则p=8a+6b+4c=8i+8j+6j+6k+4k+4i=12i+14j+10k,

　　故点A在基底i,j,k下的坐标为(12,14,10).

答案:A

6.已知a=(1,2,3),b=(3,0,-1),c=("-" 1/5 "," 1",-" 3/5),给出下列等式:①(a+b)·c=a·(b+c);②(a+b+c)2=a2+b2+c2;③(a·b)·c=a·(b·c).

其中正确的有(　　)

A.0个 B.3个

C.2个 D.1个

解析:①∵(a+b)·c=(4,2,2)·("-" 1/5 "," 1",-" 3/5)=-4/5+2-6/5=0,a·(b+c)=(1,2,3)·(14/5 "," 1",-" 8/5)=14/5+2-24/5=0,

　　∴(a+b)·c=a·(b+c).

　　②∵a·b=3+0-3=0,a·c=-1/5+2-9/5=0,b·c=-3/5+0+3/5=0,

　　∴(a+b+c)2=a2+b2+c2.

　　③∵(a·b)·c=(3+0-3)·("-" 1/5 "," 1",-" 3/5)=(0,0,0),a·(b·c)=(1,2,3)·("-" 3/5+0+3/5)=(0,0,0),

　　∴(a·b)·c=a·(b·c).故选B.

答案:B

7.若a=(x,3,1),b=(2,y,4),且a=zb,则c=(x,y,z)=　　　　　　　.

解析:由a=zb,得{■(x=2z"," @3=yz"," @1=4z"," )┤所以{■(x=1/2 "," @y=12"," @z=1/4 "." )┤

答案:(1/2 "," 12"," 1/4)

8.已知A(1,0,0),B(0,-1,1),(OA) ⃗+λ(OB) ⃗与(OB) ⃗的夹角为120°,则λ的值为　　　.

答案:-√6/6

9.已知a=(1-t,2t-1,0),b=(2,t,t),则|a-b|的最小值是　　　　　.

解析:∵a-b=(-1-t,t-1,-t),

　　∴|a-b|2=(-1-t)2+(t-1)2+(-t)2=3t2+2.

　　∴当t=0时,|a-b|取得最小值,为√2.

答案:√2

10.已知空间三点A(1,2,3),B(2,-1,5),C(3,2,-5),设a=(AB) ⃗,b=(AC) ⃗.

(1)求AB的长及|a-b|;

(2)求cos.

解(1)∵a=(AB) ⃗=(2,-1,5)-(1,2,3)=(1,-3,2),b=(AC) ⃗=(3,2,-5)-(1,2,3)=(2,0,-8),

∴a-b=(AB) ⃗-(AC) ⃗=(1,-3,2)-(2,0,-8)=(-1,-3,10).