已知|a|=2|b|≠0,且关于x的方程x^2+|a|x+a*b=0有实数根,则a与b的夹角的取值范围 (ab都是向量)这道题一直解到cos≤1/2我都会,但我不明白为什么夹角范围是[Π/3,Π]而不是[Π/3,2Π/3]呢
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![已知|a|=2|b|≠0,且关于x的方程x^2+|a|x+a*b=0有实数根,则a与b的夹角的取值范围 (ab都是向量)这道题一直解到cos≤1/2我都会,但我不明白为什么夹角范围是[Π/3,Π]而不是[Π/3,2Π/3]呢](/uploads/image/z/5246184-48-4.jpg?t=%E5%B7%B2%E7%9F%A5%7Ca%7C%3D2%7Cb%7C%E2%89%A00%2C%E4%B8%94%E5%85%B3%E4%BA%8Ex%E7%9A%84%E6%96%B9%E7%A8%8Bx%5E2%2B%7Ca%7Cx%2Ba%2Ab%3D0%E6%9C%89%E5%AE%9E%E6%95%B0%E6%A0%B9%2C%E5%88%99a%E4%B8%8Eb%E7%9A%84%E5%A4%B9%E8%A7%92%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4+%EF%BC%88ab%E9%83%BD%E6%98%AF%E5%90%91%E9%87%8F%EF%BC%89%E8%BF%99%E9%81%93%E9%A2%98%E4%B8%80%E7%9B%B4%E8%A7%A3%E5%88%B0cos%E2%89%A41%2F2%E6%88%91%E9%83%BD%E4%BC%9A%2C%E4%BD%86%E6%88%91%E4%B8%8D%E6%98%8E%E7%99%BD%E4%B8%BA%E4%BB%80%E4%B9%88%E5%A4%B9%E8%A7%92%E8%8C%83%E5%9B%B4%E6%98%AF%5B%CE%A0%EF%BC%8F3%2C%CE%A0%5D%E8%80%8C%E4%B8%8D%E6%98%AF%5B%CE%A0%2F3%2C2%CE%A0%2F3%5D%E5%91%A2)
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