2.5.5 Resolvas os vetores unitários cartesianos para suas componentes polares esféricas:
\begin{aligned} \hat{\mathbf{x}} & =\hat{\mathbf{r}} \operatorname{sen} \theta \cos \varphi+\hat{\boldsymbol{\theta}} \cos \theta \cos \varphi-\hat{\boldsymbol{\varphi}} \operatorname{sen} \varphi \\\ \hat{\mathbf{y}} & =\hat{\mathbf{r}} \operatorname{sen} \theta \operatorname{sen} \varphi+\hat{\boldsymbol{\theta}} \cos \theta \operatorname{sen} \varphi+\hat{\boldsymbol{\varphi}} \cos \varphi \\\ \hat{\mathbf{z}} & =\hat{\mathbf{r}} \cos \theta-\hat{\boldsymbol{\theta}} \operatorname{sen} \theta \end{aligned}

Passo 1

O problema pede para escrever os versores cartesianos \hat{\mathbf{x}},\hat{\mathbf{y}},\hat{\mathbf{z}} em função dos versores esféricos \hat{\mathbf{r}},\hat{\boldsymbol{\theta}},\hat{\boldsymbol{\varphi}}. O conteúdo é mudança de base em coordenadas esféricas, usando a ortogonalidade da transformação. Vamos precisar das fórmulas conhecidas dos versores esféricos em termos dos cartesianos.

Passo 2

Em coordenadas esféricas (r,\theta,\varphi) os versores esféricos são dados por
\begin{aligned} \hat{\mathbf{r}} &= \sin\theta\cos\varphi\,\hat{\mathbf{x}} \;+\;\sin\theta\sin\varphi\,\hat{\mathbf{y}}\;+\;\cos\theta\,\hat{\mathbf{z}},\\\ \hat{\boldsymbol{\theta}} &= \cos\theta\cos\varphi\,\hat{\mathbf{x}} \;+\;\cos\theta\sin\varphi\,\hat{\mathbf{y}}\;-\;\sin\theta\,\hat{\mathbf{z}},\\\ \hat{\boldsymbol{\varphi}} &= -\sin\varphi\,\hat{\mathbf{x}}\;+\;\cos\varphi\,\hat{\mathbf{y}}\,. \end{aligned}

Passo 3

Como a matriz de transformação entre bases é ortogonal, a inversa é a transposta. Logo, as componentes de \hat{\mathbf{x}} na base esférica são os produtos internos
\hat{\mathbf{x}}\cdot\hat{\mathbf{r}}=\sin\theta\cos\varphi,\quad \hat{\mathbf{x}}\cdot\hat{\boldsymbol{\theta}}=\cos\theta\cos\varphi,\quad \hat{\mathbf{x}}\cdot\hat{\boldsymbol{\varphi}}=-\sin\varphi.
Portanto
\hat{\mathbf{x}} =\sin\theta\cos\varphi\,\hat{\mathbf{r}} +\cos\theta\cos\varphi\,\hat{\boldsymbol{\theta}} -\sin\varphi\,\hat{\boldsymbol{\varphi}}.

Passo 4

De modo semelhante, para \hat{\mathbf{y}} calculamos
\hat{\mathbf{y}}\cdot\hat{\mathbf{r}}=\sin\theta\sin\varphi,\quad \hat{\mathbf{y}}\cdot\hat{\boldsymbol{\theta}}=\cos\theta\sin\varphi,\quad \hat{\mathbf{y}}\cdot\hat{\boldsymbol{\varphi}}=\cos\varphi,
levando a
\hat{\mathbf{y}} =\sin\theta\sin\varphi\,\hat{\mathbf{r}} +\cos\theta\sin\varphi\,\hat{\boldsymbol{\theta}} +\cos\varphi\,\hat{\boldsymbol{\varphi}}.

Passo 5

Finalmente, para \hat{\mathbf{z}} obtemos
\hat{\mathbf{z}}\cdot\hat{\mathbf{r}}=\cos\theta,\quad \hat{\mathbf{z}}\cdot\hat{\boldsymbol{\theta}}=-\sin\theta,\quad \hat{\mathbf{z}}\cdot\hat{\boldsymbol{\varphi}}=0,
e assim
\hat{\mathbf{z}} =\cos\theta\,\hat{\mathbf{r}} -\sin\theta\,\hat{\boldsymbol{\theta}}.

Resposta

\hat{\mathbf{x}} =\sin\theta\cos\varphi\,\hat{\mathbf{r}} +\cos\theta\cos\varphi\,\hat{\boldsymbol{\theta}} -\sin\varphi\,\hat{\boldsymbol{\varphi}},
\hat{\mathbf{y}} =\sin\theta\sin\varphi\,\hat{\mathbf{r}} +\cos\theta\sin\varphi\,\hat{\boldsymbol{\theta}} +\cos\varphi\,\hat{\boldsymbol{\varphi}},
\hat{\mathbf{z}} =\cos\theta\,\hat{\mathbf{r}} -\sin\theta\,\hat{\boldsymbol{\theta}}.

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