Determine o ângulo entre os vetores.
\mathbf{a}=\langle 4,3\rangle, \quad \mathbf{b}=\langle 2,-1\rangle
\mathbf{a}=\langle-2,5\rangle, \quad \mathbf{b}=\langle 5,12\rangle
\mathbf{a}=\langle 3,-1,5\rangle, \quad \mathbf{b}=\langle-2,4,3\rangle
R: \mathbf{6 3}{ }^{\mathbf{\circ}}, ~ \mathbf{8 1}^{\mathbf{~}} \boldsymbol{}

Passo 1

O enunciado pede para encontrar o ângulo entre três pares de vetores em \mathbb{R}^2 e \mathbb{R}^3. Para isso vamos usar produto escalar, normas de vetores e a relação que envolve o cosseno do ângulo.

Passo 2

Definição: dados dois vetores \mathbf{a} e \mathbf{b},
\mathbf{a}\cdot\mathbf{b} = \|\mathbf{a}\|\,\|\mathbf{b}\|\cos(\theta).
Logo
\theta = \operatorname{arccos}\!\biggl(\frac{\mathbf{a}\cdot\mathbf{b}}{\|\mathbf{a}\|\,\|\mathbf{b}\|}\biggr).
A norma de um vetor \mathbf{v}=(v_1,v_2) ou (v_1,v_2,v_3) é
\|\mathbf{v}\| = \sqrt{v_1^2 + v_2^2} \quad\text{ou}\quad \sqrt{v_1^2 + v_2^2 + v_3^2}.

Passo 3

Para \mathbf{a}=\langle4,3\rangle, \mathbf{b}=\langle2,-1\rangle:
\mathbf{a}\cdot\mathbf{b} = 4\cdot2 + 3\cdot(-1) = 5.
\|\mathbf{a}\| = \sqrt{4^2+3^2} = 5,\quad \|\mathbf{b}\| = \sqrt{2^2+(-1)^2} = \sqrt{5}.
\cos(\theta_1) = \frac{5}{5\sqrt{5}} = \frac{1}{\sqrt{5}}, \qquad \theta_1 = \operatorname{arccos}\!\bigl(\tfrac{1}{\sqrt{5}}\bigr)\approx63{^\circ}.

Passo 4

Para \mathbf{a}=\langle-2,5\rangle, \mathbf{b}=\langle5,12\rangle:
\mathbf{a}\cdot\mathbf{b} = (-2)\cdot5 + 5\cdot12 = 50.
\|\mathbf{a}\| = \sqrt{(-2)^2+5^2} = \sqrt{29},\quad \|\mathbf{b}\| = \sqrt{5^2+12^2} = 13.
\cos(\theta_2) = \frac{50}{13\sqrt{29}}\approx0{,}7143, \qquad \theta_2 = \operatorname{arccos}(0{,}7143)\approx44{^\circ}.

Passo 5

Para \mathbf{a}=\langle3,-1,5\rangle, \mathbf{b}=\langle-2,4,3\rangle:
\mathbf{a}\cdot\mathbf{b} = 3\cdot(-2) + (-1)\cdot4 + 5\cdot3 = 5.
\|\mathbf{a}\| = \sqrt{3^2+(-1)^2+5^2} = \sqrt{35},\quad \|\mathbf{b}\| = \sqrt{(-2)^2+4^2+3^2} = \sqrt{29}.
\cos(\theta_3) = \frac{5}{\sqrt{35}\,\sqrt{29}}\approx0{,}157, \qquad \theta_3 = \operatorname{arccos}(0{,}157)\approx81{^\circ}.

Resposta

\theta_1\approx63{^\circ},\quad \theta_2\approx44{^\circ},\quad \theta_3\approx81{^\circ}.

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