∵(AC) ⃗=(AB) ⃗+(BD) ⃗+(DC) ⃗(设AB=a),

　　∴a2=a2+2a2+a2+2·a·√2a·("-" √2/2)+2a·√2a·("-" √2/2)+2a2cos<(AB) ⃗,(DC) ⃗>,

　　∴cos<(AB) ⃗,(DC) ⃗>=1/2,

　　∴AB与CD所成的角为60°.故④正确.

　　【答案】①②④

6.在直三棱柱ABC-A1B1C1中,∠ABC=90°,∠BAC=30°,且BC=1,AA1=√6,M为CC1的中点,则cos<(AB_1 ) ⃗,(A_1 M) ⃗>的值为　　　　.

　　【解析】以B为原点,建立如图所示的空间直角坐标系,则点A1(√3,0,√6),M0,1,√6/2,A(√3,0,0),B1(0,0,√6),

　　所以(A_1 M) ⃗=("-" √3 "," 1",-" √6/2),

　　(AB_1 ) ⃗=(-√3,0,√6),

　　所以(A_1 M) ⃗·(AB_1 ) ⃗=("-" √3 "," 1",-" √6/2)·(-√3,0,√6)=0,

　　所以(A_1 M) ⃗⊥(AB_1 ) ⃗,cos<(A_1 M) ⃗,(AB_1 ) ⃗>=0.

　　【答案】0

7.如图,在四棱锥P-ABCD中,PD⊥底面ABCD,底面ABCD为正方形,PD=DC,E,F分别是AB,PB的中点.

(1)求证:EF⊥CD.

(2)在平面PAD内求一点G,使GF⊥平面PCB,并证明你的结论.

(3)求直线DB与平面DEF所成角的正弦值.

　　【解析】

　　以DA,DC,DP所在直线分别为x轴,y轴, 轴建立空间直角坐标系(如图),设AD=a,则点D(0,0,0),A(a,0,0),B(a,a,0),C(0,a,0),E(a"," a/2 "," 0),F(a/2 "," a/2 "," a/2),P(0,0,a).

　　(1)∵(EF) ⃗·(DC) ⃗=("-" a/2 "," 0"," a/2)·(0,a,0)=0,

　　∴EF⊥DC.

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