1. Encontre um vetor de norma 5 , mas com mesma direção e sentido de cada um dos vetores: (1,1,1); (-1,2,3) e \left(\frac{1}{3},-1,2\right).

Passo 1

O enunciado pede um vetor de norma 5 que seja paralelo e tenha mesmo sentido de um vetor dado \mathbf{v}. Para isso, usamos o fato de que, se \mathbf{v}\neq\mathbf{0}, então
\mathbf{w} = k\,\mathbf{v}
tem norma
\|\mathbf{w}\| = |k|\,\|\mathbf{v}\|.
Como queremos \|\mathbf{w}\|=5 e manter o mesmo sentido (isto é, k>0), basta escolher
k = \frac{5}{\|\mathbf{v}\|},
logo
\mathbf{w} = \frac{5}{\|\mathbf{v}\|}\,\mathbf{v}.

Passo 2

Para \mathbf{v}=(1,1,1), calculamos
\|\mathbf{v}\| = \sqrt{1^2+1^2+1^2} = \sqrt{3},
então
k = \frac{5}{\sqrt{3}} \quad\Longrightarrow\quad \mathbf{w} = \frac{5}{\sqrt{3}}\,(1,1,1) = \bigl(\tfrac{5}{\sqrt{3}},\;\tfrac{5}{\sqrt{3}},\;\tfrac{5}{\sqrt{3}}\bigr).

Passo 3

Para \mathbf{v}=(-1,2,3),
\|\mathbf{v}\| = \sqrt{(-1)^2+2^2+3^2} = \sqrt{14},
logo
k = \frac{5}{\sqrt{14}} \quad\Longrightarrow\quad \mathbf{w} = \frac{5}{\sqrt{14}}\,(-1,2,3) = \bigl(-\tfrac{5}{\sqrt{14}},\;\tfrac{10}{\sqrt{14}},\;\tfrac{15}{\sqrt{14}}\bigr).

Passo 4

Para \mathbf{v}=\bigl(\tfrac{1}{3},-1,2\bigr),
\|\mathbf{v}\| = \sqrt{\bigl(\tfrac{1}{3}\bigr)^{2}+(-1)^{2}+2^{2}} = \sqrt{\tfrac{1}{9}+1+4} = \sqrt{\tfrac{46}{9}} = \frac{\sqrt{46}}{3},
então
k = \frac{5}{\|\mathbf{v}\|} = \frac{5}{\frac{\sqrt{46}}{3}} = \frac{15}{\sqrt{46}} \quad\Longrightarrow\quad \mathbf{w} = \frac{15}{\sqrt{46}}\,\bigl(\tfrac{1}{3},-1,2\bigr) = \bigl(\tfrac{5}{\sqrt{46}},\;-\tfrac{15}{\sqrt{46}},\;\tfrac{30}{\sqrt{46}}\bigr).

Resposta

Para (1,1,1): \bigl(\tfrac{5}{\sqrt{3}},\;\tfrac{5}{\sqrt{3}},\;\tfrac{5}{\sqrt{3}}\bigr)
Para (-1,2,3): \bigl(-\tfrac{5}{\sqrt{14}},\;\tfrac{10}{\sqrt{14}},\;\tfrac{15}{\sqrt{14}}\bigr)
Para \bigl(\tfrac{1}{3},-1,2\bigr): \bigl(\tfrac{5}{\sqrt{46}},\;-\tfrac{15}{\sqrt{46}},\;\tfrac{30}{\sqrt{46}}\bigr)

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