问题 解答题
四棱锥P-ABCD的底面ABCD是平行四边形,
AB
=(-1,2,1),
AD
=(0,-2,3),
AP
═(8,3,2),
(1)求证:PA⊥底面ABCD;
(2)求PC的长.
答案

证明:(1)∵

AB
=(-1,2,1),
AD
=(0,-2,3),
AP
═(8,3,2),

AP
AB
=0,
AP
AD
=0

AP
AB
AP
AD

即AP⊥AB且AP⊥AD,

又∵AB∩AD=A

∴AP⊥平面ABCD;

(2)∵

AB
=(-1,2,1),
AD
=(0,-2,3),
AP
═(8,3,2),

AC
=
AB
+
AD
=(-1,0,4),
PC
=
AP
-
AC
=(9,3,-2)

|PC|=

94

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