Question

已知IaI=1,IbI=根号3,a+b=(根号3,1）,求Ia-bI ,向量a+b与向量a-b的夹角

Answer 1

let (a+b),(a-b)的夹角=x

|a+b|^2= 4 = |a|^2+|b|^2+2a.b

4 = 1+3+2a.b

a.b =0

|a-b| = |a+b| =2

(a-b).(a+b) = |a+b||a-b|cosx

|a|^2-|b|^2= |a+b||a-b|cosx

1-3= 4cosx

cosx = -1/2

x= 120度