求与向量a=(7/2,1/2),b=(1/2,7/2)的夹角相等,且模长为1的向量设所求向量c=(m,n),|c|=√(m^2+n^2)=1,设向量a和c夹角为θcosθ=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)/
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![求与向量a=(7/2,1/2),b=(1/2,7/2)的夹角相等,且模长为1的向量设所求向量c=(m,n),|c|=√(m^2+n^2)=1,设向量a和c夹角为θcosθ=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)/](/uploads/image/z/2771425-1-5.jpg?t=%E6%B1%82%E4%B8%8E%E5%90%91%E9%87%8Fa%3D%EF%BC%887%2F2%2C1%2F2%EF%BC%89%2Cb%3D%EF%BC%881%2F2%2C7%2F2%EF%BC%89%E7%9A%84%E5%A4%B9%E8%A7%92%E7%9B%B8%E7%AD%89%2C%E4%B8%94%E6%A8%A1%E9%95%BF%E4%B8%BA1%E7%9A%84%E5%90%91%E9%87%8F%E8%AE%BE%E6%89%80%E6%B1%82%E5%90%91%E9%87%8Fc%3D%28m%2Cn%29%2C%7Cc%7C%3D%E2%88%9A%EF%BC%88m%5E2%2Bn%5E2%29%3D1%2C%E8%AE%BE%E5%90%91%E9%87%8Fa%E5%92%8Cc%E5%A4%B9%E8%A7%92%E4%B8%BA%CE%B8cos%CE%B8%3Da%C2%B7c%2F%28%7Ca%7C%7Cc%7C%3D%287m%2F2%2Bn%2F2%29%2F%5B%E2%88%9A%EF%BC%8849%2F4%2B1%2F4%EF%BC%89%2A1%5D%3D%E2%88%9A2%287m%2F2%2Bn%2F2%29%2F5%2Ccos%CE%B8%3Db%C2%B7c%28%2F%7Cb%7C%7Cc%7C%29%3D%28m%2F2%2B7n%2F2%29%2F)
求与向量a=(7/2,1/2),b=(1/2,7/2)的夹角相等,且模长为1的向量设所求向量c=(m,n),|c|=√(m^2+n^2)=1,设向量a和c夹角为θcosθ=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)/
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
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),
最后一步看不懂,为什么m,n应取同号?
求与向量a=(7/2,1/2),b=(1/2,7/2)的夹角相等,且模长为1的向量设所求向量c=(m,n),|c|=√(m^2+n^2)=1,设向量a和c夹角为θcosθ=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)/
因为m=n啊所以才应取同号
建议你这么理解,在m=n之后这么进行你就不会混淆了
m^2+n^2=1,
2m^2=1,
m=±√2/2
n=m=√2/2或n=m=-√2/2
所以,向量c=(√2/2,√2/2),c=(-√2/2,-√2/2),
很假单啊 这种题 自己先草图 就知道都应该去正
如果实在想推理 可以带入推理 发现和题没有给的不符合 应为给的都是确定的值 可以求夹角检验