∴(MN) ⃗=("-" 3/4 a"," 0"," a/2).取n=(0,1,0),

　　显然n⊥平面A1D1DA,且(MN) ⃗·n=0,

　　∴(MN) ⃗⊥n.又MN⊄平面ADD1A1,

　　∴MN∥平面ADD1A1.

10如图,在四棱锥P-ABCD中,底面ABCD是边长为1的菱形,∠ABC=π/4,PA⊥底面ABCD,PA=2,点M为PA的中点,点N为BC的中点.AF⊥CD于点F,如图建立空间直角坐标系.求出平面PCD的一个法向量并证明MN∥平面PCD.

解:由题设知,在Rt△AFD中,AF=FD=√2/2,A(0,0,0),B(1,0,0),F(0"," √2/2 "," 0),D("-" √2/2 "," √2/2 "," 0),P(0,0,2),M(0,0,1),N(1"-" √2/4 "," √2/4 "," 0).

　　(MN) ⃗=(1"-" √2/4 "," √2/4 ",-" 1),(PF) ⃗=(0"," √2/2 ",-" 2),

　　(PD) ⃗=("-" √2/2 "," √2/2 ",-" 2).

　　设平面PCD的一个法向量为n=(x,y, ),

　　则{■(n"·" (PF) ⃗=0"," @n"·" (PD) ⃗=0)┤⇒{■(√2/2 y"-" 2z=0"," @"-" √2/2 x+√2/2 y"-" 2z=0"," )┤

　　令 =√2,得n=(0,4,√2).

　　因为(MN) ⃗·n=(1"-" √2/4 "," √2/4 ",-" 1)·(0,4,√2)=0,且MN⊄平面PCD,所以MN∥平面PCD.

能力提升

1已知直线l的方向向量a=(2"," 3"," 1/3),平面α的法向量为n=(6"," λ",-" 1/2),若l∥α,则λ的值是(　　)

A.4 B.-71/18 C.25/3 D.-23/6

解析:∵l∥α,∴a⊥n,即a·n=0,