{"id":34247,"date":"2022-03-29T13:00:29","date_gmt":"2022-03-29T13:00:29","guid":{"rendered":"https:\/\/toposuranos.com\/material\/?p=34247"},"modified":"2025-08-27T21:13:45","modified_gmt":"2025-08-27T21:13:45","slug":"algebra-e-projecoes-em-rn-produto-vetorial-em-r3","status":"publish","type":"post","link":"https:\/\/toposuranos.com\/material\/pt\/algebra-e-projecoes-em-rn-produto-vetorial-em-r3\/","title":{"rendered":"\u00c1lgebra e Proje\u00e7\u00f5es em Rn, Produto Vetorial em R3"},"content":{"rendered":"<style>\np, ul, ol{\ntext-align: justify;\n}\nh1{\ntext-align:center;\ntext-transform: uppercase;\n}\nh2{\ntext-align:center;\ntext-transform: uppercase;\nfont-size:24pt;\n}\nh3 { \n    text-align: center;\n    text-transform: uppercase;\n    font-size: 24px !important;\n}\n<\/style>\n<h1>\u00c1lgebra e Proje\u00e7\u00f5es em Rn, Produto Vetorial em <span class=\"katex-eq\" data-katex-display=\"false\">{\\mathbb{R}^3}<\/span><\/h1>\n<p style=\"text-align:center;\"><em><strong>Resumo:<\/strong><\/br>Esta s\u00e9rie \u00e9 a continua\u00e7\u00e3o direta da s\u00e9rie sobre o Espa\u00e7o Euclidiano de n dimens\u00f5es. Aqui revisaremos alguns conceitos de \u00e1lgebra linear que ajudam a compreender melhor o espa\u00e7o euclidiano n-dimensional, revisaremos os conceitos de proje\u00e7\u00f5es de um vetor sobre outro, demonstraremos o teorema de Pit\u00e1goras e concluiremos com uma revis\u00e3o do produto vetorial em <span class=\"katex-eq\" data-katex-display=\"false\">\\mathbb{R}^3<\/span> e sua rela\u00e7\u00e3o com os outros produtos do espa\u00e7o Euclidiano 3-dimensional. <\/p>\n<p style=\"text-align:center;\"><strong>\u00cdNDICE<\/strong><br \/>\n<a href=\"#Independencia-Lineal-Ortogonalidad-y-Proyecciones\">Independ\u00eancia Linear, Ortogonalidade e Proje\u00e7\u00f5es<\/a><br \/>\n<a href=\"#El-Teorema-de-Pitagoras-y-la-Proyecci\u00f3n-sobre-un-Subespacio\">O Teorema de Pit\u00e1goras e a Proje\u00e7\u00e3o sobre um Subespa\u00e7o<\/a><br \/>\n<a href=\"#El-Producto-Escalar-y-Vectorial-en-R3\">O Produto Escalar e Vetorial em <span class=\"katex-eq\" data-katex-display=\"false\">\\mathbb{R}^3<\/span><\/a>\n<\/p>\n<p><a name=\"Independencia-Lineal-Ortogonalidad-y-Proyecciones\"><\/a><br \/>\n<center><iframe class=\"lazyload\" width=\"560\" height=\"315\" data-src=\"https:\/\/www.youtube.com\/embed\/vtNHkaHD3aA\" title=\"YouTube video player\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture\" allowfullscreen><\/iframe><\/center><\/p>\n<h2>Independ\u00eancia Linear, Ortogonal e Proje\u00e7\u00f5es<\/h2>\n<h3>Combina\u00e7\u00e3o linear e independ\u00eancia linear<\/h3>\n<p><a href=\"https:\/\/www.youtube.com\/watch?v=vtNHkaHD3aA&#038;t=138s\" rel=\"noopener\" target=\"_blank\"><strong><span style=\"color: #ff0000;\">Um vetor n\u00e3o nulo<\/span><\/strong><\/a> <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{z}<\/span> pode ser constru\u00eddo como uma <strong>combina\u00e7\u00e3o linear<\/strong> em rela\u00e7\u00e3o a outros vetores n\u00e3o nulos <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{x}<\/span> e <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{y}<\/span> se existe um par de n\u00fameros reais <span class=\"katex-eq\" data-katex-display=\"false\">\\alpha<\/span> e <span class=\"katex-eq\" data-katex-display=\"false\">\\beta<\/span>, n\u00e3o ambos nulos simultaneamente, tais que:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\vec{z} = \\alpha \\vec{x} + \\beta\\vec{y}<\/span>\n<p>Ou seja, o vetor <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{z}<\/span> pode ser constru\u00eddo como uma soma ponderada dos vetores <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{x}<\/span> e <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{y}.<\/span>\n<p><a href=\"https:\/\/www.youtube.com\/watch?v=vtNHkaHD3aA&#038;t=609s\" rel=\"noopener\" target=\"_blank\"><strong><span style=\"color: #ff0000;\">De forma an\u00e1loga, diz-se<\/span><\/strong><\/a> que os vetores <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{x}<\/span> e <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{y}<\/span> s\u00e3o <strong>linearmente independentes<\/strong> se <\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">(\\alpha \\vec{x} + \\beta\\vec{y} = \\vec{0} ) \\longleftrightarrow (\\alpha=0 \\wedge \\beta=0 )<\/span>\n<p>A independ\u00eancia linear entre os vetores <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{x}<\/span> e <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{y}<\/span> nos diz que <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{y}<\/span> n\u00e3o pode ser obtido como um m\u00faltiplo escalar (n\u00e3o nulo) de <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{x}<\/span> nem vice-versa.<\/p>\n<p>O conceito de independ\u00eancia linear que acabamos de revisar pode ser estendido para conjuntos maiores de vetores. O conjunto de vetores n\u00e3o nulos <span class=\"katex-eq\" data-katex-display=\"false\">\\{\\vec{x}_1, \\cdots, \\vec{x}_n\\}<\/span> diz-se linearmente independente quando<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\"> \\displaystyle \\left[\\left(\\sum_{i=1}^n \\alpha_i \\vec{x}_i \\right) = \\vec{0} \\right] \\longleftrightarrow \\left[\\bigwedge_{i=1}^n (\\alpha_i = 0) \\right]<\/span>\n<h3>O \u00e2ngulo formado por dois vetores e a ortogonalidade<\/h3>\n<p><a href=\"https:\/\/www.youtube.com\/watch?v=vtNHkaHD3aA&#038;t=1289s\" rel=\"noopener\" target=\"_blank\"><strong><span style=\"color: #ff0000;\">Se recordarmos a desigualdade de Cauchy-Schwarz,<\/span><\/strong><\/a> esta nos diz que <span class=\"katex-eq\" data-katex-display=\"false\">(\\forall \\vec{x},\\vec{y}\\in\\mathbb{R}^n)(|\\vec{x}\\cdot\\vec{y}| \\leq \\|\\vec{x}\\| \\|\\vec{y}\\|).<\/span> Tendo isso em conta \u00e9 f\u00e1cil constatar que para qualquer par de vetores <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{x},\\vec{y}\\in\\mathbb{R}^n\\setminus\\{\\vec{0}\\}<\/span> cumpre-se a rela\u00e7\u00e3o:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle -1 \\leq \\frac{\\vec{x}\\cdot\\vec{y}}{\\|\\vec{x}\\|\\|\\vec{y}\\|}\\leq 1<\/span>\n<p>Agora podemos intuir uma rela\u00e7\u00e3o entre o produto escalar e o \u00e2ngulo formado pelos vetores <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{x}<\/span> e <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{y}<\/span>, porque estes geram um plano isom\u00e9trico a <span class=\"katex-eq\" data-katex-display=\"false\">\\mathbb{R}^2<\/span>. Por isso, sem perda de generalidade, podemos imagin\u00e1-los como sendo elementos de <span class=\"katex-eq\" data-katex-display=\"false\">\\mathbb{R}^2<\/span> com \u00e2ngulos em rela\u00e7\u00e3o ao eixo <span class=\"katex-eq\" data-katex-display=\"false\">\\hat{x}<\/span> de <span class=\"katex-eq\" data-katex-display=\"false\">\\theta_x<\/span> e <span class=\"katex-eq\" data-katex-display=\"false\">\\theta_y,<\/span> respectivamente, de modo que os vetores ficam escritos em forma polar como:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}{rl}\n\n\\vec{x} &amp;= \\|\\vec{x}\\|(\\cos(\\theta_x) , \\sin(\\theta_x)) \\\\ \\\\ \\vec{y} &amp;= \\|\\vec{y}\\|(\\cos(\\theta_y) , \\sin(\\theta_y))\n\n\\end{array}<\/span>\n<p>Assim podemos supor (sem perda de generalidade, outra vez) que <span class=\"katex-eq\" data-katex-display=\"false\">\\theta_x \\lt \\theta_y,<\/span> para ent\u00e3o calcular o produto escalar <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{x}\\cdot\\vec{y}.<\/span> Fazendo isso teremos o seguinte resultado:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}{rl}\\vec{x}\\cdot \\vec{y} &amp;=  \\|\\vec{x}\\|  \\|\\vec{y}\\| (\\cos(\\theta_x)\\cos(\\theta_y) + \\sin(\\theta_x)\\sin(\\theta_y)) \\\\ \\\\ &amp;=  \\|\\vec{x}\\|  \\|\\vec{y}\\| \\cos(\\theta_y-\\theta_x)\n\n\\end{array}<\/span>\n<p>Agora, tomando a diferen\u00e7a entre a posi\u00e7\u00e3o angular maior e a menor obtemos o \u00e2ngulo compreendido entre os vetores, <span class=\"katex-eq\" data-katex-display=\"false\">\\angle(\\vec{x},\\vec{y})=\\theta_y - \\theta_x.<\/span> E com isso agora podemos escrever:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\n\\displaystyle \\cos\\left(\\angle(\\vec{x},\\vec{y}) \\right) = \\frac{\\vec{x} \\cdot \\vec{y}}{\\|\\vec{x}\\|\\|\\vec{y}\\|}\n\n<\/span>\n<p>Aqui devemos destacar que <span class=\"katex-eq\" data-katex-display=\"false\">\\angle(\\vec{x},\\vec{y})\\in [0, \\pi]<\/span>\n<p>A partir disso podemos conectar a desigualdade de Cauchy-Schwarz com a geometria dos \u00e2ngulos, e al\u00e9m disso nos permite obter uma no\u00e7\u00e3o rigorosa de ortogonalidade. Dois vetores dizem-se <strong>Ortogonais<\/strong> quando sustentam entre si um \u00e2ngulo de <span class=\"katex-eq\" data-katex-display=\"false\">\\pi\/2<\/span> radianos, no sentido explicado no par\u00e1grafo anterior. Isto \u00e9 equivalente a dizer que <span class=\"katex-eq\" data-katex-display=\"false\">\\cos\\left(\\angle(\\vec{x},\\vec{y})\\right) = 0,<\/span> o que por sua vez \u00e9 equivalente a dizer que <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{x}\\cdot\\vec{y} = 0.<\/span> Por este motivo, afirmar a ortogonalidade dos vetores <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{x}<\/span> e <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{y}<\/span> \u00e9 equivalente a dizer que <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{x}\\cdot\\vec{y}=0.<\/span>\n<h4>Se dois vetores n\u00e3o nulos s\u00e3o ortogonais, ent\u00e3o s\u00e3o linearmente independentes<\/h4>\n<p><a href=\"https:\/\/www.youtube.com\/watch?v=vtNHkaHD3aA&#038;t=2365s\" rel=\"noopener\" target=\"_blank\"><strong><span style=\"color: #ff0000;\">Esta \u00e9 uma propriedade algo intuitiva dos vetores<\/span><\/strong><\/a> de <span class=\"katex-eq\" data-katex-display=\"false\">\\mathbb{R}^n<\/span> cuja demonstra\u00e7\u00e3o formal n\u00e3o \u00e9 t\u00e3o direta, e tamb\u00e9m \u00e9 uma propriedade que em ocasi\u00f5es pode gerar certa confus\u00e3o: A ortogonalidade de dois vetores implica a independ\u00eancia linear entre eles, mas a independ\u00eancia linear entre dois vetores n\u00e3o necessariamente implica sua ortogonalidade. Para ver isto \u00faltimo basta um simples contraexemplo:<\/p>\n<p>Se tomarmos os vetores <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{A}=(1,0)<\/span> e <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{B}=(1,1),<\/span> que claramente n\u00e3o s\u00e3o ortogonais porque <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{A}\\cdot\\vec{B}=1,<\/span> veremos que se fizermos<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\n\\alpha\\vec{A} + \\beta\\vec{B} = \\vec{0}\n\n<\/span>\n<p>Ent\u00e3o tem-se que<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}{rl}\n\n\\alpha + \\beta &amp;= 0 \\\\ \\beta &amp;= 0\n\n\\end{array}<\/span>\n<p>e portanto: <span class=\"katex-eq\" data-katex-display=\"false\">\\alpha = 0  \\wedge \\beta=0.<\/span> E com isso conclui-se que:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\n\\alpha\\vec{A} + \\beta\\vec{B} = \\vec{0} \\longleftrightarrow  \\alpha = 0  \\wedge \\beta=0\n\n<\/span>\n<p>O que \u00e9 equivalente a dizer que <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{A}<\/span> e <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{B}<\/span> s\u00e3o linearmente independentes. Com isso fica claro de forma muito expl\u00edcita que n\u00e3o \u00e9 verdade que a independ\u00eancia linear implique ortogonalidade. No entanto, a ortogonalidade sim implica a independ\u00eancia linear e \u00e9 o que demonstrarei formalmente a seguir, e para isso consideremos o seguinte conjunto de premissas:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\n\\mathcal{H}= \\{\\vec{x},\\vec{y}\\in \\mathbb{R}^n\\setminus\\{\\vec{0}\\}, \\vec{x}\\cdot\\vec{y}=0, \\alpha\\vec{x}+\\beta\\vec{y} = \\vec{0}\\}<\/span>\n<p>A partir disso podemos produzir o seguinte racioc\u00ednio:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}{rll}\n\n(1) &amp;\\mathcal{H}\\vdash \\vec{x},\\vec{y}\\in \\mathbb{R}^n\\setminus\\{\\vec{0}\\} &amp;{;\\;Presun\\c{c}\u00e3o}\\\\ \\\\\n\n(2) &amp;\\mathcal{H}\\vdash \\vec{x}\\cdot\\vec{y}=0 &amp;{\\;Presun\\c{c}\u00e3o} \\\\ \\\\\n\n(3) &amp;\\mathcal{H}\\vdash \\alpha\\vec{x} + \\beta\\vec{y} = \\vec{0} &amp;{\\;Presun\\c{c}\u00e3o} \\\\ \\\\\n\n(4) &amp;\\mathcal{H}\\vdash (\\alpha\\vec{x} + \\beta\\vec{y})\\cdot\\vec{x} = \\alpha\\|\\vec{x}\\|^2 + \\beta(\\vec{x}\\cdot\\vec{y}) &amp;{;\\; Bilinearidade} \\\\ \\\\\n\n(5) &amp;\\mathcal{H}\\vdash  \\alpha\\|\\vec{x}\\|^2 = 0 &amp; {;\\; De(2,3,4)} \\\\ \\\\\n\n(6) &amp;\\mathcal{H}\\vdash  \\alpha  = 0 &amp; {;\\; De(1,5)} \\\\ \\\\\n\n(7) &amp;\\mathcal{H}\\vdash (\\alpha\\vec{x} + \\beta\\vec{y})\\cdot\\vec{y} = \\alpha(\\vec{x}\\cdot\\vec{y}) + \\beta\\|\\vec{y}\\|^2 &amp; {;\\;Bilinearidade} \\\\ \\\\\n\n(8) &amp;\\mathcal{H}\\vdash \\beta\\|\\vec{y}\\|^2 = 0 &amp;{;\\;De(2,3,7)} \\\\ \\\\\n\n(9) &amp;\\mathcal{H}\\vdash \\beta = 0 &amp;{;\\;De(1,8)} \\\\ \\\\\n\n(10) &amp;\\mathcal{H}\\vdash \\alpha= 0 \\wedge \\beta = 0 &amp;{;\\;\\wedge-int(6,9)}\n\n\\end{array}<\/span>\n<p>Com isso conclu\u00edmos que<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\n\\{\\vec{x},\\vec{y}\\in \\mathbb{R}^n\\setminus\\{\\vec{0}\\}, \\vec{x}\\cdot\\vec{y}=0, \\alpha\\vec{x}+\\beta\\vec{y} = \\vec{0}\\} \\vdash \\alpha= 0 \\wedge \\beta = 0  <\/span>\n<p>Finalmente, aplicando o teorema da dedu\u00e7\u00e3o sobre esta \u00faltima express\u00e3o tem-se:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\{\\vec{x},\\vec{y}\\in \\mathbb{R}^n\\setminus\\{\\vec{0}\\}, \\vec{x}\\cdot\\vec{y}=0\\} \\vdash (\\alpha\\vec{x}+\\beta\\vec{y} = \\vec{0}) \\rightarrow (\\alpha= 0 \\wedge \\beta = 0)<\/span>\n<p>A prova com a qual se obt\u00e9m a seta na dire\u00e7\u00e3o contr\u00e1ria \u00e9 trivial.<\/p>\n<p>Ou seja: se <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{x}<\/span> e <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{y}<\/span> s\u00e3o vetores n\u00e3o nulos e ortogonais, ent\u00e3o s\u00e3o linearmente independentes.<\/p>\n<h3>A proje\u00e7\u00e3o de um vetor sobre outro<\/h3>\n<p><a href=\"https:\/\/www.youtube.com\/watch?v=vtNHkaHD3aA&#038;t=3055s\" rel=\"noopener\" target=\"_blank\"><strong><span style=\"color: #ff0000;\">Suponhamos que temos dois vetores n\u00e3o nulos<\/span><\/strong><\/a> <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{x}<\/span> e <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{y}<\/span> que sustentam entre si um \u00e2ngulo <span class=\"katex-eq\" data-katex-display=\"false\">\\angle(\\vec{x},\\vec{y})<\/span> e nos perguntamos \u00abEm que quantidade o vetor <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{x}<\/span> se encontra sobre o vetor <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{y}<\/span>?\u00bb ou \u00abDe que tamanho \u00e9 a sombra do vetor <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{x}<\/span> quando se projeta sobre a dire\u00e7\u00e3o do vetor <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{y}<\/span>?\u00bb. Esta pergunta podemos resolver atrav\u00e9s da trigonometria, e com isso definir a proje\u00e7\u00e3o de um vetor <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{x}<\/span> sobre outro <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{y},<\/span> <span class=\"katex-eq\" data-katex-display=\"false\">Proy_{\\vec{y}}(\\vec{x}),<\/span> atrav\u00e9s da express\u00e3o:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">Proy_{\\vec{y}}(\\vec{x}) = \\| \\vec{x}\\| \\cos(\\angle(\\vec{x},\\vec{y})) \\hat{y}<\/span>\n<p>Se combinarmos isso com o que vimos em par\u00e1grafos anteriores podemos escrever:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle Proy_{\\vec{y}}(\\vec{x}) = {\\| \\vec{x}\\|} \\left(\\frac{\\vec{x}\\cdot\\vec{y}}{{\\|\\vec{x}\\|} \\|\\vec{y}\\|}\\right)\\color{red}{\\hat{y}} =  \\left(\\frac{\\vec{x}\\cdot\\vec{y}}{\\|\\vec{y}\\|} \\right)\\color{red}{\\frac{\\vec{y}}{\\|\\vec{y}\\|}} = \\left(\\frac{\\vec{x}\\cdot\\vec{y}}{\\|\\vec{y}\\|^2}\\right)\\vec{y} = \\left(\\frac{\\vec{x}\\cdot\\vec{y}}{\\vec{y}\\cdot\\vec{y}}\\right)\\vec{y}<\/span>\n<p>j\u00e1 que, recordemos<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\cos(\\angle(\\vec{x},\\vec{y}))  = \\frac{\\vec{x}\\cdot\\vec{y}}{\\|\\vec{x}\\| \\|\\vec{y}\\|}<\/span>\n<p>As proje\u00e7\u00f5es s\u00e3o importantes porque nos permitem expressar os vetores em termos de qualquer base como a soma de suas proje\u00e7\u00f5es:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\vec{x} = \\displaystyle \\sum_{i=1}^n \\alpha_i \\hat{u}_i<\/span>\n<p>Onde <span class=\"katex-eq\" data-katex-display=\"false\">\\{\\vec{u}_i\\}_{i=1,\\cdots, n}<\/span> \u00e9 uma base de vetores linearmente independentes de <span class=\"katex-eq\" data-katex-display=\"false\">\\mathbb{R}^n<\/span> e os coeficientes <span class=\"katex-eq\" data-katex-display=\"false\">\\alpha_i = (\\vec{x}\\cdot\\vec{u}_i)\/\\|\\vec{u}_i\\|<\/span> s\u00e3o justamente as proje\u00e7\u00f5es sobre cada elemento da base e que constituem as coordenadas <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{x}<\/span> em rela\u00e7\u00e3o \u00e0 base <span class=\"katex-eq\" data-katex-display=\"false\">\\{\\hat{u}_i\\}_{i=1,\\cdots, n}<\/span> de <span class=\"katex-eq\" data-katex-display=\"false\">\\mathbb{R}^n.<\/span>\n<p><a name=\"El-Teorema-de-Pitagoras-y-la-Proyecci\u00f3n-sobre-un-Subespacio\"><\/a><br \/>\n<center><iframe class=\"lazyload\" width=\"560\" height=\"315\" data-src=\"https:\/\/www.youtube.com\/embed\/CGrr6IDnvjs\" title=\"YouTube video player\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture\" allowfullscreen><\/iframe><\/center><\/p>\n<h2>O Teorema de Pit\u00e1goras e a Proje\u00e7\u00e3o sobre um Subespa\u00e7o<\/h2>\n<p><a href=\"https:\/\/www.youtube.com\/watch?v=CGrr6IDnvjs&#038;t=254s\" rel=\"noopener\" target=\"_blank\"><strong><span style=\"color: #ff0000;\">O teorema de Pit\u00e1goras \u00e9 um resultado<\/span><\/strong><\/a> conhecido por todos e que conta com in\u00fameras demonstra\u00e7\u00f5es. Uma demonstra\u00e7\u00e3o poss\u00edvel deste teorema emerge justamente das mat\u00e9rias que desenvolvemos para o espa\u00e7o euclidiano com o adicional de ser v\u00e1lido para qualquer n\u00famero de dimens\u00f5es.<\/p>\n<h3>Demonstra\u00e7\u00e3o do Teorema de Pit\u00e1goras<\/h3>\n<p><a href=\"https:\/\/www.youtube.com\/watch?v=CGrr6IDnvjs&#038;t=533s\" rel=\"noopener\" target=\"_blank\"><strong><span style=\"color: #ff0000;\">Se temos um tri\u00e2ngulo ret\u00e2ngulo de catetos<\/span><\/strong><\/a> <span class=\"katex-eq\" data-katex-display=\"false\">a<\/span> e <span class=\"katex-eq\" data-katex-display=\"false\">b,<\/span> e hipotenusa <span class=\"katex-eq\" data-katex-display=\"false\">c,<\/span> o teorema de Pit\u00e1goras nos diz que <span class=\"katex-eq\" data-katex-display=\"false\">a^2+b^2=c^2.<\/span> Entendido isso, podemos representar cada cateto atrav\u00e9s de um par de vetores ortogonais <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{x}<\/span> e <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{y}<\/span> e escrever o teorema de Pit\u00e1goras da seguinte maneira:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\{\\vec{x},\\vec{y}\\in \\mathbb{R}^n\\setminus\\{\\vec{0}\\}\\} \\vdash\n\n \\vec{x}\\bot\\vec{y} \\leftrightarrow (\\|\\vec{x} + \\vec{y}\\|^2 = \\|\\vec{x}\\|^2 + \\|\\vec{y}\\|^2)<\/span>\n<p>Onde a express\u00e3o <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{x}\\bot\\vec{y}<\/span> indica que ambos os vetores s\u00e3o ortogonais, ou seja: n\u00e3o nulos e tais que <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{x}\\cdot\\vec{y}=0.<\/span> Desse modo, estabelece-se uma rela\u00e7\u00e3o de bicondicionalidade entre a ortogonalidade e a soma das magnitudes ao quadrado de dois vetores.<\/p>\n<p>Esta forma vetorial de representar o teorema de Pit\u00e1goras pode ser demonstrada atrav\u00e9s dos seguintes dois racioc\u00ednios:<\/p>\n<p><strong>Primeiro de ida:<\/strong><\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}{rll}\n\n(1) &amp; \\{\\vec{x},\\vec{y}\\in \\mathbb{R}^n\\setminus\\{\\vec{0}\\}, \\vec{x}\\bot\\vec{y}\\} \\vdash \\vec{x}\\bot\\vec{y} &amp; {;\\;Presun\\c{c}\u00e3o} \\\\ \\\\\n\n(2) &amp; \\{\\vec{x},\\vec{y}\\in \\mathbb{R}^n\\setminus\\{\\vec{0}\\}, \\vec{x}\\bot\\vec{y}\\} \\vdash \\vec{x}\\cdot\\vec{y}= 0 &amp; {;\\;De(1)} \\\\ \\\\\n\n(3) &amp; \\{\\vec{x},\\vec{y}\\in \\mathbb{R}^n\\setminus\\{\\vec{0}\\}, \\vec{x}\\bot\\vec{y}\\} \\vdash \\|\\vec{x} + \\vec{y}\\|^2 = (\\vec{x} + \\vec{y})\\cdot(\\vec{x} + \\vec{y}) = \\|\\vec{x}\\|^2 + 2(\\vec{x}\\cdot\\vec{y}) + \\|\\vec{y}\\|^2 &amp; \\\\\n\n&amp;;\\; Propriedade\\;da\\;norma\\;euclidiana\\;e\\;do\\;produto\\;escalar &amp; \\\\ \\\\\n\n(4) &amp; \\{\\vec{x},\\vec{y}\\in \\mathbb{R}^n\\setminus\\{\\vec{0}\\}, \\vec{x}\\bot\\vec{y}\\} \\vdash \\|\\vec{x} + \\vec{y}\\|^2 =  \\|\\vec{x}\\|^2  + \\|\\vec{y}\\|^2 &amp; {;\\;De(2,3)} \\\\ \\\\\n\n(5) &amp; \\{\\vec{x},\\vec{y}\\in \\mathbb{R}^n\\setminus\\{\\vec{0}\\}\\} \\vdash \\vec{x}\\bot\\vec{y} \\rightarrow ( \\|\\vec{x} + \\vec{y}\\|^2 =  \\|\\vec{x}\\|^2  + \\|\\vec{y}\\|^2) &amp; {;\\;TD(4)} \\end{array}<\/span>\n<p><strong>E agora de volta:<\/strong><\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}{rll}\n\n(1) &amp; \\{\\vec{x},\\vec{y}\\in \\mathbb{R}^n\\setminus\\{\\vec{0}\\}, \\|\\vec{x} + \\vec{y}\\|^2 =  \\|\\vec{x}\\|^2  + \\|\\vec{y}\\|^2\\} \\vdash \\|\\vec{x} + \\vec{y}\\|^2 =  \\|\\vec{x}\\|^2  + \\|\\vec{y}\\|^2 &amp; {;\\;Presun\\c{c}\u00e3o} \\\\ \\\\\n\n(2) &amp; \\{\\vec{x},\\vec{y}\\in \\mathbb{R}^n\\setminus\\{\\vec{0}\\}, \\|\\vec{x} + \\vec{y}\\|^2 =  \\|\\vec{x}\\|^2  + \\|\\vec{y}\\|^2\\} \\vdash \\|\\vec{x} + \\vec{y}\\|^2 =  \\|\\vec{x}\\|^2 +2(\\vec{x}\\cdot\\vec{y}) + \\|\\vec{y}\\|^2 &amp;  \\\\\n\n&amp;;\\; Propriedade\\;da\\;norma\\;euclidiana\\;e\\;do\\;produto\\;escalar &amp;\\\\ \\\\\n\n(3) &amp; \\{\\vec{x},\\vec{y}\\in \\mathbb{R}^n\\setminus\\{\\vec{0}\\}, \\|\\vec{x} + \\vec{y}\\|^2 =  \\|\\vec{x}\\|^2  + \\|\\vec{y}\\|^2\\} \\vdash  \\vec{x}\\cdot\\vec{y}=0 &amp; {;\\;De(1,2)} \\\\ \\\\\n\n(4) &amp; \\{\\vec{x},\\vec{y}\\in \\mathbb{R}^n\\setminus\\{\\vec{0}\\}, \\|\\vec{x} + \\vec{y}\\|^2 =  \\|\\vec{x}\\|^2  + \\|\\vec{y}\\|^2\\} \\vdash  \\vec{x}\\bot\\vec{y} &amp; {;\\;De(3)} \\\\ \\\\\n\n(5) &amp; \\{\\vec{x},\\vec{y}\\in \\mathbb{R}^n\\setminus\\{\\vec{0}\\}\\} \\vdash (\\|\\vec{x} + \\vec{y}\\|^2 =  \\|\\vec{x}\\|^2  + \\|\\vec{y}\\|^2) \\rightarrow  \\vec{x}\\bot\\vec{y} &amp; {;\\;TD(4)} \\end{array}<\/span>\n<p><strong>E finalmente, juntando ambos racioc\u00ednios tem-se o que se queria demonstrar:<\/strong><\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\{\\vec{x},\\vec{y}\\in \\mathbb{R}^n\\setminus\\{\\vec{0}\\}\\} \\vdash   \\vec{x}\\bot\\vec{y} \\leftrightarrow (\\|\\vec{x} + \\vec{y}\\|^2 = \\|\\vec{x}\\|^2 + \\|\\vec{y}\\|^2)<\/span>\n<h3>A Proje\u00e7\u00e3o de um vetor sobre um Subespa\u00e7o de <span class=\"katex-eq\" data-katex-display=\"false\">\\mathbb{R}^n<\/span><\/h3>\n<p><a href=\"https:\/\/www.youtube.com\/watch?v=CGrr6IDnvjs&#038;t=1545s\" rel=\"noopener\" target=\"_blank\"><strong><span style=\"color: #ff0000;\">Consideremos um subespa\u00e7o<\/span><\/strong><\/a> <span class=\"katex-eq\" data-katex-display=\"false\">H<\/span> de <span class=\"katex-eq\" data-katex-display=\"false\">\\mathbb{R}^n<\/span> formado por uma base de vetores unit\u00e1rios <span class=\"katex-eq\" data-katex-display=\"false\">\\{\\hat{v}_1, \\cdots, \\hat{v}_k\\}.<\/span> Se tomarmos um vetor <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{x}\\in\\mathbb{R}^n\\setminus\\{\\vec{0}\\},<\/span> define-se a proje\u00e7\u00e3o do vetor <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{x}<\/span> sobre o espa\u00e7o <span class=\"katex-eq\" data-katex-display=\"false\">H<\/span> atrav\u00e9s da express\u00e3o:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">Proy_{H}(\\vec{x}) = \\displaystyle \\sum_{j=1}^k (\\vec{x} \\cdot \\hat{v}_j)\\hat{v}_j<\/span>\n<p>Que um conjunto seja ortonormal quer dizer que todos os seus elementos s\u00e3o ortogonais entre si e cada um tem norma igual \u00e0 unidade.<\/p>\n<p>Isto \u00e9, por assim dizer, a sombra que projeta um vetor sobre cada uma das componentes do subespa\u00e7o <span class=\"katex-eq\" data-katex-display=\"false\">H<\/span> de <span class=\"katex-eq\" data-katex-display=\"false\">\\mathbb{R}^n<\/span> <\/p>\n<p><center><img decoding=\"async\" src=\"data:image\/gif;base64,R0lGODlhAQABAIAAAAAAAP\/\/\/yH5BAEAAAAALAAAAAABAAEAAAIBRAA7\" data-src=\"https:\/\/blogger.googleusercontent.com\/img\/a\/AVvXsEga986LBrInk-B_9gUKPe01TF10dNECXU54KK1bSf3mAPakWE-FqdqyPbb0TVy88OfGxQmJRd-yW4dwAfcC21i2dM0KZqQjPe_Qx0M5OUz4f_P6IipJQ6PcxtkOmcO7-GqRiGZ-3StQpzy8FMIfPYE89Wae6JZIC2Jk9dSTPFTK1L4TsnpkcdpV1Dbr\" width=\"578\" height=\"591\" class=\"alignnone size-full lazyload\" \/><noscript><img decoding=\"async\" src=\"https:\/\/blogger.googleusercontent.com\/img\/a\/AVvXsEga986LBrInk-B_9gUKPe01TF10dNECXU54KK1bSf3mAPakWE-FqdqyPbb0TVy88OfGxQmJRd-yW4dwAfcC21i2dM0KZqQjPe_Qx0M5OUz4f_P6IipJQ6PcxtkOmcO7-GqRiGZ-3StQpzy8FMIfPYE89Wae6JZIC2Jk9dSTPFTK1L4TsnpkcdpV1Dbr\" width=\"578\" height=\"591\" class=\"alignnone size-full lazyload\" \/><\/noscript><\/center><\/p>\n<h3>Dist\u00e2ncia entre um Ponto ou Vetor de <span class=\"katex-eq\" data-katex-display=\"false\">\\mathbb{R}^n<\/span> e um Subespa\u00e7o de <span class=\"katex-eq\" data-katex-display=\"false\">\\mathbb{R}^n<\/span><\/h3>\n<p><a href=\"https:\/\/www.youtube.com\/watch?v=CGrr6IDnvjs&#038;t=1974s\" rel=\"noopener\" target=\"_blank\"><strong><span style=\"color: #ff0000;\">A partir da proje\u00e7\u00e3o de um vetor<\/span><\/strong><\/a> <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{x}\\in\\mathbb{R}^n\\setminus\\{\\vec{0}\\}<\/span> sobre um subespa\u00e7o <span class=\"katex-eq\" data-katex-display=\"false\">H<\/span> de <span class=\"katex-eq\" data-katex-display=\"false\">\\mathbb{R}^n<\/span> pode-se construir um vetor da forma<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\vec{x} - Proy_{H}(\\vec{x})<\/span>\n<p>O vetor formado desta maneira ser\u00e1 um vetor que une um ponto do subespa\u00e7o <span class=\"katex-eq\" data-katex-display=\"false\">H<\/span> com o ponto de coordenadas <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{x},<\/span> que sai ortogonalmente ao subespa\u00e7o <span class=\"katex-eq\" data-katex-display=\"false\">H.<\/span> Isto n\u00e3o \u00e9 dif\u00edcil de provar: se tomarmos um vetor qualquer <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{z}\\in H<\/span> e calcularmos o produto escalar <span class=\"katex-eq\" data-katex-display=\"false\">(\\vec{x}-Proy_{H}(\\vec{x}))\\cdot \\vec{z},<\/span> basta ver que o resultado desta opera\u00e7\u00e3o \u00e9 zero. Fa\u00e7amos as contas para verificar se de fato isso \u00e9 assim:<\/p>\n<p>Se <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{z}\\in H,<\/span> ent\u00e3o ser\u00e1 da forma<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\vec{z}=\\displaystyle \\sum_{j=1}^k \\beta_j\\hat{v}_j<\/span>\n<p>Onde <span class=\"katex-eq\" data-katex-display=\"false\">\\{\\hat{v}_j\\}_{j=1}^k<\/span> \u00e9 uma base ortonormal de <span class=\"katex-eq\" data-katex-display=\"false\">H<\/span> e <span class=\"katex-eq\" data-katex-display=\"false\">\\beta_j \\in\\mathbb{R}<\/span> s\u00e3o os coeficientes de <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{z}<\/span> em <span class=\"katex-eq\" data-katex-display=\"false\">H.<\/span> Tendo isso em conta, o c\u00e1lculo do produto escalar <span class=\"katex-eq\" data-katex-display=\"false\">(\\vec{x}-Proy_{H}(\\vec{x}))\\cdot \\vec{z},<\/span> dar\u00e1:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}{rl} (\\vec{x}-Proy_{H}(\\vec{x}))\\cdot \\vec{z} &amp;= \\left(\\vec{x} - \\displaystyle \\sum_{j=1}^k (\\vec{x} \\cdot \\hat{v}_j)\\hat{v}_j \\right) \\cdot \\displaystyle \\sum_{j=1}^k \\beta_j\\hat{v}_j \\\\ \\\\ &amp;= \\vec{x} \\cdot \\displaystyle \\sum_{j=1}^k \\beta_j\\hat{v}_j - \\displaystyle \\sum_{j=1}^k (\\vec{x} \\cdot \\hat{v}_j)\\hat{v}_j \\cdot \\displaystyle \\sum_{j=1}^k \\beta_j\\hat{v}_j \\end{array}<\/span>\n<p>Mas como <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{x}<\/span> \u00e9 um vetor de <span class=\"katex-eq\" data-katex-display=\"false\">\\mathbb{R}^n<\/span> do qual <span class=\"katex-eq\" data-katex-display=\"false\">H<\/span> \u00e9 subespa\u00e7o, \u00e9 poss\u00edvel encontrar um conjunto de <span class=\"katex-eq\" data-katex-display=\"false\">n-k<\/span> vetores ortonormais entre si e ao mesmo tempo ortonormais a todos os vetores de <span class=\"katex-eq\" data-katex-display=\"false\">H,<\/span> digamos <span class=\"katex-eq\" data-katex-display=\"false\">\\{\\hat{v}_{k+1}, \\cdots, \\hat{v}_n\\},<\/span> de modo tal que junto \u00e0 base de <span class=\"katex-eq\" data-katex-display=\"false\">H<\/span> formem uma base para <span class=\"katex-eq\" data-katex-display=\"false\">\\mathbb{R}^n<\/span> e se possa escrever<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\vec{x} = \\displaystyle  \\sum_{j=1}^k (\\vec{x}\\cdot\\hat{v}_j )\\hat{v}_j + \\sum_{j=k+1}^n \\alpha_j \\hat{v}_j <\/span>\n<p>De modo que o desenvolvimento acima segue da forma:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}{rl}\n\n(\\vec{x}-Proy_{H}(\\vec{x}))\\cdot \\vec{z} &amp;= \\displaystyle \\left( \\sum_{j=1}^k (\\vec{x}\\cdot\\hat{v}_j )\\hat{v}_j + \\sum_{j=k+1}^n \\alpha_j \\hat{v}_j\\right) \\cdot  \\sum_{j=1}^k \\beta_j\\hat{v}_j -  \\sum_{j=1}^k (\\vec{x} \\cdot \\hat{v}_j)\\hat{v}_j \\cdot  \\sum_{j=1}^k \\beta_j\\hat{v}_j \\\\ \\\\\n\n&amp;=  \\displaystyle \\sum_{j=1}^k (\\vec{x}\\cdot\\hat{v}_j )\\hat{v}_j \\cdot \\sum_{j=1}^k \\beta_j\\hat{v}_j + \\underbrace{\\color{red}{\\sum_{j=k+1}^n \\alpha_j \\hat{v}_j \\cdot \\sum_{j=1}^k \\beta_j\\hat{v}_j}}_{(*)} - \\sum_{j=1}^k (\\vec{x} \\cdot \\hat{v}_j)\\hat{v}_j \\cdot  \\sum_{j=1}^k \\beta_j\\hat{v}_j \\\\ \\\\\n\n&amp;=  \\displaystyle \\sum_{j=1}^k (\\vec{x}\\cdot\\hat{v}_j )\\hat{v}_j \\cdot \\sum_{j=1}^k \\beta_j\\hat{v}_j  - \\sum_{j=1}^k (\\vec{x} \\cdot \\hat{v}_j)\\hat{v}_j \\cdot  \\sum_{j=1}^k \\beta_j\\hat{v}_j \\\\ \\\\\n\n&amp;= 0  \\end{array}<\/span>\n<p>(*) Soma zero porque <span class=\"katex-eq\" data-katex-display=\"false\">\\{v_j\\}_{j=1}^n<\/span> \u00e9 uma base ortonormal de <span class=\"katex-eq\" data-katex-display=\"false\">\\mathbb{R}^n.<\/span>\n<p>A partir disso podemos demonstrar que a dist\u00e2ncia entre o subespa\u00e7o <span class=\"katex-eq\" data-katex-display=\"false\">H<\/span> e o vetor <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{x}<\/span> \u00e9 dada por:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\|\\vec{x} - Proy_{H}(\\vec{x})\\|<\/span>\n<h4>Demonstra\u00e7\u00e3o<\/h4>\n<p><a href=\"https:\/\/www.youtube.com\/watch?v=CGrr6IDnvjs&#038;t=2995s\" rel=\"noopener\" target=\"_blank\"><strong><span style=\"color: #ff0000;\">Para demonstrar este resultado mostrar-se-\u00e1<\/span><\/strong><\/a> que para todo <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{z}\\in H<\/span> sempre se cumprir\u00e1 que <span class=\"katex-eq\" data-katex-display=\"false\">\\|\\vec{x} - Proy_{H}(\\vec{x})\\| \\leq \\|\\vec{x} - \\vec{z}\\|,<\/span> para isso utilizaremos o teorema de Pit\u00e1goras da seguinte maneira:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}{rl} \\|\\vec{x} - \\vec{z}\\|^2 &amp;= \\| \\left(\\vec{x} -Proy_{H}(\\vec{x}) \\right) + \\left(Proy_{H}(\\vec{x}) - \\vec{z}\\right)\\|^2 \\\\ \\\\ &amp;= \\| \\vec{x} -Proy_{H}(\\vec{x}) \\|^2 + \\|Proy_{H}(\\vec{x}) - \\vec{z}\\|^2 \\\\ \\\\ \\end{array}<\/span>\n<p>Esta \u00faltima igualdade obt\u00e9m-se porque os vetores <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{x} -Proy_{H}(\\vec{x})<\/span> e <span class=\"katex-eq\" data-katex-display=\"false\">Proy_{H}(\\vec{x}) - \\vec{z}<\/span> s\u00e3o ortogonais. E portanto:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\|\\vec{x} - Proy_{H}(\\vec{x})\\|^2 \\leq \\|\\vec{x} - \\vec{z}\\|^2<\/span>\n<p>que \u00e9 o que se queria demonstrar.<\/p>\n<p>J\u00e1 com este resultado em nossas m\u00e3os, podemos dizer que a dist\u00e2ncia entre um ponto <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{x}\\in\\mathbb{R}^n<\/span> e um subespa\u00e7o <span class=\"katex-eq\" data-katex-display=\"false\">H<\/span> de <span class=\"katex-eq\" data-katex-display=\"false\">\\mathbb{R}^n<\/span> gerado pelos vetores ortonormais <span class=\"katex-eq\" data-katex-display=\"false\">\\{\\hat{v}_1, \\cdots, \\hat{v}_k\\}<\/span> \u00e9 dada por:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">dist(\\vec{x},H) =\\left\\|\\vec{x} - Proy_{H}(\\vec{x})\\right\\|= \\left\\|\\vec{x} - \\displaystyle \\sum_{j=1}^k (\\vec{x} \\cdot \\hat{v}_j)\\hat{v}_j\\right\\|<\/span>\n<p><a name=\"El-Producto-Escalar-y-Vectorial-en-R3\"><\/a><br \/>\n<center><iframe class=\"lazyload\" width=\"560\" height=\"315\" data-src=\"https:\/\/www.youtube.com\/embed\/uei6y2tniOc\" title=\"YouTube video player\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture\" allowfullscreen><\/iframe><\/center><\/p>\n<h2>O Produto Escalar e Vetorial em <span class=\"katex-eq\" data-katex-display=\"false\">\\mathbb{R}^3<\/span><\/h2>\n<p><strong><a href=\"https:\/\/www.youtube.com\/watch?v=uei6y2tniOc&#038;t=242s\" rel=\"noopener\" target=\"_blank\"><span style=\"color: #ff0000;\">Agora mudaremos um pouco nosso enfoque<\/span><\/a><\/strong> para nos centrarmos nos vetores de <span class=\"katex-eq\" data-katex-display=\"false\">\\mathbb{R}^3.<\/span> Aqui, al\u00e9m das opera\u00e7\u00f5es que j\u00e1 revisamos em geral para <span class=\"katex-eq\" data-katex-display=\"false\">\\mathbb{R}^n,<\/span> \u00e9 poss\u00edvel tamb\u00e9m o produto vetorial que, a partir do produto de dois vetores, d\u00e1 como resultado outro vetor. Este \u00e9 um produto exclusivo de <span class=\"katex-eq\" data-katex-display=\"false\">\\mathbb{R}^3<\/span> (e possivelmente de <span class=\"katex-eq\" data-katex-display=\"false\">\\mathbb{R}^7<\/span>, cujo caso n\u00e3o analisaremos aqui). Geralmente, os vetores da base can\u00f4nica de <span class=\"katex-eq\" data-katex-display=\"false\">\\mathbb{R}^3<\/span> s\u00e3o representados pelas letras <span class=\"katex-eq\" data-katex-display=\"false\">\\hat{x}, \\hat{y}, \\hat{z}<\/span> ou como <span class=\"katex-eq\" data-katex-display=\"false\">\\hat{\\imath}, \\hat{\\jmath}, \\hat{k}<\/span>. A prefer\u00eancia de uma ou outra \u00e9 pessoal.<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}{rl} \\hat{\\imath} = \\hat{x}&amp;=(1,0,0)\\\\ \\hat{\\jmath} =\\hat{y}&amp;=(0,1,0)\\\\ \\hat{k} =\\hat{z}&amp;=(0,0,1)\\\\ \\end{array}<\/span>\n<p>Assim, se temos um vetor da forma <span class=\"katex-eq\" data-katex-display=\"false\">(a,b,c),<\/span> pode ser escrito em forma alg\u00e9brica da seguinte maneira:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">(a,b,c) = a\\hat{x} + b\\hat{y} + c\\hat{z}<\/span>\n<h3>O produto vetorial em <span class=\"katex-eq\" data-katex-display=\"false\">\\mathbb{R}^3<\/span><\/h3>\n<p><a href=\"https:\/\/www.youtube.com\/watch?v=uei6y2tniOc&#038;t=330s\" rel=\"noopener\" target=\"_blank\"><strong><span style=\"color: #ff0000;\">Sejam <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{x}=(x_1,x_2,x_3)<\/span> e <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{y}=(y_1,y_2,y_3)<\/span> vetores de <span class=\"katex-eq\" data-katex-display=\"false\">\\mathbb{R}^3.<\/span><\/span><\/strong><\/a> Define-se o produto vetorial de <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{x}<\/span> com <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{y},<\/span> <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{x}\\times\\vec{y}<\/span> atrav\u00e9s de:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\"> \\begin{array}{rl} \\vec{x}\\times\\vec{y} &amp;= \\left|\\begin{array}{ccc} \\hat{x} &amp; \\hat{y} &amp; \\hat{z} \\\\ x_1 &amp; x_2 &amp; x_3 \\\\ y_1 &amp; y_2 &amp; y_3 \\end{array}\\right| \\\\ \\\\ &amp;=\\hat{x}x_2y_3 + \\hat{y}x_3y_1 + \\hat{z} x_1y_2 - \\left( \\hat{z} x_2 y_1 + \\hat{y} x_1 y_3 + \\hat{x}x_3y_2\\right) \\\\ \\\\ &amp;=\\hat{x}(x_2y_3 - x_3y_2) + \\hat{y}(x_3y_1 - x_1y_3) + \\hat{z}(x_1y_2 - x_2y_1) \\end{array}<\/span>\n<h3>A Identidade de Lagrange<\/h3>\n<p><a href=\"https:\/\/www.youtube.com\/watch?v=uei6y2tniOc&#038;t=1399s\" rel=\"noopener\" target=\"_blank\"><strong><span style=\"color: #ff0000;\">Para o caso dos vetores de <span class=\"katex-eq\" data-katex-display=\"false\">\\mathbb{R}^3<\/span><\/span><\/strong><\/a> podemos reconhecer tr\u00eas tipos de \u00abprodutos\u00bb: O escalar <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{x}\\cdot\\vec{y},<\/span> o vetorial <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{x}\\times\\vec{y},<\/span> e o das normas <span class=\"katex-eq\" data-katex-display=\"false\">\\|\\vec{x}\\|\\|\\vec{y}\\|.<\/span> Estes tr\u00eas produtos est\u00e3o relacionados entre si atrav\u00e9s da identidade de Lagrange<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\|\\vec{x}\\times\\vec{y}\\|^2  = \\|\\vec{x}\\|^2\\|\\vec{y}\\|^2- (\\vec{x}\\cdot\\vec{y})^2 <\/span>\n<h4>Demonstra\u00e7\u00e3o da identidade de Lagrange<\/h4>\n<p>Sejam <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{x}=(x_1,x_2,x_3)<\/span> e <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{y}=(y_1,y_2,y_3)<\/span> vetores de <span class=\"katex-eq\" data-katex-display=\"false\">\\mathbb{R}^3,<\/span> ent\u00e3o tem-se que:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\"> \\begin{array}{rl} \\vec{x}\\times\\vec{y} &amp;=(x_2y_3 - x_3y_2) \\hat{x} + (x_3y_1 - x_1y_3)\\hat{y} + (x_1y_2 - x_2y_1)\\hat{z} \\end{array}<\/span>\n<p>De modo que:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\"> \\begin{array}{rl}\n\n\\|\\vec{x}\\times\\vec{y}\\|^2 &amp;=(x_2y_3 - x_3y_2)^2 + (x_3y_1 - x_1y_3)^2 + (x_1y_2 - x_2y_1)^2 \\\\ \\\\\n\n&amp;= \\color{green}{x_2^2y_3^2 - 2x_2x_3y_3y_2 + x_3^2y_2^2} + \\cdots\\\\ \\\\\n\n&amp;\\cdots + \\color{blue}{x_3^2y_1^2 - 2x_3x_1y_1y_3 + x_1^2y_3^2} + \\cdots \\\\ \\\\\n\n&amp;\\cdots + \\color{red}{x_1^2y_2^2 - 2x_1x_2y_2y_1 + x_2^2y_1^2} \\end{array}<\/span>\n<p>Por outro lado:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\"> \\begin{array}{rl}\n\n\\|\\vec{x}\\|^2 \\|\\vec{y}\\|^2 - (\\vec{x}\\cdot\\vec{y})^2 &amp;= (x_1^2 + x_2^2 + x_3^2)(y_1^2+y_2^2 + y_3^2) - (x_1y_1 + x_2y_2 + x_3 y_3)^2 \\\\ \\\\ \\\\\n\n&amp;=  {x_1^2y_1^2} + \\color{red}{x_1^2y_2^2} + \\color{blue}{x_1^2y_3^2} + \\cdots \\\\ \\\\\n\n&amp;\\cdots + \\color{red}{x_2^2y_1^2} +  {x_2^2y_2^2} + \\color{green}{x_2^2y_3^2} + \\cdots \\\\ \\\\\n\n&amp;\\cdots + \\color{blue}{x_3^2y_1^2} + \\color{green}{x_3^2y_2^2} +  {x_3^2y_3^2} + \\cdots \\\\ \\\\\n\n&amp;\\cdots - \\left[ {x_1^2y_1^2} +  {x_2^2y_2^2} +  {x_3^2y_3^2} + \\right. \\cdots \\\\ \\\\\n\n&amp;\\cdots + 2\\left(\\color{red}{x_1x_2y_1y_2} + \\color{blue}{x_1x_3y_1y_3} + \\color{green}{x_2x_3y_2y_3} \\right)\\left.\\right] \\\\ \\\\ \\\\\n\n&amp;= \\color{red}{x_1^2y_2^2 - 2x_1x_2y_2y_1 + x_2^2y_1^2} + \\cdots \\\\ \\\\\n\n&amp; \\cdots + \\color{blue}{x_1^2y_3^2 - 2x_1x_3y_3y_1 + x_3^2y_1^2} + \\cdots \\\\ \\\\\n\n&amp; \\cdots + \\color{green}{x_2^2y_3^2 - 2x_2x_3y_3y_2 + x_3^2y_2^2}\n\n\\end{array}<\/span>\n<p>Finalmente, comparando as express\u00f5es em cores tem-se o que se queria demonstrar.<\/p>\n<h3>O Produto Vetorial e o \u00e2ngulo entre vetores<\/h3>\n<p><a href=\"https:\/\/www.youtube.com\/watch?v=uei6y2tniOc&#038;t=1954s\" rel=\"noopener\" target=\"_blank\"><strong><span style=\"color: #ff0000;\">Anteriormente vimos que existe uma estreita rela\u00e7\u00e3o<\/span><\/strong><\/a> entre o \u00e2ngulo sustentado por dois vetores e o resultado do produto escalar, isto \u00e9 dado pela rela\u00e7\u00e3o <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{x}\\cdot\\vec{y} = \\|\\vec{x}\\|\\|\\vec{y}\\|\\cos(\\angle(\\vec{x},\\vec{y})).<\/span> Acontece que algo semelhante ocorre com o produto vetorial e \u00e9 dado pela seguinte rela\u00e7\u00e3o:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\|\\vec{x}\\times\\vec{y}\\| = \\|\\vec{x}\\|\\|\\vec{y}\\| \\sin(\\angle(\\vec{x},\\vec{y}))<\/span>\n<p>Esta express\u00e3o \u00e9 um resultado direto da identidade de Lagrange que se demonstrou acima, a demonstra\u00e7\u00e3o fica mais ou menos assim:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}{rl} \\|\\vec{x}\\times\\vec{y}\\|^2 &amp;= \\|\\vec{x}\\|^2\\|\\vec{y}\\|^2 - (\\vec{x}\\cdot\\vec{y})^2 \\\\ \\\\ &amp;= \\|\\vec{x}\\|^2\\|\\vec{y}\\|^2 - (\\|\\vec{x}\\|\\|\\vec{y}\\|\\cos(\\angle(\\vec{x},\\vec{y})))^2 \\\\ \\\\ &amp;= \\|\\vec{x}\\|^2\\|\\vec{y}\\|^2 - \\|\\vec{x}\\|^2\\|\\vec{y}\\|^2\\cos^2(\\angle(\\vec{x},\\vec{y})) \\\\ \\\\ &amp;= \\|\\vec{x}\\|^2\\|\\vec{y}\\|^2 (1 - \\cos^2(\\angle(\\vec{x},\\vec{y}))) \\\\ \\\\ &amp;= \\|\\vec{x}\\|^2\\|\\vec{y}\\|^2 \\sin^2(\\angle(\\vec{x},\\vec{y})) \\end{array}<\/span>\n<p>Finalmente, tomando ra\u00edzes chegamos a:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\|\\vec{x}\\times\\vec{y}\\| = \\|\\vec{x}\\|\\|\\vec{y}\\|\\; |\\sin(\\angle(\\vec{x},\\vec{y}))|<\/span>\n<p>Mas recordemos que <span class=\"katex-eq\" data-katex-display=\"false\">\\angle(\\vec{x},\\vec{y})\\in[0,\\pi],<\/span> e nesse intervalo de valores a fun\u00e7\u00e3o seno \u00e9 sempre n\u00e3o-negativa, de modo que podemos retirar o valor absoluto e chegamos ao que se queria demonstrar.<\/p>\n<p>A partir desta express\u00e3o podemos intuir que o resultado da opera\u00e7\u00e3o <span class=\"katex-eq\" data-katex-display=\"false\">\\|\\vec{x}\\times\\vec{y}\\|<\/span> nos d\u00e1 como resultado a \u00e1rea gerada pelos vetores <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{x}<\/span> e <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{y}.<\/span>\n","protected":false},"excerpt":{"rendered":"<p>\u00c1lgebra e Proje\u00e7\u00f5es em Rn, Produto Vetorial em Resumo:Esta s\u00e9rie \u00e9 a continua\u00e7\u00e3o direta da s\u00e9rie sobre o Espa\u00e7o Euclidiano de n dimens\u00f5es. Aqui revisaremos alguns conceitos de \u00e1lgebra linear que ajudam a compreender melhor o espa\u00e7o euclidiano n-dimensional, revisaremos os conceitos de proje\u00e7\u00f5es de um vetor sobre outro, demonstraremos o teorema de Pit\u00e1goras e [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":34241,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"iawp_total_views":2,"footnotes":""},"categories":[1118,571],"tags":[],"class_list":["post-34247","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-calculo-multivariavel","category-matematica-pt"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v27.4 - https:\/\/yoast.com\/product\/yoast-seo-wordpress\/ -->\n<title>\u00c1lgebra e Proje\u00e7\u00f5es em Rn, Produto Vetorial em R3 - toposuranos.com\/material<\/title>\n<meta name=\"description\" content=\"Explore proje\u00e7\u00f5es e vetores de Rn: 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