求与向量a=（7/2,1/2）,b=（1/2,7/2）的夹角相等,且模长为1的向量

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设所求向量c=(m,n),
|c|=√（m^2+n^2)=1,
设向量a和c夹角为θ
cosθ=a·c/(|a||c|=(7m/2+n/2)/[√（49/4+1/4）*1]=√2(7m/2+n/2)/5,
cosθ=b·c(/|b||c|)=(m/2+7n/2)/√[(1/4+49/4)*1]=√2(m/2+7n/2)/5,
√2(7m/2+n/2)/5=√2(m/2+7n/2)/5,
m=n,m^2+n^2=1,
m=±√2/2,
n=±√2/2,
m,n应取同号
则向量c=(√2/2,√2/2),c=(-√2/2,-√2/2),
======以下答案可供参考======
供参考答案1:
设所求向量c=(m,n),
|c|=√（m^2+n^2)=1,
设向量a和c夹角为θ
cosθ=a·c/(|a||c|=(7m/2+n/2)/[√（49/4+1/4）*1]=√2(7m/2+n/2)/5，
cosθ=b·c(/|b||c|)=(m/2+7n/2)/√[(1/4+49/4)*1]=√2(m/2+7n/2)/5,
√2(7m/2+n/2)/5=√2(m/2+7n/2)/5,
m=n,m^2+n^2=1,
m=±√2/2，
n=±√2/2,
m,n应取同号
则向量c=(√2/2,√2/2), c=(-√2/2,-√2/2),
供参考答案2:
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