Question

a b均为向量 |a|=1 向量axb=1/2 ， (a-b)x(a+b)=1/2

求 a b的夹角

a-b 与a+b的夹角的余弦值

Answer 1

|a|=1, a.b=1/2, (a-b).(a+b)=1/2
To find:
(1)a b的夹角
(2) a-b 与a+b的夹角的余弦值
(1)
(a-b).(a+b)=1/2
|a|^2-|b|^2 = 1/2
1-|b|^2=1/2
|b| = √2/2
let a b的夹角=x
a.b = 1/2
|a||b|cosx = 1/2
(√2/2)cosx =1/2
cosx =√2/2
x= π/4
(2)
|a+b|^2 =|a|^2+|b|^2+2a.b= 1+1/2 + 2(1/2) = 5/2
|a+b| = √10/2
|a-b|^2 =|a|^2+|b|^2-2a.b= 1+1/2 - 2(1/2) = 1/2
|a-b| = √2/2
let a-b 与a+b的夹角 = y
(a-b).(a+b)=1/2
|a-b||a+b|cosy = 1/2
(√10/2)(√2/2)cosy =1/2
cosy = √5/5