(1)设=(x0,y0,z0),设z轴上一点为(0,0,a)(a≠0),则由题意得:
| (3,1,5)•(x0,y0,z0)=9 | (1,2,-3)•(x0,y0,z0)=-4 | (0,0,a)•(x0,y0,z0)=0(a≠0) |
| | , 解得,即=(,-,0). (2)令设=(x1,y1,z1),设z轴上一点为(0,0,a)(a≠0),则由题意, 知(x1,y1,z1)=λ(0,0,a)=(0,0,λa)(a≠0), 所以x1=0,y1=0,z1=λa,即=(0,0,λa)(a≠0), 又•=9,•=-4,即 | (3,1,5)•(0,0,λa)=9 | (1,2,-3)•(0,0,λa)=-4 |
| | ⇒,显然矛盾. ∴不存在满足题意的向量,使得与z轴共线. |