题目内容
已知正三棱锥P-ABC,点P,A,B,C都在半径为
的球面上.若PA,PB,PC两两相互垂直,则球心到截面ABC的距离为________.
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本题主要考查球的概念与性质.解题的突破口为解决好点P到截面ABC的距离.
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由已知条件可知,以PA,PB,PC为棱的正三棱锥可以补充成球的内接正方体,故而PA2+PB2+PC2=
, 由已知PA=PB=PC, 得到PA=PB=PC=2, 因为VP-ABC=VA-PBC⇒
h·S△ABC=
PA·S△PBC, 得到h=
,故而球心到截面ABC的距离为R-h=
.
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由已知条件可知,以PA,PB,PC为棱的正三棱锥可以补充成球的内接正方体,故而PA2+PB2+PC2=
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