{"id":5295,"date":"2010-11-13T01:28:47","date_gmt":"2010-11-13T01:28:47","guid":{"rendered":"https:\/\/www.acasinhadamatematica.pt\/?p=5295"},"modified":"2022-01-12T01:30:59","modified_gmt":"2022-01-12T01:30:59","slug":"determine-a-amplitude-do-angulo-dos-dois-vectores","status":"publish","type":"post","link":"https:\/\/www.acasinhadamatematica.pt\/?p=5295","title":{"rendered":"Determine a amplitude do \u00e2ngulo dos dois vetores"},"content":{"rendered":"<p><ul id='GTTabs_ul_5295' class='GTTabs' style='display:none'>\n<li id='GTTabs_li_0_5295' class='GTTabs_curr'><a  id=\"5295_0\" onMouseOver=\"GTTabsShowLinks('Enunciado'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Enunciado<\/a><\/li>\n<li id='GTTabs_li_1_5295' ><a  id=\"5295_1\" onMouseOver=\"GTTabsShowLinks('Resolu\u00e7\u00e3o'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Resolu\u00e7\u00e3o<\/a><\/li>\n<\/ul>\n\n<div class='GTTabs_divs GTTabs_curr_div' id='GTTabs_0_5295'>\n<span class='GTTabs_titles'><b>Enunciado<\/b><\/span><\/p>\n<p>Determine (se necess\u00e1rio apresente o resultado aproximado \u00e0s d\u00e9cimas) a amplitude do \u00e2ngulo de $\\overrightarrow{u}$ com $\\overrightarrow{v}$, sabendo que:<\/p>\n<ol>\n<li>$\\overrightarrow{u}.\\overrightarrow{v}=50\\sqrt{2}$\u00a0e $\\left\\| \\overrightarrow{u} \\right\\|=\\left\\| \\overrightarrow{v} \\right\\|=10$<\/li>\n<li>$\\overrightarrow{u}.\\overrightarrow{v}=-10\\sqrt{3}$ e $\\left\\| \\overrightarrow{u} \\right\\|=4$ e $\\left\\| \\overrightarrow{v} \\right\\|=5$<\/li>\n<li>$\\overrightarrow{u}(1,2,3)$ e $\\overrightarrow{v}(-1,1,-1)$<\/li>\n<li>$\\overrightarrow{u}(\\frac{2}{3},\\frac{2}{3},1)$ e $\\overrightarrow{v}(-\\frac{\\sqrt{2}}{2},\\frac{\\sqrt{2}}{2},0)$<\/li>\n<\/ol>\n<p><div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_next'><a href='#GTTabs_ul_5295' onClick='GTTabs_show(1,5295)'>Resolu\u00e7\u00e3o &gt;&gt;<\/a><\/span><\/div><\/div>\n\n<div class='GTTabs_divs' id='GTTabs_1_5295'>\n<span class='GTTabs_titles'><b>Resolu\u00e7\u00e3o<\/b><\/span><!--more--><\/p>\n<ol>\n<li>Ora,<br \/>\n\\[\\begin{array}{*{35}{l}}<br \/>\n\\cos (\\widehat{\\vec{u}\\,\\vec{v}}) &amp; = &amp; \\frac{\\vec{u}.\\vec{v}}{\\left\\| {\\vec{u}} \\right\\|.\\left\\| {\\vec{v}} \\right\\|}\u00a0 \\\\<br \/>\n{} &amp; = &amp; \\frac{50\\sqrt{2}}{10\\times 10}\u00a0 \\\\<br \/>\n{} &amp; = &amp; \\frac{\\sqrt{2}}{2}\u00a0 \\\\<br \/>\n\\end{array}\\]<br \/>\nLogo, $\\widehat{\\vec{u}\\,\\vec{v}}=45{}^\\text{o}$.<br \/>\n\u00ad<\/li>\n<li>Ora,<br \/>\n\\[\\begin{array}{*{35}{l}}<br \/>\n\\cos (\\widehat{\\vec{u}\\,\\vec{v}}) &amp; = &amp; \\frac{\\vec{u}.\\vec{v}}{\\left\\| {\\vec{u}} \\right\\|.\\left\\| {\\vec{v}} \\right\\|}\u00a0 \\\\<br \/>\n{} &amp; = &amp; \\frac{-10\\sqrt{3}}{4\\times 5}\u00a0 \\\\<br \/>\n{} &amp; = &amp; -\\frac{\\sqrt{3}}{2}\u00a0 \\\\<br \/>\n\\end{array}\\]<br \/>\nLogo, $\\widehat{\\vec{u}\\,\\vec{v}}=150{}^\\text{o}$.<br \/>\n\u00ad<\/li>\n<li>Ora,<br \/>\n\\[\\begin{array}{*{35}{l}}<br \/>\n\\cos (\\widehat{\\vec{u}\\,\\vec{v}}) &amp; = &amp; \\frac{\\vec{u}.\\vec{v}}{\\left\\| {\\vec{u}} \\right\\|.\\left\\| {\\vec{v}} \\right\\|}\u00a0 \\\\<br \/>\n{} &amp; = &amp; \\frac{(1,2,3).(-1,1,-1)}{\\sqrt{{{1}^{2}}+{{2}^{2}}+{{3}^{2}}}\\times \\sqrt{{{(-1)}^{2}}+{{1}^{2}}+{{(-1)}^{2}}}}\u00a0 \\\\<br \/>\n{} &amp; = &amp; \\frac{1\\times (-1)+2\\times 1+3\\times (-1)}{\\sqrt{14}\\times \\sqrt{3}}\u00a0 \\\\<br \/>\n{} &amp; = &amp; -\\frac{2}{\\sqrt{14}\\times \\sqrt{3}}\u00a0 \\\\<br \/>\n{} &amp; = &amp; -\\frac{2\\sqrt{42}}{42}\u00a0 \\\\<br \/>\n{} &amp; = &amp; -\\frac{\\sqrt{42}}{21}\u00a0 \\\\<br \/>\n\\end{array}\\]<br \/>\nLogo, $\\widehat{\\vec{u}\\,\\vec{v}}=108,0{}^\\text{o}$.<br \/>\n\u00ad<\/li>\n<li>Ora,<br \/>\n\\[\\begin{array}{*{35}{l}}<br \/>\n\\cos (\\widehat{\\vec{u}\\,\\vec{v}}) &amp; = &amp; \\frac{\\vec{u}.\\vec{v}}{\\left\\| {\\vec{u}} \\right\\|.\\left\\| {\\vec{v}} \\right\\|}\u00a0 \\\\<br \/>\n{} &amp; = &amp; \\frac{(\\frac{2}{3},\\frac{2}{3},1).(-\\frac{\\sqrt{2}}{2},\\frac{\\sqrt{2}}{2},0)}{\\sqrt{{{\\left( \\frac{2}{3} \\right)}^{2}}+{{\\left( \\frac{2}{3} \\right)}^{2}}+{{1}^{2}}}\\times \\sqrt{{{\\left( -\\frac{\\sqrt{2}}{2} \\right)}^{2}}+{{\\left( \\frac{\\sqrt{2}}{2} \\right)}^{2}}+{{0}^{2}}}}\u00a0 \\\\<br \/>\n{} &amp; = &amp; \\frac{-2\\sqrt{2}+2\\sqrt{2}+0}{\\sqrt{\\frac{17}{9}}\\times \\sqrt{1}}\u00a0 \\\\<br \/>\n{} &amp; = &amp; 0\u00a0 \\\\<br \/>\n\\end{array}\\]<br \/>\nLogo, $\\widehat{\\vec{u}\\,\\vec{v}}=90{}^\\text{o}$.<\/li>\n<\/ol>\n<div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_prev'><a href='#GTTabs_ul_5295' onClick='GTTabs_show(0,5295)'>&lt;&lt; Enunciado<\/a><\/span><\/div><\/div>\n\n","protected":false},"excerpt":{"rendered":"<p>Enunciado Resolu\u00e7\u00e3o Enunciado Determine (se necess\u00e1rio apresente o resultado aproximado \u00e0s d\u00e9cimas) a amplitude do \u00e2ngulo de $\\overrightarrow{u}$ com $\\overrightarrow{v}$, sabendo que: $\\overrightarrow{u}.\\overrightarrow{v}=50\\sqrt{2}$\u00a0e $\\left\\| \\overrightarrow{u} \\right\\|=\\left\\| \\overrightarrow{v} \\right\\|=10$ $\\overrightarrow{u}.\\overrightarrow{v}=-10\\sqrt{3}$ e $\\left\\| \\overrightarrow{u} \\right\\|=4$ e&#46;&#46;&#46;<\/p>\n","protected":false},"author":1,"featured_media":19380,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[98,97,110],"tags":[422,67,111],"series":[],"class_list":["post-5295","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-11--ano","category-aplicando","category-geometria-analitica","tag-11-o-ano","tag-geometria","tag-vectores"],"views":1791,"jetpack_featured_media_url":"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2010\/11\/Produto_escalar.png","jetpack_sharing_enabled":true,"jetpack_likes_enabled":true,"_links":{"self":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/5295","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=5295"}],"version-history":[{"count":0,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/5295\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/media\/19380"}],"wp:attachment":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=5295"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=5295"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=5295"},{"taxonomy":"series","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fseries&post=5295"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}