Dados os vetores \vec{u}=(0,1,2), \vec{v}=(2,4,-6). Calcule:
(a) \|\vec{u}+\vec{v}\|
(b) \|\vec{u}-2 \vec{v}\|
(c) \left\|-\vec{u}+\frac{1}{2} \vec{v}\right\|
(d) \|\vec{u}-\vec{v}\|

Passo 1

O enunciado dá os vetores \vec{u}=(0,1,2) e \vec{v}=(2,4,-6) e pede várias normas de combinações lineares. Vamos usar a definição de norma euclidiana em \mathbb{R}^3: se \vec{w}=(w_1,w_2,w_3), então
\|\vec{w}\| = \sqrt{w_1^2 + w_2^2 + w_3^2}.

Passo 2

Calcular \vec{u} + \vec{v}:
\vec{u} + \vec{v} = (0+2,\;1+4,\;2+(-6)) = (2,5,-4).
Logo,
\|\vec{u}+\vec{v}\| = \sqrt{2^2 + 5^2 + (-4)^2} = \sqrt{4 + 25 + 16} = \sqrt{45} = 3\sqrt{5}.

Passo 3

Calcular \vec{u} - 2\vec{v}:
2\vec{v} = (4,8,-12), \quad \vec{u} - 2\vec{v} = (0-4,\;1-8,\;2-(-12)) = (-4,-7,14).
Logo,
\|\vec{u}-2\vec{v}\| = \sqrt{(-4)^2 + (-7)^2 + 14^2} = \sqrt{16 + 49 + 196} = \sqrt{261} = 3\sqrt{29}.

Passo 4

Calcular -\vec{u} + \tfrac12\vec{v}:
-\vec{u} = (0,-1,-2), \quad \tfrac12\vec{v} = (1,2,-3),
-\vec{u} + \tfrac12\vec{v} = (0+1,\;-1+2,\;-2+(-3)) = (1,1,-5).
Logo,
\Bigl\|-\vec{u}+\tfrac12\vec{v}\Bigr\| = \sqrt{1^2 + 1^2 + (-5)^2} = \sqrt{1 + 1 + 25} = \sqrt{27} = 3\sqrt{3}.

Passo 5

Calcular \vec{u} - \vec{v}:
\vec{u} - \vec{v} = (0-2,\;1-4,\;2-(-6)) = (-2,-3,8).
Logo,
\|\vec{u}-\vec{v}\| = \sqrt{(-2)^2 + (-3)^2 + 8^2} = \sqrt{4 + 9 + 64} = \sqrt{77}.

Resposta

(a) 3\sqrt{5}
(b) 3\sqrt{29}
(c) 3\sqrt{3}
(d) \sqrt{77}

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