Question

已知向量a=(cosa,sina), b=(cosβ，sinβ), 且a,b满足关系式｜ka+b｜=根号3 ｜a-kb｜(k＞0)

1：求证；（a+b）垂直与（a-b)

Answer 1

ka+b=(kcosa+cosβ,ksina+sinβ)
|ka+b|^2=(kcosa+cosβ)^2+(ksina+sinβ)^2
=k^2+1+2k(cosacosβ+sinasinβ)
=(k^2+1)+2kcos(a-β)

a-kb=(cosa-kcosβ,sina-ksinβ)
｜a-kb｜^2=(cosa-kcosβ)^2+(sina-ksinβ)^2
=1+k^2-2kcos(a-β)
｜ka+b｜=根号3 ｜a-kb｜
｜ka+b｜^2=3 ｜a-kb｜^2
(k^2+1)+2kcos(a-β)=3(k^2+1)-2kcos(a-β)
4kcos(a-β)=2(k^2+1)
cos(a-β)=(k^2+1)/2k
-1<=cos(a-β)<=1
k＞0 k^2+1>=2k (k^2+1)/2k>=1
所以 (k^2+1)/2k=1 k=1
cos(a-β)=1
a+b=(cosa+cosβ,sina+sinβ)
a-b=(cosa-cosβ,sina-sinβ)
(a+b)*(a-b)=cos^2a-cos^2β+sin^2a-sin^2β=1-1=0
（a+b）垂直与（a-b)

Answer 2

(a+b)·(a-b)=|a|^2-|b|^2=1-1=0，所以得证。

Answer 3

内积((a+b),(a-b))=(a,a)+(a,b)-(b,a)-(b,b)=a^2-b^2=0
故他们垂直