Question

已知向量a=(mcosa,-根号3),b=(1,n+sina),且a⊥b

若m=-根号3，且a∈(0,π/2),求实数n的取值范围
过程谢谢

Answer 1

a=(-√3cosa,-√3)
b=(1,n+sina)
a⊥b
a·b=-√3cosa-√3(n+sina)=-√3(cosa+n+sina)=0
cosa+n+sina=0
n+√2[sin(a+π/4)=0
n=-√2[sin(a+π/4)]
a∈(0,π/2)
(a+π/4)∈(π/4,3π/4)
sin(a+π/4)∈(√2,1]
-√2[sin(a+π/4)]∈[-√2,-2)
n∈[-√2,-2)