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
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![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](/uploads/image/z/5351636-20-6.jpg?t=%E5%B7%B2%E7%9F%A5%E5%90%91%E9%87%8Fa%3D%28cosa%2Csina%2C1%29+b%3D%28cosb%2Csinb%2C1%29%E4%B8%94a+b%E6%BB%A1%E8%B6%B3%E2%94%82ka%2Bb%E2%94%82%3D%E6%A0%B9%E5%8F%B73%E2%94%82a-kb%E2%94%82%28k%EF%BC%9E0%29%E2%91%A0%E7%94%A8k%E8%A1%A8%E7%A4%BAa%C2%B7b%EF%BC%9B%E2%91%A1%E8%8B%A5a-b%3D%CF%80%2F2%2C%E6%B1%82k%E5%80%BC)
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