题目
空间向量的直角坐标运算律 设a=(a1,a2,a3),b=(b1,b2,b3),则 (1)a+b=_____________________________; (2)a-b=_________________________________________; (3)λa=______________________(λ∈R); (4)a·b=________________________; (5)a∥b________________________________; (6)a⊥b________________________.
答案: (1)(a1+b1,a2+b2,a3+b3) (2)(a1-b1,a2-b2,a3-b3) (3)(λa1,λa2,λa3) (4)a1b1+a2b2+a3b3 (5)a1=λb1,a2=λb2,a3=λb3(λ∈R) (6)a1b1+a2b2+a3b3=0