Exercício 13. Calcule \operatorname{proj}_{\vec{u}} \vec{v} nos seguintes casos:
\vec{v}=(1,2,3) e \vec{u}=\left(\frac{1}{3}, \frac{1}{5}, \frac{1}{7}\right).

Passo 1

O enunciado pede a projeção de \vec{v} sobre \vec{u}, que é um caso de projeção ortogonal em \mathbb{R}^3. Vamos usar a fórmula
\mathrm{proj}_{\vec{u}}\vec{v} = \frac{\vec{v}\!\cdot\!\vec{u}}{\vec{u}\!\cdot\!\vec{u}}\,\vec{u},
portanto precisamos calcular o produto interno \vec{v}\!\cdot\!\vec{u} e \vec{u}\!\cdot\!\vec{u}.

Passo 2

Calcular \vec{v}\!\cdot\!\vec{u} com \vec{v}=(1,2,3) e \vec{u}=\bigl(\tfrac{1}{3},\tfrac{1}{5},\tfrac{1}{7}\bigr):
\vec{v}\!\cdot\!\vec{u} =1\cdot\tfrac{1}{3}+2\cdot\tfrac{1}{5}+3\cdot\tfrac{1}{7} =\tfrac{1}{3}+\tfrac{2}{5}+\tfrac{3}{7} =\tfrac{122}{105}.

Passo 3

Calcular \vec{u}\!\cdot\!\vec{u}:
\vec{u}\!\cdot\!\vec{u} =\bigl(\tfrac{1}{3}\bigr)^{2}+\bigl(\tfrac{1}{5}\bigr)^{2}+\bigl(\tfrac{1}{7}\bigr)^{2} =\tfrac{1}{9}+\tfrac{1}{25}+\tfrac{1}{49} =\tfrac{1891}{11025}.

Passo 4

Obter o escalar \displaystyle\frac{\vec{v}\!\cdot\!\vec{u}}{\vec{u}\!\cdot\!\vec{u}}:
\frac{\tfrac{122}{105}}{\tfrac{1891}{11025}} =\frac{122}{105}\times\frac{11025}{1891} =\frac{12810}{1891} =\frac{210}{31}.

Passo 5

Multiplicar esse escalar por \vec{u}:
\mathrm{proj}_{\vec{u}}\vec{v} =\frac{210}{31}\,\bigl(\tfrac{1}{3},\tfrac{1}{5},\tfrac{1}{7}\bigr) =\Bigl(\tfrac{70}{31},\,\tfrac{42}{31},\,\tfrac{30}{31}\Bigr).

Resposta

\displaystyle \mathrm{proj}_{\vec{u}}\vec{v} =\Bigl(\tfrac{70}{31},\,\tfrac{42}{31},\,\tfrac{30}{31}\Bigr).

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