AB=|(AB) ⃗|=√(1^2+"(-" 3")" ^2+2^2 )=√14.

　　∴|a-b|=√("(-" 1")" ^2+"(-" 3")" ^2+10^2 )=√110.

　　(2)由(1)知|(AC) ⃗|=√(2^2+0^2+"(-" 8")" ^2 )=2√17,

　　∴cos=(a"·" b)/("|" a"||" b"|" )=(1×2+"(-" 3")" ×0+2×"(-" 8")" )/(√14 "·" 2√17)=-√238/34.

★11.已知A(1,0,0),B(0,1,0),C(0,0,2).

(1)若(DB) ⃗∥(AC) ⃗,(DC) ⃗∥(AB) ⃗,求点D的坐标.

(2)问是否存在实数α,β,使得(AC) ⃗=α(AB) ⃗+β(BC) ⃗成立?若存在,求出α,β的值;若不存在,说明理由.

解(1)设D(x,y,z),则(DB) ⃗=(-x,1-y,-z),(AC) ⃗=(-1,0,2),(DC) ⃗=(-x,-y,2-z),(AB) ⃗=(-1,1,0).

　　因为(DB) ⃗∥(AC) ⃗,(DC) ⃗∥(AB) ⃗,所以存在实数m,n,使得(DB) ⃗=m(AC) ⃗,(DC) ⃗=n(AB) ⃗,

　　所以{■("(-" x"," 1"-" y",-" z")" =m"(-" 1"," 0"," 2")," @"(-" x",-" y"," 2"-" z")" =n"(-" 1"," 1"," 0")," )┤

　　解得{■(x="-" 1"," @y=1"," @z=2"," )┤

　　即D(-1,1,2).

　　(2)依题意(AB) ⃗=(-1,1,0),(AC) ⃗=(-1,0,2),(BC) ⃗=(0,-1,2).

　　假设存在实数α,β,使得(AC) ⃗=α(AB) ⃗+β(BC) ⃗成立,

　　则有(-1,0,2)=α(-1,1,0)+β(0,-1,2)=(-α,α-β,2β),

　　所以{■(α=1"," @α"-" β=0"," @2β=2"," )┤故存在α=β=1,使得(AC) ⃗=α(AB) ⃗+β(BC) ⃗成立.

★12.已知(OA) ⃗=(1,2,3),(OB) ⃗=(2,1,2),(OP) ⃗=(1,1,2),O为坐标原点,点Q在直线OP上运动,则(QA) ⃗·(QB) ⃗有没有最小值?如果有,求点Q的坐标;如果没有,请说明理由.

解由O,P,Q三点共线得(OQ) ⃗=x(OP) ⃗=(x,x,2x)(x∈R),

则(QA) ⃗=(1-x,2-x,3-2x),(QB) ⃗=(2-x,1-x,2-2x),∴(QA) ⃗·(QB) ⃗=(1-x)(2-x)+(2-x)(1-x)+(3-2x)(2-2x)=6(x"-" 4/3)^2-2/3,

　　故当x=4/3 时,(QA) ⃗·(QB) ⃗取得最小值,是-2/3,

　　此时(OQ) ⃗=(4/3 "," 4/3 "," 8/3),

　　即Q点坐标为(4/3 "," 4/3 "," 8/3).