已知向量a=(cosa,sina,1) b=(cosb,sinb,1)且a b满足│ka+b│=根号3│a-kb│(k＞0)①用k表示a·b；②若a-b=π/2,求k值

已知向量a=(cosa,sina,1) b=(cosb,sinb,1)且a b满足│ka+b│=根号3│a-kb│(k＞0)①用k表示a·b；②若a-b=π/2,求k值
已知向量a=(cosa,sina,1) b=(cosb,sinb,1)且a b满足│ka+b│=根号3│a-kb│(k＞0)
①用k表示a·b；②若a-b=π/2,求k值

已知向量a=(cosa,sina,1) b=(cosb,sinb,1)且a b满足│ka+b│=根号3│a-kb│(k＞0)①用k表示a·b；②若a-b=π/2,求k值
由题意,|a|=sqrt(2),|b|=sqrt(2)
1
|ka+b|^2=(ka+b) dot (ka+b)=k^2|a|^2+|b|^2+2k(a dot b)=2k^2+2+2k(a dot b)
|a-kb|^2=(a-kb) dot (a-kb)=|a|^2+k^2|b|^2-2k(a dot b)=2+2k^2-2k(a dot b)
即：2k^2+2+2k(a dot b)=6+6k^2-6k(a dot b),即：8k(a dot b)=4k^2+4
即：a dot b=(k^2+1)/(2k)
2
向量和角都混了：A-B=π/2,则：cosA=cos(π/2+B)=-sinB,sinA=sin(π/2+B)=cosB
a dot b=(-sinB,cosB,1) dot (cosB,sinB,1)=-sinBcosB+sinBcosB+1=1
即：(k^2+1)/(2k)=1,即：k^2+1=2k,故：k=1