解析:如图,以点C1为坐标原点,C1B1,C1A1,C1C所在的直线分别为x轴、y轴、z轴,建立空间直角坐标系,

　　不妨设BC=CA=CC1=1,可知点A(0,1,1),N(0"," 1/2 "," 0),B(1,0,1),M(1/2 "," 1/2 "," 0).

　　∴(AN) ⃗=(0",-" 1/2 ",-" 1),(BM) ⃗=("-" 1/2 "," 1/2 ",-" 1).

　　∴cos<(AN) ⃗,(BM) ⃗>=((AN) ⃗"·" (BM) ⃗)/("|" (AN) ⃗"||" (BM) ⃗"|" )=√30/10.

　　根据(AN) ⃗与(BM) ⃗的夹角及AN与BM所成角的关系可知,BM与AN所成角的余弦值为 √30/10.

答案:C

4.在正方体ABCD-A1B1C1D1中,二面角A-BD1-B1的大小为(　　)

A.90° B.60° C.120° D.45°

解析:如图所示,以C为原点建立空间直角坐标系C-xyz,设正方体的边长为a,则A(a,a,0),B(a,0,0),D1(0,a,a),B1(a,0,a),

　　∴(BA) ⃗=(0,a,0),(BD_1 ) ⃗=(-a,a,a),(BB_1 ) ⃗=(0,0,a).

　　设平面ABD1的法向量为n=(x,y,z),

　　则n·(BA) ⃗=(x,y,z)·(0,a,0)=ay=0,n·(BD_1 ) ⃗=(x,y,z)·(-a,a,a)=-ax+ay+az=0.

　　∵a≠0,

　　∴y=0,x=z.

　　令x=z=1,则n=(1,0,1),

　　同理,平面B1BD1的法向量m=(-1,-1,0).

　　cos=(n"·" m)/("|" n"||" m"|" )=-1/2,

　　而二面角A-BD1-B1为钝角,故为120°.

答案:C