已知向量OA=(k,k),向量OB=(m-√3m,m+√3m),其中k>0,m>0,当k=m=1时,(1)当k=m=1时,证明向量OA⊥向量OB(2)求向量OA,OB的夹角的大小(3)设向量AB的模=√3,求OA模+OB模的最大值
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![已知向量OA=(k,k),向量OB=(m-√3m,m+√3m),其中k>0,m>0,当k=m=1时,(1)当k=m=1时,证明向量OA⊥向量OB(2)求向量OA,OB的夹角的大小(3)设向量AB的模=√3,求OA模+OB模的最大值](/uploads/image/z/6161261-5-1.jpg?t=%E5%B7%B2%E7%9F%A5%E5%90%91%E9%87%8FOA%3D%28k%2Ck%29%2C%E5%90%91%E9%87%8FOB%3D%28m-%E2%88%9A3m%2Cm%2B%E2%88%9A3m%29%2C%E5%85%B6%E4%B8%ADk%3E0%2Cm%3E0%2C%E5%BD%93k%3Dm%3D1%E6%97%B6%2C%281%29%E5%BD%93k%3Dm%3D1%E6%97%B6%EF%BC%8C%E8%AF%81%E6%98%8E%E5%90%91%E9%87%8FOA%E2%8A%A5%E5%90%91%E9%87%8FOB%282%29%E6%B1%82%E5%90%91%E9%87%8FOA%2COB%E7%9A%84%E5%A4%B9%E8%A7%92%E7%9A%84%E5%A4%A7%E5%B0%8F%283%29%E8%AE%BE%E5%90%91%E9%87%8FAB%E7%9A%84%E6%A8%A1%3D%E2%88%9A3%EF%BC%8C%E6%B1%82OA%E6%A8%A1%2BOB%E6%A8%A1%E7%9A%84%E6%9C%80%E5%A4%A7%E5%80%BC)
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