(2)设点G(x,0, ),则(FG) ⃗=(x"-" a/2 ",-" a/2 "," z"-" a/2),

　　(FG) ⃗·(CB) ⃗=(x"-" a/2 ",-" a/2 "," z"-" a/2)·(a,0,0)

　　=a(x"-" a/2)=0,∴x=a/2,

　　(FG) ⃗·(CP) ⃗=(x"-" a/2 ",-" a/2 "," z"-" a/2)·(0,-a,a)

　　=a^2/2+a(z"-" a/2)=0,∴ =0.

　　∴点G的坐标为(a/2 "," 0"," 0),即G为AD的中点.

　　(3)设平面DEF的法向量为n=(x,y, ).

　　则{■(n"·" (DF) ⃗=0"," @n"·" (DE) ⃗=0"," )┤即{■(a/2 "(" x+y+z")" =0"," @ax+a/2 y=0"," )┤

　　取x=1,则y=-2, =1,∴n=(1,-2,1).

　　cos<(BD) ⃗,n>=((BD) ⃗"·" n)/("|" (BD) ⃗" " n"|" )=a/(√2 a"·" √6)=√3/6,

　　∴直线DB与平面DEF所成角的正弦值为√3/6.

拓展提升(水平二)

8.如图,在长方体ABCD-A1B1C1D1中,AD=AA1=1,AB=2,E是棱AB的中点,则点E到平面ACD1的距离为(　　).

　　A.1/2 B.√2/2 C.1/3 D.1/6

　　【解析】如图,以D为坐标原点,直线DA,DC,DD1分别为x轴,y轴, 轴建立空间直角坐标系,

　　则点D1(0,0,1),E(1,1,0),A(1,0,0),C(0,2,0),

　　所以(D_1 E) ⃗=(1,1,-1),(AC) ⃗=(-1,2,0),(AD_1 ) ⃗=(-1,0,1).

　　设平面ACD1的法向量为n=(a,b,c),

　　则{■(n"·" (AC) ⃗=0"," @n"·" (AD_1 ) ⃗=0"," )┤即{■("-" a+2b=0"," @"-" a+c=0"," )┤得{■(a=2b"," @a=c"," )┤

　　令a=2,则n=(2,1,2).

　　所以点E到平面ACD1的距离为d=("|" (D_1 E) ⃗"·" n"|" )/("|" n"|" )=1/3.

　　【答案】C

9.如图,已知平面ABCD⊥平面ABEF,四边形ABCD是正方形,四边形ABEF是矩形,且AF=1/2AD=a,G是EF的中点,则GB与平面AGC所成角的正弦值为(　　).