∵点M在(AC_1 ) ⃗上且(AM) ⃗=1/2 (MC_1 ) ⃗,

　　∴(x-a,y,z)=1/2(-x,a-y,a-z),

　　∴x=2/3 a,y=a/3,z=a/3,得M(2a/3 "," a/3 "," a/3),

　　∴|(MN) ⃗|=√((a"-" 2/3 a)^2+(a"-" a/3)^2+(a/2 "-" a/3)^2 )=√21/6 a.

答案:√21/6 a

7.设A(2,3,1),B(4,1,2),C(6,3,7),D(-5,-4,8),则点D到平面ABC的距离为　　　　　.

解析:设平面ABC的法向量n=(x,y,z).

∵n·(AB) ⃗=0,n·(AC) ⃗=0,

∴{■("(" x"," y"," z")·(" 2",-" 2"," 1")" =0"," @"(" x"," y"," z")·(" 4"," 0"," 6")" =0"," )┤

即{■(2x"-" 2y+z=0"," @4x+6z=0"," )┤∴{■(x="-" 3/2 z"," @y="-" z"." )┤

令z=-2,则n=(3,2,-2).

又∵(AD) ⃗=(-7,-7,7),

∴点D到平面ABC的距离为d=("|" (AD) ⃗"·" n"|" )/("|" n"|" )=("|" 3×"(-" 7")" +2×"(-" 7")-" 2×7"|" )/√(3^2+2^2+"(-" 2")" ^2 )=49/√17=(49√17)/17.

答案:(49√17)/17

8.在长方体ABCD-A1B1C1D1中,AA1=AB=2,AD=1,点F,G分别是AB,CC1的中点,则点D1到直线GF的距离为　　　　　.

解析:

如图所示,以D为坐标原点,直线DA,DC,DD1分别为x轴、y轴、z轴建立空间直角坐标系,则D1(0,0,2),F(1,1,0),G(0,2,1),

　　∴(GF) ⃗=(1,-1,-1),(GD_1 ) ⃗=(0,-2,1),

　　∴((GD_1 ) ⃗"·" (GF) ⃗)/("|" (GF) ⃗"|" )=(2"-" 1)/√3=1/√3,|(GD_1 ) ⃗|=√5,

∴点D1到直线GF的距离