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 1

Temos 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 2

Para \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 3

Para \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 4

Para \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).

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