- 简单向量问题已知向量a=(cos3/2x,sin3/2x),b=
- 已知向量a=(c3/2x,sin3/2x),b=(cosx/2,-sinx/2)
且x∈［0,π/2］求：
1 a*b和Ⅰa+bⅠ
- 已知向量a=(cos3/2x,sin3/2x),b=(cosx/2,-sinx/2)
且x∈［0,π/2］求：
1 a*b和Ⅰa+bⅠ
a*b=(cos3/2x,sin3/2x)*(cosx/2,-sinx/2)
=cos3/2xcosx/2-sin3/2xsinx/2
=cos(3/2x+x/2)
=cos2x
|a+b|=|(cos3/2x,sin3/2x)+ (cosx/2,-sinx/2)|
=|(cos3/2x+cosx/2),(sin3/2x-sinx/2)|
=√[(cos3/2x+cosx/2)^2+(sin3/2x-sinx/2)^2]
=√[2+2(cos3/2xcosx/2-sin3/2xsinx/2)]
=√(2+2cosx)