Question

已知向量a=(cosb,sinb),向量b=(√3 ,-1),求|2a-b|的取值范围

Answer 1

向量a=(cosb,sinb),向量b=(√3 ,-1),

∴ |a|=√(cos²b+sin²b)=1

|b|=√(3+1)=2

|2a-b|²=4a²+b²-4a.b

=4+4-4(√3cosb-sinb)

=8+4(sinb-√3cosb)

=8+8[sinb*(1/2)-cosb*(√3/2)]

=8+8[sinbcos(π/3)-cosb*sin(π/3)]

=8+8sin(b-π/3)

∴|2a-b|²的最大值是16,最小值是0

即 |2a-b|的范围是[0,4]