已知球O的半径为1,A,B,C三点都在球面上,且每两点间的球面距离为π2,则球心O到平面ABC的距离为
\u5df2\u77e5\u7403O\u7684\u534a\u5f84\u4e3a1\uff0cA\u3001B\u3001C\u4e09\u70b9\u90fd\u5728\u7403\u9762\u4e0a\uff0c\u4e14\u6bcf\u4e24\u70b9\u95f4\u7684\u7403\u9762\u8ddd\u79bb\u5747\u4e3a\u03c0/2\uff0c\u5219\u7403\u5fc3O\u5230\u5e73\u9762ABC\u7684\u8ddd\u79bb\u4e3a\u591a\u5c11A\u3001B\u3001C\u4e09\u70b9\u90fd\u5728\u7403\u9762\u4e0a\uff0c\u4e14\u6bcf\u4e24\u70b9\u95f4\u7684\u7403\u9762\u8ddd\u79bb\u5747\u4e3a\u03c0/2
\u6240\u4ee5\uff1a\u2220AOB=\u2220AOC=\u2220BOC=90º
\u6240\u4ee5\uff1a\u4e09\u89d2\u5f62ABC\u662f\u7b49\u8fb9\u4e09\u89d2\u5f62
\u8fc7O\u4f5cOD\u5782\u76f4\u4e8e\u9762ABC
D\u70b9\u6b63\u597d\u843d\u5728\u4e09\u89d2\u5f62ABC\u7684\u4e2d\u5fc3
\u5df2\u77e5\u7403O\u7684\u534a\u5f84\u4e3a1
\u6240\u4ee5\uff1aAB=AC=BC=\u221a2
AD=\u221a6 / 3
OD=\u221a3 / 3
\u7403\u5fc3O\u4e0eA\uff0cB\uff0cC\u4e09\u70b9\u6784\u6210\u6b63\u4e09\u68f1\u9525O-ABC\uff0c\u5982\u56fe\u6240\u793a\uff0c\u5df2\u77e5OA=OB=OC=R=1\uff0c\u2220AOB=\u2220BOC=\u2220AOC=90\u00b0\uff0c\u7531\u6b64\u53ef\u5f97AO\u22a5\u9762BOC\uff0e\u2235 S \u25b3BOC = 1 2 \uff0c S \u25b3ABC = 3 2 \uff0e\u2234\u7531V A-BOC =V O-ABC \uff0c\u5f97 h= 3 3 \uff0e\u6545\u7b54\u6848\u4e3a\uff1a 3 3
解:球心O与A,B,C三点构成正三棱锥O-ABC,如图所示,已知OA=OB=OC=R=1,∠AOB=∠BOC=∠AOC=90°,
由此可得AO⊥面BOC.
∵S△BOC=
1 |
2 |
|