{"id":6368,"date":"2010-12-12T20:04:03","date_gmt":"2010-12-12T20:04:03","guid":{"rendered":"https:\/\/www.acasinhadamatematica.pt\/?p=6368"},"modified":"2022-01-21T18:55:30","modified_gmt":"2022-01-21T18:55:30","slug":"um-octaedro","status":"publish","type":"post","link":"https:\/\/www.acasinhadamatematica.pt\/?p=6368","title":{"rendered":"Um octaedro"},"content":{"rendered":"<p><ul id='GTTabs_ul_6368' class='GTTabs' style='display:none'>\n<li id='GTTabs_li_0_6368' class='GTTabs_curr'><a  id=\"6368_0\" onMouseOver=\"GTTabsShowLinks('Enunciado'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Enunciado<\/a><\/li>\n<li id='GTTabs_li_1_6368' ><a  id=\"6368_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_6368'>\n<span class='GTTabs_titles'><b>Enunciado<\/b><\/span><\/p>\n<p><a href=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2010\/12\/pag185-45.jpg\"><img loading=\"lazy\" decoding=\"async\" data-attachment-id=\"6369\" data-permalink=\"https:\/\/www.acasinhadamatematica.pt\/?attachment_id=6369\" data-orig-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2010\/12\/pag185-45.jpg\" data-orig-size=\"337,347\" 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\/pag185-45.jpg\" class=\"alignright wp-image-6369\" title=\"Octaedro\" src=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2010\/12\/pag185-45-291x300.jpg\" alt=\"\" width=\"240\" height=\"247\" srcset=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2010\/12\/pag185-45-291x300.jpg 291w, https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2010\/12\/pag185-45-145x150.jpg 145w, https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2010\/12\/pag185-45.jpg 337w\" sizes=\"auto, (max-width: 240px) 100vw, 240px\" \/><\/a>Na figura est\u00e1 representado, em referencial o.n. Oxyz, um octaedro [ABCDEF].<\/p>\n<p>Sabe-se que:<\/p>\n<ul>\n<li>O v\u00e9rtice B tem coordenadas (1,0,1).<\/li>\n<li>O v\u00e9rtice E tem coordenadas (0,1,1).<\/li>\n<li>O v\u00e9rtice F pertence ao plano xOy.<\/li>\n<li>O v\u00e9rtice A tem coordenadas (1,1,2).<\/li>\n<\/ul>\n<ol>\n<li>Mostre que a reta definida pela condi\u00e7\u00e3o $x=y=z$ \u00e9 perpendicular ao plano ACD.<\/li>\n<li>Determine uma equa\u00e7\u00e3o da superf\u00edcie esf\u00e9rica que cont\u00e9m os seis v\u00e9rtices do octaedro.<\/li>\n<li>Seja $\\alpha $ o plano definido pelo eixo Oz e pelo ponto A.<br \/>\nA sec\u00e7\u00e3o produzida no octaedro pelo plano $\\alpha $ \u00e9 um quadril\u00e1tero. Caracterize esse quadril\u00e1tero e determine o seu per\u00edmetro.<\/li>\n<\/ol>\n<p><div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_next'><a href='#GTTabs_ul_6368' onClick='GTTabs_show(1,6368)'>Resolu\u00e7\u00e3o &gt;&gt;<\/a><\/span><\/div><\/div>\n\n<div class='GTTabs_divs' id='GTTabs_1_6368'>\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\/pag185-45.jpg\"><img loading=\"lazy\" decoding=\"async\" data-attachment-id=\"6369\" data-permalink=\"https:\/\/www.acasinhadamatematica.pt\/?attachment_id=6369\" data-orig-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2010\/12\/pag185-45.jpg\" data-orig-size=\"337,347\" 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\/pag185-45.jpg\" class=\"alignright wp-image-6369\" title=\"Octaedro\" src=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2010\/12\/pag185-45-291x300.jpg\" alt=\"\" width=\"240\" height=\"247\" srcset=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2010\/12\/pag185-45-291x300.jpg 291w, https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2010\/12\/pag185-45-145x150.jpg 145w, https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2010\/12\/pag185-45.jpg 337w\" sizes=\"auto, (max-width: 240px) 100vw, 240px\" \/><\/a>Como $x=y=z\\Leftrightarrow \\frac{x-0}{1}=\\frac{y-0}{1}=\\frac{z-0}{1}$, ent\u00e3o o vetor $\\vec{r}\\,(1,1,1)$\u00a0 \u00e9 diretor da reta dada.\n<p>Ora, $A\\,(1,1,2)$, $C\\,(2,1,1)$ e $D\\,(1,2,1)$.<br \/>\nLogo, $\\overrightarrow{AC}=(1,0,-1)$ e $\\overrightarrow{AD}=(0,1,-1)$.<\/p>\n<p>Vejamos se o vetor\u00a0\u00a0$\\vec{r}$\u00a0 \u00e9 perpendicular aos vetores $\\overrightarrow{AC}$ e $\\overrightarrow{AD}$:<\/p>\n<p>$\\vec{r}.\\overrightarrow{AC}=(1,1,1).(1,0,-1)=1+0-1=0$<br \/>\n$\\vec{r}.\\overrightarrow{AD}=(1,1,1).(0,1,-1)=0+1-1=0$<\/p>\n<p>Como o vetor\u00a0\u00a0$\\vec{r}$\u00a0 \u00e9 perpendicular aos vetores $\\overrightarrow{AC}$ e $\\overrightarrow{AD}$, ent\u00e3o a reta dada \u00e9 perpendicular ao plano ACD, pois \u00e9 perpendicular a duas retas concorrentes (AC e AD) desse plano.<br \/>\n\u00ad<\/p>\n<\/li>\n<li>\n<p>O centro dessa superf\u00edcie esf\u00e9rica \u00e9 o ponto $G\\,(1,1,1)$ e o raio \u00e9 $r=\\overline{GA}=1$.<\/p>\n<p>Logo, uma equa\u00e7\u00e3o dessa superf\u00edcie esf\u00e9rica \u00e9 ${{(x-1)}^{2}}+{{(y-1)}^{2}}+{{(z-1)}^{2}}=1$.<br \/>\n\u00ad<\/p>\n<\/li>\n<li>\n<p>Esse quadril\u00e1tero \u00e9 [AHFIA], de diagonais [AF] e [HI], sendo H e I os pontos m\u00e9dios dos segmentos [CD] e [BE], respetivamente.<\/p>\n<p>Os lados desse quadril\u00e1tero s\u00e3o iguais e as diagonais s\u00e3o perpendiculares, mas com comprimentos diferentes. (Note que $\\overline{AF}=2$ e $\\overline{HI}=\\overline{BC}=\\sqrt{2}$)<br \/>\nLogo, trata-se de um losango.<\/p>\n<p>O ponto m\u00e9dio do segmento [CD] \u00e9 $H\\,(\\frac{2+1}{2},\\frac{1+2}{2},\\frac{1+1}{2})=(\\frac{3}{2},\\frac{3}{2},1)$.<\/p>\n<p>Portanto, o per\u00edmetro desse losango \u00e9: \\[P = 4 \\times \\overline {AH}\u00a0 = 4 \\times \\sqrt {{{(1 &#8211; \\frac{3}{2})}^2} + {{(1 &#8211; \\frac{3}{2})}^2} + {{(2 &#8211; 1)}^2}}\u00a0 = 4 \\times \\sqrt {\\frac{1}{4} + \\frac{1}{4} + 1}\u00a0 = 4\\frac{{\\sqrt 3 }}{{\\sqrt 2 }} = 2\\sqrt 6 \\]<\/p>\n<\/li>\n<\/ol>\n<div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_prev'><a href='#GTTabs_ul_6368' onClick='GTTabs_show(0,6368)'>&lt;&lt; Enunciado<\/a><\/span><\/div><\/div>\n\n","protected":false},"excerpt":{"rendered":"<p>Enunciado Resolu\u00e7\u00e3o Enunciado Na figura est\u00e1 representado, em referencial o.n. Oxyz, um octaedro [ABCDEF]. Sabe-se que: O v\u00e9rtice B tem coordenadas (1,0,1). O v\u00e9rtice E tem coordenadas (0,1,1). O v\u00e9rtice F pertence ao plano&#46;&#46;&#46;<\/p>\n","protected":false},"author":1,"featured_media":20832,"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],"series":[],"class_list":["post-6368","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"],"views":3845,"jetpack_featured_media_url":"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2010\/12\/11V1Pag185-45_520x245.png","jetpack_sharing_enabled":true,"jetpack_likes_enabled":true,"_links":{"self":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/6368","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=6368"}],"version-history":[{"count":0,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/6368\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/media\/20832"}],"wp:attachment":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=6368"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=6368"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=6368"},{"taxonomy":"series","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fseries&post=6368"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}