{"id":6404,"date":"2010-12-20T01:49:01","date_gmt":"2010-12-20T01:49:01","guid":{"rendered":"https:\/\/www.acasinhadamatematica.pt\/?p=6404"},"modified":"2022-01-21T23:40:47","modified_gmt":"2022-01-21T23:40:47","slug":"um-octaedro-2","status":"publish","type":"post","link":"https:\/\/www.acasinhadamatematica.pt\/?p=6404","title":{"rendered":"Um octaedro"},"content":{"rendered":"<p><ul id='GTTabs_ul_6404' class='GTTabs' style='display:none'>\n<li id='GTTabs_li_0_6404' class='GTTabs_curr'><a  id=\"6404_0\" onMouseOver=\"GTTabsShowLinks('Enunciado'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Enunciado<\/a><\/li>\n<li id='GTTabs_li_1_6404' ><a  id=\"6404_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_6404'>\n<span class='GTTabs_titles'><b>Enunciado<\/b><\/span><\/p>\n<p><a href=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2010\/12\/pag-189-64.jpg\"><img loading=\"lazy\" decoding=\"async\" data-attachment-id=\"6405\" data-permalink=\"https:\/\/www.acasinhadamatematica.pt\/?attachment_id=6405\" data-orig-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2010\/12\/pag-189-64.jpg\" data-orig-size=\"382,400\" data-comments-opened=\"1\" data-image-meta=\"{&quot;aperture&quot;:&quot;0&quot;,&quot;credit&quot;:&quot;&quot;,&quot;camera&quot;:&quot;HP pstc4380&quot;,&quot;caption&quot;:&quot;&quot;,&quot;created_timestamp&quot;:&quot;0&quot;,&quot;copyright&quot;:&quot;&quot;,&quot;focal_length&quot;:&quot;0&quot;,&quot;iso&quot;:&quot;0&quot;,&quot;shutter_speed&quot;:&quot;0&quot;,&quot;title&quot;:&quot;&quot;}\" data-image-title=\"Octaedro\" data-image-description=\"\" data-image-caption=\"\" data-large-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2010\/12\/pag-189-64.jpg\" class=\"alignright wp-image-6405 size-medium\" title=\"Octaedro\" src=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2010\/12\/pag-189-64-286x300.jpg\" alt=\"\" width=\"286\" height=\"300\" srcset=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2010\/12\/pag-189-64-286x300.jpg 286w, https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2010\/12\/pag-189-64-143x150.jpg 143w, https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2010\/12\/pag-189-64.jpg 382w\" sizes=\"auto, (max-width: 286px) 100vw, 286px\" \/><\/a>Considere o referencial o.n. (O,x,y,z) e o octaedro regular representado na figura.<\/p>\n<p>As arestas [AC], [CD], [DE] e [EA] est\u00e3o contidas no plano xOy e o v\u00e9rtice B pertence ao eixo das cotas. O ponto C tem coordenadas $(2,2,0)$.<\/p>\n<ol>\n<li>Prove que o ponto B tem as coordenadas $(0,0,2\\sqrt{2})$.<\/li>\n<li>Determine uma equa\u00e7\u00e3o do plano ACB.<\/li>\n<li>Considere o plano de equa\u00e7\u00e3o $x+y-2z=4$. Determine a sua intersec\u00e7\u00e3o com o plano xOy e mostre que o ponto C pertence a essa intersec\u00e7\u00e3o.<\/li>\n<li>Determine o \u00e2ngulo que a recta CB faz com CE.<\/li>\n<li>Prove que o plano mediador de [CB] passa em A.<\/li>\n<li>Identifique e escreva uma equa\u00e7\u00e3o do lugar geom\u00e9trico dos pontos do espa\u00e7o definido por $\\overrightarrow{BP}\\,.\\,\\overrightarrow{PF}=0$.<\/li>\n<\/ol>\n<p><div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_next'><a href='#GTTabs_ul_6404' onClick='GTTabs_show(1,6404)'>Resolu\u00e7\u00e3o &gt;&gt;<\/a><\/span><\/div><\/div>\n\n<div class='GTTabs_divs' id='GTTabs_1_6404'>\n<span class='GTTabs_titles'><b>Resolu\u00e7\u00e3o<\/b><\/span><!--more--><\/p>\n<ol>\n<li><a href=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2010\/12\/pag-189-64.jpg\"><img loading=\"lazy\" decoding=\"async\" data-attachment-id=\"6405\" data-permalink=\"https:\/\/www.acasinhadamatematica.pt\/?attachment_id=6405\" data-orig-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2010\/12\/pag-189-64.jpg\" data-orig-size=\"382,400\" data-comments-opened=\"1\" data-image-meta=\"{&quot;aperture&quot;:&quot;0&quot;,&quot;credit&quot;:&quot;&quot;,&quot;camera&quot;:&quot;HP pstc4380&quot;,&quot;caption&quot;:&quot;&quot;,&quot;created_timestamp&quot;:&quot;0&quot;,&quot;copyright&quot;:&quot;&quot;,&quot;focal_length&quot;:&quot;0&quot;,&quot;iso&quot;:&quot;0&quot;,&quot;shutter_speed&quot;:&quot;0&quot;,&quot;title&quot;:&quot;&quot;}\" data-image-title=\"Octaedro\" data-image-description=\"\" data-image-caption=\"\" data-large-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2010\/12\/pag-189-64.jpg\" class=\"alignright wp-image-6405 size-medium\" title=\"Octaedro\" src=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2010\/12\/pag-189-64-286x300.jpg\" alt=\"\" width=\"286\" height=\"300\" srcset=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2010\/12\/pag-189-64-286x300.jpg 286w, https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2010\/12\/pag-189-64-143x150.jpg 143w, https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2010\/12\/pag-189-64.jpg 382w\" sizes=\"auto, (max-width: 286px) 100vw, 286px\" \/><\/a>Sendo o octaedro regular, temos $A\\,(2,-2,0)$ e $\\overline{AC}=4$.\n<p>Logo, $\\overline{OB}=\\sqrt{{{\\overline{BC}}^{2}}-{{\\overline{OC}}^{2}}}=\\sqrt{{{4}^{2}}-{{\\left( \\sqrt{{{2}^{2}}+{{2}^{2}}} \\right)}^{2}}}=2\\sqrt{2}$.<\/p>\n<p>Ent\u00e3o, $B\\,(0,0,2\\sqrt{2})$.<br \/>\n\u00ad<\/p>\n<\/li>\n<li>Sendo $A\\,(2,-2,0)$, $B\\,(0,0,2\\sqrt{2})$ e $C\\,(2,2,0)$, ent\u00e3o $\\overrightarrow{AB}=(-2,2,2\\sqrt{2})$ e $\\overrightarrow{AC}=(0,4,0)$.\n<p>Comecemos por determinar um vetor $\\vec{n}=(a,b,c)$ \u00a0normal ao plano ACB, isto \u00e9 , um vetor $\\vec{n}=(a,b,c)$\u00a0 tal que $\\vec{n}\\bot \\overrightarrow{AB}\\wedge \\vec{n}\\bot \\overrightarrow{AC}$ .<\/p>\n<p>Ora,<\/p>\n<p>$\\begin{array}{*{35}{l}}<br \/>\n\\left\\{ \\begin{array}{*{35}{l}}<br \/>\n\\vec{n}.\\overrightarrow{AB}=0\u00a0 \\\\<br \/>\n\\vec{n}.\\overrightarrow{AC}=0\u00a0 \\\\<br \/>\n\\end{array} \\right. &amp; \\Leftrightarrow\u00a0 &amp; \\left\\{ \\begin{array}{*{35}{l}}<br \/>\n-2a+2b+2\\sqrt{2}c=0\u00a0 \\\\<br \/>\n4b=0\u00a0 \\\\<br \/>\n\\end{array} \\right. &amp; \\Leftrightarrow\u00a0 &amp; \\left\\{ \\begin{array}{*{35}{l}}<br \/>\na=\\sqrt{2}c\u00a0 \\\\<br \/>\nb=0\u00a0 \\\\<br \/>\n\\end{array} \\right.\u00a0 \\\\<br \/>\n\\end{array}$.<\/p>\n<p>Logo, $\\overrightarrow{{{n}_{1}}}=(\\sqrt{2},0,1)$, por exemplo, \u00e9 um vetor normal ao plano ACB.<\/p>\n<p>Sendo assim, a equa\u00e7\u00e3o do plano ACB \u00e9 da forma $\\sqrt{2}x+z+d=0$.<br \/>\nComo o ponto C \u00e9 um ponto desse plano, ent\u00e3o $0+2\\sqrt{2}+d=0\\Leftrightarrow d=-2\\sqrt{2}$.<\/p>\n<p>Logo,\u00a0$\\sqrt{2}x+z-2\\sqrt{2}=0$ \u00e9 uma equa\u00e7\u00e3o do plano ACB.<br \/>\n\u00ad<\/p>\n<\/li>\n<li>O plano xOy pode ser definido por $z=0$.\n<p>Ora,<br \/>\n$\\begin{array}{*{35}{l}}<br \/>\n\\left\\{ \\begin{array}{*{35}{l}}<br \/>\nx+y-2z=4\u00a0 \\\\<br \/>\nz=0\u00a0 \\\\<br \/>\n\\end{array} \\right. &amp; \\Leftrightarrow\u00a0 &amp; \\left\\{ \\begin{array}{*{35}{l}}<br \/>\nx+y=4\u00a0 \\\\<br \/>\nz=0\u00a0 \\\\<br \/>\n\\end{array} \\right. &amp; \\Leftrightarrow\u00a0 &amp; \\begin{matrix}<br \/>\nx+y=4 &amp; \\wedge\u00a0 &amp; z=0\u00a0 \\\\<br \/>\n\\end{matrix}\u00a0 \\\\<br \/>\n\\end{array}$.<\/p>\n<p>Portanto, $\\begin{matrix}<br \/>\nx+y=4 &amp; \\wedge\u00a0 &amp; z=0\u00a0 \\\\<br \/>\n\\end{matrix}$ define a reta de intersec\u00e7\u00e3o dos dois planos.<\/p>\n<p>O ponto C pertence a essa reta, pois as suas coordenadas verificam a condi\u00e7\u00e3o anterior: $\\begin{matrix}<br \/>\n2+2=4 &amp; \\wedge\u00a0 &amp; 0=0\u00a0 \\\\<br \/>\n\\end{matrix}$.<br \/>\n\u00ad<\/p>\n<\/li>\n<li>Como $B\\,(0,0,2\\sqrt{2})$, $C\\,(2,2,0)$ e $E(-2,-2,0)$, ent\u00e3o $\\overrightarrow{CB}=(-2,-2,2\\sqrt{2})$ e $\\overrightarrow{CE}=(-4,-4,0)$.\n<p>Logo, \\[\\cos (\\overrightarrow{CB}\\overset{\\hat{\\ }}{\\mathop{{}}}\\,\\overrightarrow{CE})=\\frac{(-2,-2,2\\sqrt{2}).(-4,-4,0)}{\\sqrt{4+4+8}\\times \\sqrt{16+16}}=\\frac{8+8+0}{4\\times 4\\sqrt{2}}=\\frac{\\sqrt{2}}{2}\\]<\/p>\n<p>Portanto, a amplitude do \u00e2ngulo entre as retas CB e CE \u00e9 $\\alpha =\\overrightarrow{CB}\\overset{\\hat{\\ }}{\\mathop{{}}}\\,\\overrightarrow{CE}=45{}^\\text{o}$.<br \/>\n(Note que [EBCF] \u00e9 um quadrado, logo as suas diagonais fazem \u00e2ngulos de 45\u00ba com os seus lados.)<br \/>\n\u00ad<\/p>\n<\/li>\n<li>O ponto A pertence ao plano mediador do segmento [CB], pois $\\overline{AB}=\\overline{AC}$, visto as arestas de um octaedro regular serem geometricamente iguais.<br \/>\n\u00ad<\/li>\n<li>A condi\u00e7\u00e3o $\\overrightarrow{BP}\\,.\\,\\overrightarrow{PF}=0$ define a superf\u00edcie esf\u00e9rica de di\u00e2metro [BF].\n<p>Com efeito, temos: \\[\\begin{array}{*{35}{l}}<br \/>\n\\overrightarrow{BP}\\,.\\,\\overrightarrow{PF}=0 &amp; \\Leftrightarrow\u00a0 &amp; (x,y,z-2\\sqrt{2}).(x,y,z+2\\sqrt{2})=0\u00a0 \\\\<br \/>\n{} &amp; \\Leftrightarrow\u00a0 &amp; {{x}^{2}}+{{y}^{2}}+{{z}^{2}}-{{(2\\sqrt{2})}^{2}}=0\u00a0 \\\\<br \/>\n{} &amp; \\Leftrightarrow\u00a0 &amp; {{x}^{2}}+{{y}^{2}}+{{z}^{2}}=8\u00a0 \\\\<br \/>\n\\end{array}\\]<\/p>\n<\/li>\n<\/ol>\n<div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_prev'><a href='#GTTabs_ul_6404' onClick='GTTabs_show(0,6404)'>&lt;&lt; Enunciado<\/a><\/span><\/div><\/div>\n\n","protected":false},"excerpt":{"rendered":"<p>Enunciado Resolu\u00e7\u00e3o Enunciado Considere o referencial o.n. (O,x,y,z) e o octaedro regular representado na figura. As arestas [AC], [CD], [DE] e [EA] est\u00e3o contidas no plano xOy e o v\u00e9rtice B pertence ao eixo&#46;&#46;&#46;<\/p>\n","protected":false},"author":1,"featured_media":20848,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_newsletter_access":"","_jetpack_dont_email_post_to_subs":false,"_jetpack_newsletter_tier_id":0,"_jetpack_memberships_contains_paywalled_content":false,"_jetpack_feature_clip_id":0,"_jetpack_memberships_contains_paid_content":false,"footnotes":"","jetpack_post_was_ever_published":false},"categories":[98,97,110],"tags":[422,67,119,120],"series":[],"class_list":["post-6404","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-interseccao-de-planos","tag-plano-mediador"],"views":4148,"jetpack_featured_media_url":"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2010\/12\/11V1Pag189-64_520x245.png","jetpack_sharing_enabled":true,"jetpack_likes_enabled":true,"_links":{"self":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/6404","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=6404"}],"version-history":[{"count":0,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/6404\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/media\/20848"}],"wp:attachment":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=6404"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=6404"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=6404"},{"taxonomy":"series","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fseries&post=6404"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}