-A=(1,5,-4). Calculamos \overrightarrow{AB}\cdot\overrightarrow{AC}=0\cdot1+2\cdot5+0\cdot(-4)=10, \|\overrightarrow{AB}\|=2, \quad \|\overrightarrow{AC}\|=\sqrt{1^2+5^2+(-4)^2}=\sqrt{42}, logo \cos\alpha=\frac{10}{2\sqrt{42}}=\frac{5}{\sqrt{42}}, \alpha = \operatorname{arccos}\!\bigl(\tfrac{5}{\sqrt{42}}\bigr). Passo 3Para \beta=\angle B, definimos \overrightarrow{BA}=A-B=(0,-2,0), \quad \overrightarrow{BC}=C-B=(1,3,-4). Então \overrightarrow{BA}\cdot\overrightarrow{BC}=0\cdot1+(-2)\cdot3+0\cdot(-4)=-6, \|\overrightarrow{BA}\|=2, \quad \|\overrightarrow{BC}\|=\sqrt{1^2+3^2+(-4)^2}=\sqrt{26}, portanto \cos\beta=\frac{-6}{2\sqrt{26}}=-\frac{3}{\sqrt{26}}, \beta = \operatorname{arccos}\!\bigl(-\tfrac{3}{\sqrt{26}}\bigr). Passo 4Para \gamma=\angle C, tomamos \overrightarrow{CA}=A-C=(-1,-5,4), \quad \overrightarrow{CB}=B-C=(-1,-3,4). Calculamos \overrightarrow{CA}\cdot\overrightarrow{CB}=(-1)(-1)+(-5)(-3)+4\cdot4=32, \|\overrightarrow{CA}\|=\sqrt{(-1)^2+(-5)^2+4^2}=\sqrt{42}, \quad \|\overrightarrow{CB}\|=\sqrt{(-1)^2+(-3)^2+4^2}=\sqrt{26}, e daí \cos\gamma=\frac{32}{\sqrt{42}\,\sqrt{26}}=\frac{16}{\sqrt{273}}, \gamma = \operatorname{arccos}\!\bigl(\tfrac{16}{\sqrt{273}}\bigr). Resposta \displaystyle \alpha = \operatorname{arccos}\!\bigl(\tfrac{5}{\sqrt{42}}\bigr),\quad \beta = \operatorname{arccos}\!\bigl(-\tfrac{3}{\sqrt{26}}\bigr),\quad \gamma = \operatorname{arccos}\!\bigl(\tfrac{16}{\sqrt{273}}\bigr). MOSTRAR SOLUÇÃO COMPLETAExercícios relacionadosSabendo que o vetor \vec{v}=(1,-1,2) forma um ângulo de 60^{\circ} com o vetor \overrightarrow{A B} determinado pelos pontos A(1,-2,3) e B(0, m, 4), calcule o valor de m. Resp.: m=-3 / 2. Sabendo que o vetor \vec{v}=(1,-1,2) forma um ângulo de 60^{\circ} com o vetor \overrightarrow{A B} determinado pelos pontos A(1,-2,3) e B(0, m, 4), calcule o valor de m. Resp.: m=-3 / 2. Prove que o triângulo de vértices A(0,2,3), B(-2,3,5) e C(4,4,6) é um triângulo retângulo. O ângulo de 90^{\circ} está em que vértice?