设A(cosα,sinα).B(cos(2∏/3+α ),sin(2 ∏ /3+α)), c(cos(4∏ /3+α ),sin(4 ∏/3+α )),求证向量OA+OB+OC=0
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![设A(cosα,sinα).B(cos(2∏/3+α ),sin(2 ∏ /3+α)), c(cos(4∏ /3+α ),sin(4 ∏/3+α )),求证向量OA+OB+OC=0](/uploads/image/z/1170467-35-7.jpg?t=%E8%AE%BEA%28cos%CE%B1%2Csin%CE%B1%29.B%28cos%282%E2%88%8F%2F3%2B%CE%B1%26%23160%3B%29%2Csin%282%26%23160%3B%E2%88%8F%26%23160%3B%2F3%2B%CE%B1%29%29%2C%26%23160%3B+c%28cos%284%E2%88%8F%26%23160%3B%2F3%2B%CE%B1%26%23160%3B%26%23160%3B%29%2Csin%284%26%23160%3B%E2%88%8F%2F3%2B%CE%B1%26%23160%3B%29%29%2C%E6%B1%82%E8%AF%81%E5%90%91%E9%87%8FOA%2BOB%2BOC%3D0)
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设A(cosα,sinα).B(cos(2∏/3+α ),sin(2 ∏ /3+α)), c(cos(4∏ /3+α ),sin(4 ∏/3+α )),求证向量OA+OB+OC=0
设A(cosα,sinα).B(cos(2∏/3+α ),sin(2 ∏ /3+α)), c(cos(4∏ /3+α ),sin(4 ∏/3+α )),求证向量OA+OB+OC=0
设A(cosα,sinα).B(cos(2∏/3+α ),sin(2 ∏ /3+α)), c(cos(4∏ /3+α ),sin(4 ∏/3+α )),求证向量OA+OB+OC=0
OB=[cos(2∏/3+α ),sin(2 ∏ /3+α)]
=[-1/2cosα-√3/2sinα,√3/2cosα-1/2sinα]
OC=[cos(4∏ /3+α ),sin(4 ∏/3+α )]
=[-1/2cosα+√3/2sinα,-√3/2cosα+1/2sinα]
0A=(cosα,sinα)
所以:OA+OB+OC=(-1/2cosα-√3/2sinα-1/2cosα+√3/2sinα+cosα,√3/2cosα-1/2sinα-√3/2cosα+1/2sinα+sinα)=(0,0)=0
向量OA,OB,OC两两夹角为120°,且这三个向量模长相等,根据力学知识,一定有向量OA+OB+OC=0
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