答案:60°

9.设空间两个单位向量(OA) ⃗=(m,n,0),(OB) ⃗=(0,n,p)与(OC) ⃗=(1,1,1)的夹角都等于 π/4,求cos∠AOB.

解:由题意得,{■(m^2+n^2=1"," @n^2+p^2=1"," @cos π/4=(m+n)/(√(m^2+n^2 )×√3) "," @cos π/4=(n+p)/(√(n^2+p^2 )×√3) "," )┤

　　解得n=(√6±√2)/4,

　　故cos∠AOB=((OA) ⃗"·" (OB) ⃗)/("|" (OA) ⃗"||" (OB) ⃗"|" )=n2=(2±√3)/4.

★10.在正方体ABCD - A1B1C1D1中,M是AA1的中点,问当点N位于AB何处时,MN⊥MC1?

解:以A为坐标原点,棱AB,AD,AA1所在直线分别为x轴、y轴、z轴,建立空间直角坐标系,设正方体的棱长为a,则M(0"," 0"," a/2),C1(a,a,a),N(x,0,0),

　　所以(MC_1 ) ⃗=(a"," a"," a/2),(MN) ⃗=(x"," 0",-" a/2).

　　因为MN⊥MC1,

　　所以(MN) ⃗·(MC_1 ) ⃗=xa-a^2/4=0,得x=a/4.

　　所以点N的坐标为(a/4 "," 0"," 0),即N为AB的四等分点且靠近点A时,MN⊥MC1.