{"id":1442,"date":"2015-04-25T17:11:01","date_gmt":"2015-04-25T16:11:01","guid":{"rendered":"http:\/\/www.unimath.fr\/?p=1442"},"modified":"2015-04-29T19:59:13","modified_gmt":"2015-04-29T18:59:13","slug":"1442","status":"publish","type":"post","link":"http:\/\/www.unimath.fr\/?p=1442","title":{"rendered":"Exercice \u00e0 trous : Orthogonalit\u00e9 dans l’espace"},"content":{"rendered":"

\"exercice-trous-tetraedre\"<\/a><\/p>\n

Pour le sujet :\u00a0Cliquer ici<\/a><\/p>\n

OABC est un t\u00e9tra\u00e8dre.
\nLes triangles AOB, AOC et BOC sont rectangles en O.
\nI est le projet\u00e9 orthogonal de O sur (BC)
\net H celui de O sur (AI).<\/p>\n

Question 1 :<\/strong> Montrer que (AI) est une hauteur du triangle ABC.
\n<\/strong>Premi\u00e8re \u00e9tape :<\/strong>\u00a0D’apr\u00e8s l’\u00e9nonc\u00e9, on a :\u00a0(OA) orthogonale aux deux droites \u00a0[peekaboo_link name=\u00a0\u00bbquestion1″]…[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestion1″](OB) et (OC)\u00a0<\/span>[\/peekaboo_content]donc (OA) est orthogonale \u00e0 deux\u00a0[peekaboo_link name=\u00a0\u00bbquestion3″]…[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestion3″]droites s\u00e9cantes\u00a0<\/span>[\/peekaboo_content] du plan (OBC) donc (OA) est orthogonale au plan [peekaboo_link name=\u00a0\u00bbquestion4″]…[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestion4″]\u00a0(OBC)\u00a0<\/span>[\/peekaboo_content]On en d\u00e9duit donc que (OA) est [peekaboo_link name=\u00a0\u00bbquestion5″]…[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestion5″]orthogonale\u00a0<\/span>[\/peekaboo_content] \u00e0 (BC)\u00a0car [peekaboo_link name=\u00a0\u00bbquestion6″]…[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestion6″](OA) est orthogonale au plan (OBC) et donc orthogonale \u00e0 toute droite de ce plan\u00a0<\/span>[\/peekaboo_content]<\/p>\n

Deuxi\u00e8me \u00e9tape :<\/strong>\u00a0D’apr\u00e8s l’\u00e9nonc\u00e9, (BC) et (OI) sont [peekaboo_link name=\u00a0\u00bbquestion7″]…[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestion7″]orthogonales (ou perpendiculaires car s\u00e9cantes) \u00a0<\/span>[\/peekaboo_content] donc (BC) est orthogonale aux deux droites (OI) et (OA) \u00a0donc (BC) est orthogonale au plan\u00a0[peekaboo_link name=\u00a0\u00bbquestion10″]…[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestion10″](OAI)\u00a0<\/span>[\/peekaboo_content] car\u00a0[peekaboo_link name=\u00a0\u00bbquestion11″]…[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestion11″](BC) est othogonale \u00e0 (OI) et (OA) qui sont deux droites s\u00e9cantes du plan (OAI)\u00a0<\/span>[\/peekaboo_content]<\/p>\n

Conclusion :\u00a0(BC) est orthogonale au plan (OAI)\u00a0\u00a0et donc [peekaboo_link name=\u00a0\u00bbquestion15″]…[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestion15″]orthogonale\u00a0<\/span>[\/peekaboo_content]\u00a0\u00e0 toute [peekaboo_link name=\u00a0\u00bbquestion16″]…[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestion16″]<\/span>droite\u00a0<\/span>[\/peekaboo_content] de ce plan.
\n<\/span>On en d\u00e9duit donc que [peekaboo_link name=\u00a0\u00bbquestion17″]…[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestion17″](BC) est orthogonale \u00e0 la droite (AI) ce qui prouve que (AI) est une hauteur du triangle ABC\u00a0<\/span>[\/peekaboo_content]<\/p>\n

Question 2 :<\/strong> Montrer que (OH) est perpendiculaire au plan (ABC).
\n<\/strong>Pour cela, on va montrer que (OH) est orthogonale \u00e0 [peekaboo_link name=\u00a0\u00bbquestion18″]…[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestion18″]deux droites s\u00e9cantes du plan (ABC)\u00a0<\/span>[\/peekaboo_content]<\/p>\n

D’apr\u00e8s l’\u00e9nonc\u00e9, la droite (OH) est orthogonale \u00e0 la droite [peekaboo_link name=\u00a0\u00bbquestion19″]…[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestion19″](AI)\u00a0<\/span>[\/peekaboo_content]
\nOn a d\u00e9montr\u00e9 que la droite (BC) \u00e9tait orthogonale au plan\u00a0(OAI)\u00a0donc orthogonale \u00e0 toute [peekaboo_link name=\u00a0\u00bbquestion22″]…[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestion22″]droite de ce plan\u00a0<\/span>[\/peekaboo_content]et donc (BC) est orthogonale \u00e0 la droite [peekaboo_link name=\u00a0\u00bbquestion23″]…[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestion23″](OH)\u00a0<\/span>[\/peekaboo_content]
\nConclusion : la droite (OH) est orthogonale \u00e0 la droite (BC) et \u00e0 la droite (AI) et donc (OH) est orthogonale au plan (ABC) car [peekaboo_link name=\u00a0\u00bbquestion27″]…[\/peekaboo_link][peekaboo_content name=\u00a0\u00bbquestion27″](OH) est orthogonale \u00e0 (BC) et (AI) qui sont deux droites s\u00e9cantes du plan (ABC)\u00a0<\/span>[\/peekaboo_content]<\/p>\n","protected":false},"excerpt":{"rendered":"

Pour le sujet :\u00a0Cliquer ici OABC est un t\u00e9tra\u00e8dre. Les triangles AOB, AOC et BOC sont rectangles en O. I est le projet\u00e9 orthogonal de O sur (BC) et H celui de O sur (AI). Question 1 : Montrer que (AI) est une hauteur du triangle ABC. Premi\u00e8re \u00e9tape :\u00a0D’apr\u00e8s l’\u00e9nonc\u00e9, on a :\u00a0(OA) orthogonale […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":[],"categories":[1],"tags":[],"_links":{"self":[{"href":"http:\/\/www.unimath.fr\/index.php?rest_route=\/wp\/v2\/posts\/1442"}],"collection":[{"href":"http:\/\/www.unimath.fr\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"http:\/\/www.unimath.fr\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"http:\/\/www.unimath.fr\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"http:\/\/www.unimath.fr\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=1442"}],"version-history":[{"count":31,"href":"http:\/\/www.unimath.fr\/index.php?rest_route=\/wp\/v2\/posts\/1442\/revisions"}],"predecessor-version":[{"id":1480,"href":"http:\/\/www.unimath.fr\/index.php?rest_route=\/wp\/v2\/posts\/1442\/revisions\/1480"}],"wp:attachment":[{"href":"http:\/\/www.unimath.fr\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=1442"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"http:\/\/www.unimath.fr\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=1442"},{"taxonomy":"post_tag","embeddable":true,"href":"http:\/\/www.unimath.fr\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=1442"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}