(2,0-0,4 cada)
Dois vetores são dados por \mathbf{a}=3,0 \mathbf{i}+5,0 \mathbf{j} e \mathbf{b}=2,0 \mathbf{i}+4,0 \mathbf{j}. Determine (a) \mathbf{a} \times \mathbf{b}, (b) \mathbf{a} \cdot \mathbf{b}, (c) (\mathbf{a}+\mathbf{b}) \cdot \mathbf{b} e (d) módulo de a e modulo de \mathbf{b}(\mathrm{e}) ângulo entre \mathbf{a} \mathbf{e} \mathbf{b}.

Passo 1

O enunciado dá dois vetores no plano
\mathbf{a}=3,0\,\mathbf{i}+5,0\,\mathbf{j}
\mathbf{b}=2,0\,\mathbf{i}+4,0\,\mathbf{j}
e pede:
(a) produto vetorial \mathbf{a}\times\mathbf{b}
(b) produto escalar \mathbf{a}\cdot\mathbf{b}
(c) (\mathbf{a}+\mathbf{b})\cdot\mathbf{b}
(d) módulos \|\mathbf{a}\| e \|\mathbf{b}\|
(e) ângulo entre \mathbf{a} e \mathbf{b}
Vamos usar as definições clássicas de produto vetorial em 2D (com componente k), produto escalar, norma e fórmula do cosseno do ângulo.

Passo 2

Para o produto vetorial em 2D definimos
\mathbf{a}\times\mathbf{b}=(a_xb_y - a_yb_x)\,\mathbf{k}.
Substituindo a_x=3,0, a_y=5,0, b_x=2,0, b_y=4,0:
a_xb_y - a_yb_x = 3,0\cdot4,0 - 5,0\cdot2,0 = 12,0 - 10,0 = 2,0.
Logo
\mathbf{a}\times\mathbf{b}=2,0\,\mathbf{k}.

Passo 3

O produto escalar é
\mathbf{a}\cdot\mathbf{b}=a_xb_x + a_yb_y.
Substituindo:
\mathbf{a}\cdot\mathbf{b}=3,0\cdot2,0 + 5,0\cdot4,0 = 6,0 + 20,0 = 26,0.

Passo 4

Primeiro somamos
\mathbf{a}+\mathbf{b}=(3,0+2,0)\,\mathbf{i} + (5,0+4,0)\,\mathbf{j} = 5,0\,\mathbf{i} + 9,0\,\mathbf{j}.
Em seguida
(\mathbf{a}+\mathbf{b})\cdot\mathbf{b} = 5,0\cdot2,0 + 9,0\cdot4,0 = 10,0 + 36,0 = 46,0.

Passo 5

A norma de um vetor \mathbf{v}=(v_x,v_y) é
\|\mathbf{v}\| = \sqrt{v_x^2 + v_y^2}.
Portanto
\|\mathbf{a}\| = \sqrt{3,0^2 + 5,0^2} = \sqrt{9,0 + 25,0} = \sqrt{34,0}\approx5,83,
\|\mathbf{b}\| = \sqrt{2,0^2 + 4,0^2} = \sqrt{4,0 + 16,0} = \sqrt{20,0}\approx4,47.

Passo 6

O ângulo \theta entre \mathbf{a} e \mathbf{b} satisfaz
\cos\theta = \frac{\mathbf{a}\cdot\mathbf{b}}{\|\mathbf{a}\|\|\mathbf{b}\|} = \frac{26,0}{\sqrt{34,0}\,\sqrt{20,0}} = \frac{26,0}{\sqrt{680,0}} = \frac{26,0}{26,076\dots} \approx0,9971.
Logo
\theta = \operatorname{arccos}(0,9971)\approx0,0768\text{ rad}\approx4,4^\circ.

Resposta

(a) \mathbf{a}\times\mathbf{b}=2,0\,\mathbf{k}
(b) \mathbf{a}\cdot\mathbf{b}=26,0
(c) (\mathbf{a}+\mathbf{b})\cdot\mathbf{b}=46,0
(d) \|\mathbf{a}\|=\sqrt{34,0}\approx5,83, \|\mathbf{b}\|=\sqrt{20,0}\approx4,47
(e) \theta\approx4,4^\circ

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