A,B,C是半径为1的球面上的三点
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B\u3001C\u4e24\u70b9\u95f4\u7684\u7403\u9762\u8ddd\u79bb\u4e3a\u03c0/3\uff0c\u53ef\u77e5\u2220BOC=60\u00b0\uff0cBC=1
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V(O-ABC)=1/3*1*sqrt(3)/4=sqrt(3)/12
S(ABC)=1/2|ABxAC|
=1/2|(sqrt(3)/2,,1/2,sqrt(3)/2)|
=7/8
d=(sqrt(3)/4)/(7/8)=2sqrt(3)/7
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绛旓細璁綩(0,0,0),A(0,0,1),B(1,0,0),C(1/2,sqrt(3)/2,0)V(O-ABC)=1/3*1*sqrt(3)/4=sqrt(3)/12 S(ABC)=1/2|ABxAC| =1/2|(sqrt(3)/2,,1/2,sqrt(3)/2)| =7/8 d=(sqrt(3)/4)/(7/8)=2sqrt(3)/7 ...
绛旓細閫氳繃宸茬煡A銆丅銆C鏄崐寰勪负1鐨勭悆闈笂鐨涓夌偣鍙煡OA=OB=OC=1銆侭銆丆涓ょ偣闂寸殑鐞冮潰璺濈涓合/3锛屽彲鐭モ垹BOC=60掳锛孊C=1 鐐笰涓嶣銆丆涓ょ偣闂寸殑鐞冮潰璺濈鍧囦负蟺/2锛屽彲鐭モ垹AOB=鈭燗OC=90掳 浜﹀彲鎺ㄥ嚭OA鈯ラ潰OBC銆侽鍒版埅闈BC鐨勮窛绂伙紝鍗宠繃O鍚戔柍ABC鍋氱殑鍨傜嚎璺濈锛岄氳繃鈻抽潰绉彲浠ョ畻鍑篛鍒版埅闈BC鐨勮窛绂讳负锛...
绛旓細鐞鍗婂緞R=1 瑙扐OB=蟺/2,AB=2sin(蟺/4)=鏍瑰彿2,AC=AB=鏍瑰彿2 瑙払OC=蟺/3,BC=2sin(蟺/6)=1 sin(瑙払AC/2)=(1/2)/(鏍瑰彿2)=1/(2(鏍瑰彿2))=(鏍瑰彿2)/4 cos(瑙払AC/2)=(1-((鏍瑰彿2)/4)^2)^(1/2)=(1/4)(鏍瑰彿(14)sin(瑙払AC)=2sin(瑙払AC/2)cos(瑙払AC/2)=(1/4)(...
绛旓細鍏堟眰鍑哄鸡闀夸负1锛60搴﹀渾蹇冭鍗婂緞绛変簬寮﹂暱锛孉B=AC=BC=1锛屽湪绛夎竟涓夎褰BC涓叾楂樹负鈭3/2锛屽唴蹇冧笌椤剁偣璺濈涓衡垰3/3锛岀悆蹇僌鍒板钩闈BC鐨勮窛绂=鈭1^2-(鈭3/3)^2=鈭6/3.
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绛旓細鐞冨績O涓A锛孊锛孋涓夌偣鏋勬垚姝ｄ笁妫遍敟O-ABC锛屽凡鐭A=OB=OC=R=1锛屸垹AOB=鈭燘OC=鈭燗OC=90掳锛岀敱姝ゅ彲寰桝O鈯ラ潰BOC锛庘埖 S 鈻矪OC = 1 2 锛 S 鈻矨BC = 3 2 锛庤鐞冨績鍒伴潰鐨勮窛绂讳负h锛岀敱V A-BOC =V O-ABC 锛屽緱 h= 3 3 锛庢墍浠ョ悆蹇僌 鍒板钩闈...
