【答案】①②③

7.已知正方体ABCD-A'B'C'D',E,F分别为CC',BD的中点.求证:A'F⊥平面BDE.

　　【解析】不妨设正方体的棱长为2,建立空间直角坐标系D-xy (如图),则点B(2,2,0),D(0,0,0),E(0,2,1),A'(2,0,2),F(1,1,0).

　　(A"'" F) ⃗=(-1,1,-2),(DB) ⃗=(2,2,0),(DE) ⃗=(0,2,1).

　　∵(A"'" F) ⃗·(DB) ⃗=0,(A"'" F) ⃗·(DE) ⃗=0,

　　∴(A"'" F) ⃗⊥(DB) ⃗,(A"'" F) ⃗⊥(DE) ⃗.

　　又DB∩DE=D,∴A'F⊥平面BDE.

拓展提升(水平二)

8.如图,已知△PAC是等腰三角形,△ABC是以AC为斜边的等腰直角三角形,E,F,O为PA,PB,AC的中点,PO⊥平面ABC,AC=16,PA=PC=10,点M在平面ABC内,且FM⊥平面BOE,以O为坐标原点,以OB,OC,OP所在直线分别为x轴,y轴, 轴,建立空间直角坐标系O -xy ,则点M的坐标为(　　).

　　A.(4",-" 9/4 "," 0) B.(3,-4,0)

　　C.(3,-9,0) D.(3",-" 9/4 "," 0)

　　【解析】由题意得点O(0,0,0),A(0,-8,0),B(8,0,0),P(0,0,6),因为E,F分别为PA,PB的中点,所以点E(0,-4,3),F(4,0,3).

　　设点M(x,y,0),

　　可得(OB) ⃗=(8,0,0),(OE) ⃗=(0,-4,3),(FM) ⃗=(x-4,y,-3).

　　因为FM⊥平面BOE,

　　所以{■((FM) ⃗"·" (OB) ⃗=0"," @(FM) ⃗"·" (OE) ⃗=0"," )┤所以{■(8"(" x"-" 4")" =0"," @"-" 4y"-" 9=0"," )┤

　　解得{■(x=4"," @y="-" 9/4 "." )┤

　　所以点M的坐标为(4",-" 9/4 "," 0).

　　【答案】A

9.已知点A(0,0,0),B(1,1,1),C(1"," 1/2 "," 1),D(1/2 "," 1"," 1),E(1"," 1"," 1/2),则直线AB与平面CDE的位置关系是.

　　【解析】(AB) ⃗=(1,1,1),(CD) ⃗=("-" 1/2 "," 1/2 "," 0),(CE) ⃗=(0"," 1/2 ",-" 1/2),因为(AB) ⃗·(CD) ⃗=(1,1,1)·("-" 1/2 "," 1/2 "," 0)=0, (AB) ⃗·(CE) ⃗=(1,1,1)·(0"," 1/2 ",-" 1/2)=0,所以AB⊥CD,AB⊥CE.又CD∩CE=C,所以AB⊥平面CDE.

　　【答案】垂直

10.如图,在长方形ABCD中,AB=2, BC=1,E为DC的中点,F为线段EC(端点除外)上一动点.现将△AFD沿AF折起,使平面ABD⊥平面ABC,在平面ABD内过点D作DK⊥AB,K为垂足,设AK=t,则t的取值范围是　　　　.

【解析】