Considere o espaço vetorial V=\mathbb{R}^{2} e os vetores u=\left[\begin{array}{c}-1 \\\\ 3\end{array}\right], v=\left[\begin{array}{l}2 \\\\ 5\end{array}\right], w=\left[\begin{array}{c}-1 \\\\ 2\end{array}\right] \in \mathbb{R}^{2}, calcule \|2 u-v\| e d(u+v, v-w) nos seguintes casos :
\left\langle\left[\begin{array}{l}x_{1} \\\ x_{2}\end{array}\right],\[\begin{array}{l}y_{1} \\\ y_{2}\end{array}\right]\right\rangle=3 x_{1} y_{1}+2 x_{2} y_{2}]

Passo 1

O enunciado dá o espaço vetorial V=\mathbb{R}^{2} com produto interno
\langle (x_{1},x_{2}),(y_{1},y_{2})\rangle = 3x_{1}y_{1} + 2x_{2}y_{2},
e os vetores
u = \begin{pmatrix}-1\\\ 3\end{pmatrix},\quad v = \begin{pmatrix}2\\\ 5\end{pmatrix},\quad w = \begin{pmatrix}-1\\\ 2\end{pmatrix}.
Precisa-se calcular \|2u - v\| e a distância d(u+v,\,v-w), onde
\|x\| = \sqrt{\langle x,x\rangle}, \quad d(a,b)=\|a-b\|.

Passo 2

Calcular o vetor 2u - v:
2u = 2\begin{pmatrix}-1\\\ 3\end{pmatrix} = \begin{pmatrix}-2\\\ 6\end{pmatrix}, \quad 2u - v = \begin{pmatrix}-2\\\ 6\end{pmatrix} - \begin{pmatrix}2\\\ 5\end{pmatrix} = \begin{pmatrix}-4\\\ 1\end{pmatrix}.

Passo 3

Encontrar \|2u - v\| computando o produto interno de \begin{pmatrix}-4\\\ 1\end{pmatrix} consigo mesmo:
\big\langle \begin{pmatrix}-4\\\ 1\end{pmatrix}, \begin{pmatrix}-4\\\ 1\end{pmatrix}\big\rangle = 3\cdot(-4)\cdot(-4) + 2\cdot1\cdot1 = 3\cdot16 + 2\cdot1 = 48 + 2 = 50.
Logo
\|2u - v\| = \sqrt{50} = 5\sqrt{2}.

Passo 4

Calcular os vetores u+v e v-w:
u+v = \begin{pmatrix}-1\\\ 3\end{pmatrix} + \begin{pmatrix}2\\\ 5\end{pmatrix} = \begin{pmatrix}1\\\ 8\end{pmatrix}, \quad v-w = \begin{pmatrix}2\\\ 5\end{pmatrix} - \begin{pmatrix}-1\\\ 2\end{pmatrix} = \begin{pmatrix}3\\\ 3\end{pmatrix}.

Passo 5

Formar a diferença (u+v) - (v-w) e calcular sua norma:
(u+v) - (v-w) = \begin{pmatrix}1\\\ 8\end{pmatrix} - \begin{pmatrix}3\\\ 3\end{pmatrix} = \begin{pmatrix}-2\\\ 5\end{pmatrix}.
O produto interno de \begin{pmatrix}-2\\\ 5\end{pmatrix} por si mesmo é
\big\langle \begin{pmatrix}-2\\\ 5\end{pmatrix},\begin{pmatrix}-2\\\ 5\end{pmatrix}\big\rangle = 3\cdot(-2)\cdot(-2) + 2\cdot5\cdot5 = 3\cdot4 + 2\cdot25 = 12 + 50 = 62.
Portanto
d(u+v,\,v-w) = \sqrt{62}.

Resposta

\|2u - v\| = 5\sqrt{2}
d(u+v,\,v-w) = \sqrt{62}

Exercícios relacionados