已知SA,SB,SC是不在同一平面内的三条射线，且 ∠ASB＝ ∠BSC＝ ∠ASC＝π/3,SA＝2√3,求点A到平面SBC的...
\u5df2\u77e5SA\uff0cSB\uff0cSC\u662f\u4e0d\u5728\u540c\u4e00\u5e73\u9762\u7684\u4e09\u6761\u5c04\u7ebf\uff0c\u4e14\u2220ASB=\u2220BSC=\u2220ASC=60\u00b0 SA=2\u221a3\u3002\u6c42\u70b9A\u5230\u5e73\u9762SBC\u7684\u8ddd\u79bb\u30022\u6839\u53f72
\u53d6BC\u7684\u4e2d\u70b9D\uff0c\u7136\u540e\u8fde\u63a5AD,SD\uff0c
\u9996\u5148SD\u22a5BC,AD\u22a5BC\uff0c
\u63a5\u7740\u6839\u636e\u5df2\u77e5\u7684\u6570\u91cf\u5173\u7cfb\u8bbeSA=2a\uff0c\u628aSD,AD\u8868\u793a\u51fa\u6765\uff0c
\u6700\u540e\u7528\u52fe\u80a1\u5b9a\u7406\u8bc1\u660eAD\u22a5SD\uff0c
\u5373\u5f97\u5230\u5e73\u9762ABC\u5782\u76f4\u5e73\u9762SBC
则S-ABC为正四面体
若正四面体的棱长为a，则高为√6a/3
所以点A到平面SBC的距离 = 2√3 * √6/3 = 2√2
即求A到SBC面的中点的距离
设中点为o，so的连线交bc于q
sq=3，oq=1，bc=2√3,ob=2，ab=2√3,所以ao=2√2
绛旓細浠SB=SC=SA=2鈭3 鍒橲-ABC涓烘鍥涢潰浣 鑻ユ鍥涢潰浣撶殑妫遍暱涓篴锛屽垯楂樹负鈭6a/3 鎵浠ョ偣A鍒板钩闈BC鐨勮窛绂 = 2鈭3 * 鈭6/3 = 2鈭2
绛旓細璇佹槑锛氳SC=a锛SA=b锛屽垯AB=b锛SB=2b锛庡張AC2=a2+b2-2abcos45掳=a2+b2-2abBC2=a2+(2b)2-22bacos60掳=a2+2b2-2ab锛庘埓AB2+AC2=b2+a2+b2-2ab=a2+2b2-2ab=BC2鈭粹垹BAC=90掳锛屽嵆AB鈯C锛庡張AB鈯A锛屼笖AC鈭㏒A=A锛屸埓AB鈯ュ钩闈AC锛屽張SC?骞抽潰SAC锛屸埓AB鈯C锛
绛旓細鍏堜綔浜岄潰瑙払-SA-C鐨勫钩闈㈣锛屾牴鎹粰瀹氱殑鏉′欢锛屽湪妫盨涓婂彇涓鐐筆锛屽垎鍒槸鍦ㄤ袱涓钩闈㈠唴浣滅洿绾夸笌妫卞瀭鐩磋瘉鏄庯細鍦⊿A涓婂彇涓鐐筆杩嘝浣淧R鈯A浜SC浜嶳杩嘝浣淧Q鈯A浜SB浜嶲鈭粹垹QPR涓轰簩闈㈣B-SA-C鐨勫钩闈㈣璁綪S=a鈭碘垹PSQ=45掳锛屸垹SPQ=90掳鈭碢Q=a锛孲Q= a鍚岀悊PR= a锛孲R= a鈭碘垹PSQ=60掳锛孲R=S...
绛旓細鈭SA=SB=SC,鈭燗SB=鈭燗SC=60掳 鈭粹柍SAB涓庘柍SAC鏄涓夎褰 鈭碅B=SA=AC 鈻矨BC鏄瓑鑵颁笁瑙掑舰 鈭礝鏄疊C涓偣 鈭碅O鈯C AO^2=AC^2-OC^2 鑰屸垹BSC=90掳 鍒欌柍SBC鏄瓑鑵扮洿瑙掍笁瑙掑舰 鈭礝鏄疊C涓偣 鈭碨O=1/2BC=OC SO^2=OC^2 鍒橝O^2+SO^2=AC^2=AS^2 鍗矨O鈯O 鍙堚埖AO鈯C,SO鈭〣C=O ...
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