{"id":425,"date":"2019-09-16T14:12:21","date_gmt":"2019-09-16T14:12:21","guid":{"rendered":"http:\/\/localhost\/?p=425"},"modified":"2019-09-16T14:18:49","modified_gmt":"2019-09-16T14:18:49","slug":"rotacoes-e-reflexoes","status":"publish","type":"post","link":"http:\/\/localhost\/index.php\/2019\/09\/16\/rotacoes-e-reflexoes\/","title":{"rendered":"Rota\u00e7\u00f5es e reflex\u00f5es"},"content":{"rendered":"

$\\newcommand{\\rot}[1]{\\mbox{Rot}(#1)}\\newcommand{\\refl}[1]{\\mbox{Ref}(#1)}\\newcommand{\\sen}[1]{\\mbox{sen}\\,#1}\\newcommand{\\R}{\\mathbb R}$<\/p>\n

Rota\u00e7\u00f5es<\/h3>\n
\n
\n

Seja $\\varphi$ um \u00e2ngulo. Denotaremos por $\\rot\\varphi$ a rota\u00e7\u00e3o do plano $\\R^2$ pela origem com \u00e2ngulo $\\varphi$ no sentido contr\u00e1rio aos ponteiros do rel\u00f3gio. Vamos calcular a forma matricial de $\\rot\\varphi$. \u00a0Seja $v=(v_1,v_2)\\in\\R^2$. Seja $\\alpha$ o \u00e2ngulo entre o vetor $v$ e o eixo $x$. Seja $v_0$ o vetor unit\u00e1rio na dire\u00e7\u00e3o de $v$. Ent\u00e3o $v$ pode ser escrito como
\n$$
\nv=\\|v\\|v_0=\\|v\\|(\\cos\\alpha,\\sen\\alpha).
\n$$<\/p>\n

Se $v’=v\\rot\\varphi$, ent\u00e3o
\n\\begin{align*}
\nv’=&\\|v’\\|(\\cos(\\alpha+\\varphi),\\sen{(\\alpha+\\varphi)})=\\\\&\\|v\\|(\\cos\\alpha\\cos\\varphi-\\sen\\alpha\\sen\\varphi,\\sen\\alpha\\cos\\varphi+\\cos\\alpha\\sen\\varphi)=\\\\&\\|v\\|(\\cos\\alpha,\\sen\\alpha)
\n\\begin{pmatrix}
\n\\cos\\varphi & \\sen\\varphi\\\\
\n-\\sen\\varphi & \\cos\\varphi
\n\\end{pmatrix}.
\n\\end{align*}
\nObtemos ent\u00e3o que a rota\u00e7\u00e3o $\\rot\\varphi$ \u00e9 uma transforma\u00e7\u00e3o linear com matriz
\n$$
\nA_\\varphi=\\begin{pmatrix}
\n\\cos\\varphi & \\sen\\varphi\\\\
\n-\\sen\\varphi & \\cos\\varphi
\n\\end{pmatrix}.
\n$$
\nComo $\\det A_\\varphi=\\cos^2\\varphi+\\sen^2\\varphi=1$, a transforma\u00e7\u00e3o $\\rot\\varphi$ \u00e9 invert\u00edvel e claramente $(\\rot\\varphi)^{-1}=\\rot{-\\varphi}$.<\/p>\n

\n

Lema.<\/strong> Temos que<\/p>\n

    \n
  1. $\\rot\\varphi\\rot\\psi=\\rot{\\varphi+\\psi}$;<\/li>\n
  2. $\\rot 0=\\mbox{id}$;<\/li>\n
  3. $(\\rot\\varphi)^{-1}=\\rot{-\\varphi}$.<\/li>\n<\/ol>\n<\/div>\n
    Demonstra\u00e7\u00e3o.<\/strong> Afirma\u00e7\u00e3o 2 \u00e9 clara e afirma\u00e7\u00e3o 3 segue do argumento acima. Provamos afirma\u00e7\u00e3o 1. Seja $v\\in\\R^2$,
    \n\\begin{align*}&v\\rot\\varphi\\rot\\psi=v
    \n\\begin{pmatrix}
    \n\\cos\\varphi & \\sen\\varphi\\\\
    \n-\\sen\\varphi & \\cos\\varphi
    \n\\end{pmatrix}
    \n\\begin{pmatrix}
    \n\\cos\\psi & \\sen\\psi\\\\
    \n-\\sen\\psi & \\cos\\psi
    \n\\end{pmatrix}
    \n=\\\\&v
    \n\\begin{pmatrix}
    \n\\cos\\varphi\\cos\\psi-\\sen\\varphi\\sen\\psi & \\cos\\varphi\\sen\\psi+\\sen\\varphi\\cos\\psi\\\\
    \n-\\cos\\varphi\\sen\\psi-\\sen\\varphi\\cos\\psi &\u00a0\\cos\\varphi\\cos\\psi-\\sen\\varphi\\sen\\psi
    \n\\end{pmatrix}=\\\\&
    \nv
    \n\\begin{pmatrix}
    \n\\cos(\\varphi+\\psi) & \\sen(\\varphi+\\psi)\\\\
    \n-\\sen(\\varphi+\\psi) &\\cos(\\varphi+\\psi)
    \n\\end{pmatrix}=v\\rot{\\varphi+\\psi}.
    \n\\end{align*}<\/div>\n

     <\/p>\n

    Reflex\u00f5es<\/h3>\n

    Consideramos a reflex\u00e3o pelo eixo que passe pela origem e tem \u00e2ngulo $\\varphi$ com o eixo $x$. Denotamos essa reflex\u00e3o por $\\refl\\varphi$. Dado $v=(v_1,v_2)\\in\\R^2$, seja $\\alpha$ o \u00e2ngulo de $v$ com o eixo $x$, e temos que
    \n\\begin{align*}
    \nv\\refl\\varphi=&v\\rot{-2(\\alpha-\\varphi)}=v\\rot{-2\\alpha}\\rot{2\\varphi}=\\\\&(v_1,-v_2)\\rot{2\\varphi}=\\\\&
    \n(v_1,v_2)
    \n\\begin{pmatrix}
    \n1 & 0 \\\\
    \n0 & -1
    \n\\end{pmatrix}
    \n\\begin{pmatrix}
    \n\\cos 2\\varphi & \\sen 2\\varphi\\\\
    \n-\\sen 2\\varphi & \\cos 2\\varphi
    \n\\end{pmatrix}=\\\\&
    \nv\\begin{pmatrix}
    \n\\cos 2\\varphi & \\sen 2\\varphi\\\\
    \n\\sen 2\\varphi & -\\cos 2\\varphi
    \n\\end{pmatrix}.
    \n\\end{align*}
    \nObtemos ent\u00e3o que $\\refl\\varphi$ \u00e9 uma transforma\u00e7\u00e3o linear com a matriz
    \n$$
    \nB_\\varphi=\\begin{pmatrix}
    \n\\cos 2\\varphi & \\sen 2\\varphi\\\\
    \n\\sen 2\\varphi & -\\cos 2\\varphi
    \n\\end{pmatrix}.
    \n$$
    \nComo $\\det B_\\varphi=-\\cos^22\\varphi-\\sen^22\\varphi=-1$, a transforma\u00e7\u00e3o $\\refl\\varphi$ \u00e9 invert\u00edvel. De fato,
    \n$$
    \n\\begin{pmatrix}
    \n\\cos 2\\varphi & \\sen 2\\varphi\\\\
    \n\\sen 2\\varphi & -\\cos 2\\varphi
    \n\\end{pmatrix}\\begin{pmatrix}\\cos 2\\varphi & \\sen 2\\varphi\\\\\\sen 2\\varphi & -\\cos 2\\varphi\\end{pmatrix}=\\begin{pmatrix}1 & 0 \\\\ 0 & 1 \\end{pmatrix},
    \n$$<\/p>\n

    que implica que $(\\refl\\varphi)^{-1}=\\refl\\varphi$.<\/p>\n

     <\/p>\n

    Composi\u00e7\u00e3o de reflex\u00f5es e rota\u00e7\u00f5es<\/h3>\n

    J\u00e1 vimos que a composi\u00e7\u00e3o de 2 rota\u00e7\u00f5es \u00e9 uma rota\u00e7\u00e3o. No entanto, a composi\u00e7\u00e3o de 2 reflex\u00f5es n\u00e3o \u00e9 geralmente uma reflex\u00e3o. De fato
    \n\\begin{align*}
    \n&v\\refl\\varphi\\refl\\psi
    \n=v\\begin{pmatrix}
    \n\\cos 2\\varphi & \\sen 2\\varphi\\\\
    \n\\sen 2\\varphi & -\\cos 2\\varphi
    \n\\end{pmatrix}\\begin{pmatrix}
    \n\\cos 2\\psi & \\sen 2\\psi\\\\
    \n\\sen 2\\psi & -\\cos 2\\psi
    \n\\end{pmatrix}=\\\\&
    \nv\\begin{pmatrix}
    \n\\cos 2\\varphi\\cos 2\\psi+\\sen 2\\varphi\\sen 2\\psi & \\cos 2\\varphi\\sen 2\\psi-\\sen 2\\varphi\\cos 2\\psi\\\\
    \n\\sen 2\\varphi\\cos 2\\psi-\\cos 2\\varphi\\sen 2\\psi & \\sen 2\\varphi\\sen 2\\psi+\\cos 2\\varphi\\cos 2\\psi\\end{pmatrix}=\\\\&
    \nv\\begin{pmatrix}
    \n\\cos 2(\\psi-\\varphi) & \\sen 2(\\psi-\\varphi)\\\\
    \n-\\sen 2(\\psi-\\varphi) &\u00a0\\cos 2(\\psi-\\varphi).
    \n\\end{pmatrix}=v\\rot{2(\\psi-\\varphi)}.
    \n\\end{align*}<\/p>\n

    Obtemos ent\u00e3o que a composi\u00e7\u00e3o de 2 reflex\u00f5es \u00e9 uma rota\u00e7\u00e3o.<\/p>\n

     <\/p>\n

    \n

    Lema.<\/strong> Sejam $\\varphi$ e $\\psi$ \u00e2ngulos. Ent\u00e3o<\/p>\n

      \n
    1. $\\refl\\varphi\\refl\\psi=\\rot{2(\\psi-\\varphi)}$<\/li>\n
    2. $\\rot\\varphi\\refl\\psi=\\refl{\\psi-\\varphi\/2}$;<\/li>\n
    3. $\\refl\\varphi\\rot\\psi=\\refl{\\varphi+\\psi\/2}$.<\/li>\n<\/ol>\n<\/div>\n
      Demonstra\u00e7\u00e3o.<\/strong> Afirma\u00e7\u00e3o 1. \u00e9 consequ\u00eancia do argumento acima. Demonstramos 2. Temos de 1. que $\\refl{\\vartheta}\\refl{\\psi}=\\rot{2(\\psi-\\vartheta)}$. Usando que
      \n$\\refl{\\psi}^{-1}=\\refl{\\psi}$, obtemos que $\\refl{\\vartheta}=\\rot{2(\\psi-\\vartheta)}\\refl{\\psi}$. Agora substitu\u00edmos $\\varphi=2(\\psi-\\vartheta)$ e $\\vartheta=\\psi-\\varphi\/2$ e obtemos 2. A demonstracao de 3. \u00e9 similar.<\/div>\n<\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"

      $\\newcommand{\\rot}[1]{\\mbox{Rot}(#1)}\\newcommand{\\refl}[1]{\\mbox{Ref}(#1)}\\newcommand{\\sen}[1]{\\mbox{sen}\\,#1}\\newcommand{\\R}{\\mathbb R}$ Rota\u00e7\u00f5es Seja $\\varphi$ um \u00e2ngulo. Denotaremos por $\\rot\\varphi$ a rota\u00e7\u00e3o do plano $\\R^2$ pela origem com \u00e2ngulo $\\varphi$ no sentido contr\u00e1rio aos ponteiros do rel\u00f3gio. Vamos calcular a forma matricial de $\\rot\\varphi$. \u00a0Seja $v=(v_1,v_2)\\in\\R^2$. Seja $\\alpha$ o \u00e2ngulo entre o vetor $v$ e o eixo $x$. Seja $v_0$ o vetor unit\u00e1rio na dire\u00e7\u00e3o … Continue reading Rota\u00e7\u00f5es e reflex\u00f5es<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":[],"categories":[1],"tags":[],"_links":{"self":[{"href":"http:\/\/localhost\/index.php\/wp-json\/wp\/v2\/posts\/425"}],"collection":[{"href":"http:\/\/localhost\/index.php\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"http:\/\/localhost\/index.php\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"http:\/\/localhost\/index.php\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"http:\/\/localhost\/index.php\/wp-json\/wp\/v2\/comments?post=425"}],"version-history":[{"count":11,"href":"http:\/\/localhost\/index.php\/wp-json\/wp\/v2\/posts\/425\/revisions"}],"predecessor-version":[{"id":436,"href":"http:\/\/localhost\/index.php\/wp-json\/wp\/v2\/posts\/425\/revisions\/436"}],"wp:attachment":[{"href":"http:\/\/localhost\/index.php\/wp-json\/wp\/v2\/media?parent=425"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"http:\/\/localhost\/index.php\/wp-json\/wp\/v2\/categories?post=425"},{"taxonomy":"post_tag","embeddable":true,"href":"http:\/\/localhost\/index.php\/wp-json\/wp\/v2\/tags?post=425"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}