{"id":5314,"date":"2010-11-13T02:25:32","date_gmt":"2010-11-13T02:25:32","guid":{"rendered":"https:\/\/www.acasinhadamatematica.pt\/?p=5314"},"modified":"2022-01-12T11:53:53","modified_gmt":"2022-01-12T11:53:53","slug":"averigue-se-os-vectores-sao-perpendiculares","status":"publish","type":"post","link":"https:\/\/www.acasinhadamatematica.pt\/?p=5314","title":{"rendered":"Averigue se os vetores s\u00e3o perpendiculares"},"content":{"rendered":"<p><ul id='GTTabs_ul_5314' class='GTTabs' style='display:none'>\n<li id='GTTabs_li_0_5314' class='GTTabs_curr'><a  id=\"5314_0\" onMouseOver=\"GTTabsShowLinks('Enunciado'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Enunciado<\/a><\/li>\n<li id='GTTabs_li_1_5314' ><a  id=\"5314_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_5314'>\n<span class='GTTabs_titles'><b>Enunciado<\/b><\/span><\/p>\n<p>Averigue se os vetores $\\overrightarrow{u}$ e $\\overrightarrow{v}$ s\u00e3o perpendiculares:<\/p>\n<ol>\n<li>$\\vec{u}(1,-3,2)$ \u00a0e $\\vec{v}(2,4,5)$<\/li>\n<li>$\\vec{u}(\\sqrt{3}-1,4,-1)$ \u00a0e $\\vec{v}(\\sqrt{3}+1,1,6)$<\/li>\n<li>$\\vec{u}(\\frac{2}{3},-\\frac{3}{2},\\frac{5}{7})$ \u00a0e $\\vec{v}(-\\frac{3}{2},\\frac{2}{3},\\frac{7}{5})$<\/li>\n<li>$\\vec{u}(-5,\\alpha ,3)$\u00a0 e\u00a0\u00a0$\\vec{v}(2\\alpha ,10,0)$<\/li>\n<\/ol>\n<p><div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_next'><a href='#GTTabs_ul_5314' onClick='GTTabs_show(1,5314)'>Resolu\u00e7\u00e3o &gt;&gt;<\/a><\/span><\/div><\/div>\n\n<div class='GTTabs_divs' id='GTTabs_1_5314'>\n<span class='GTTabs_titles'><b>Resolu\u00e7\u00e3o<\/b><\/span><!--more--><\/p>\n<ol>\n<li>Ora, $\\vec{u}.\\vec{v}=(1,-3,2).(2,4,5)=1\\times 2-3\\times 4+2\\times 5=0$.<br \/>\nLogo os vetores s\u00e3o perpendiculares.<br \/>\n\u00ad<\/li>\n<li>Ora, $\\vec{u}.\\vec{v}=(\\sqrt{3}-1,4,-1).(\\sqrt{3}+1,1,6)=(\\sqrt{3}-1)\\times (\\sqrt{3}+1)+4\\times 1-1\\times 6=3-1+4-6=0$.<br \/>\nLogo os vetores s\u00e3o perpendiculares.<br \/>\n\u00ad<\/li>\n<li>Ora, $\\vec{u}.\\vec{v}=(\\frac{2}{3},-\\frac{3}{2},\\frac{5}{7}).(-\\frac{3}{2},\\frac{2}{3},\\frac{7}{5})=\\frac{2}{3}\\times (-\\frac{3}{2})-\\frac{3}{2}\\times \\frac{2}{3}+\\frac{5}{7}\\times \\frac{7}{5}=-1-1+1=-1$.<br \/>\nLogo os vetores n\u00e3o s\u00e3o perpendiculares.<br \/>\n\u00ad<\/li>\n<li>Ora, $\\vec{u}.\\vec{v}=(-5,\\alpha ,3).(2\\alpha ,10,0)=-10\\alpha +10\\alpha +0=0$.<br \/>\nLogo os vetores s\u00e3o perpendiculares.<\/li>\n<\/ol>\n<p><strong><br \/>\nNote<\/strong>:<\/p>\n<p>Da defini\u00e7\u00e3o de produto escalar de dois vetores, $\\overrightarrow{u}.\\overrightarrow{v}=\\left\\| \\overrightarrow{u} \\right\\|.\\left\\| \\overrightarrow{v} \\right\\|.\\cos (\\widehat{\\overrightarrow{u}\\,\\overrightarrow{v}})$, decorre:<\/p>\n<blockquote>\n<p>\\[\\begin{matrix}<br \/>\n\\overrightarrow{u}.\\overrightarrow{v}=0 &amp; \\Leftrightarrow\u00a0 &amp; \\overrightarrow{u}=0 &amp; \\vee\u00a0 &amp; \\overrightarrow{v}=0 &amp; \\vee\u00a0 &amp; \\overrightarrow{u}\\bot \\overrightarrow{v}\u00a0 \\\\<br \/>\n\\end{matrix}\\]<\/p>\n<\/blockquote>\n<div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_prev'><a href='#GTTabs_ul_5314' onClick='GTTabs_show(0,5314)'>&lt;&lt; Enunciado<\/a><\/span><\/div><\/div>\n\n","protected":false},"excerpt":{"rendered":"<p>Enunciado Resolu\u00e7\u00e3o Enunciado Averigue se os vetores $\\overrightarrow{u}$ e $\\overrightarrow{v}$ s\u00e3o perpendiculares: $\\vec{u}(1,-3,2)$ \u00a0e $\\vec{v}(2,4,5)$ $\\vec{u}(\\sqrt{3}-1,4,-1)$ \u00a0e $\\vec{v}(\\sqrt{3}+1,1,6)$ $\\vec{u}(\\frac{2}{3},-\\frac{3}{2},\\frac{5}{7})$ \u00a0e $\\vec{v}(-\\frac{3}{2},\\frac{2}{3},\\frac{7}{5})$ $\\vec{u}(-5,\\alpha ,3)$\u00a0 e\u00a0\u00a0$\\vec{v}(2\\alpha ,10,0)$ Resolu\u00e7\u00e3o &gt;&gt; Resolu\u00e7\u00e3o &lt;&lt; Enunciado<\/p>\n","protected":false},"author":1,"featured_media":19382,"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-5314","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":1704,"jetpack_featured_media_url":"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2010\/11\/Averigue_se_os_vetores_sao_perpendiculares.png","jetpack_sharing_enabled":true,"jetpack_likes_enabled":true,"_links":{"self":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/5314","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=5314"}],"version-history":[{"count":0,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/5314\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/media\/19382"}],"wp:attachment":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=5314"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=5314"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=5314"},{"taxonomy":"series","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fseries&post=5314"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}