若平面向量a,b,c两两所成的角为120度,|a|=|b|=2|c|=2,则|a+b+c|=?
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![若平面向量a,b,c两两所成的角为120度,|a|=|b|=2|c|=2,则|a+b+c|=?](/uploads/image/z/3768518-38-8.jpg?t=%E8%8B%A5%E5%B9%B3%E9%9D%A2%E5%90%91%E9%87%8Fa%2Cb%2Cc%E4%B8%A4%E4%B8%A4%E6%89%80%E6%88%90%E7%9A%84%E8%A7%92%E4%B8%BA120%E5%BA%A6%2C%7Ca%7C%3D%7Cb%7C%3D2%7Cc%7C%3D2%2C%E5%88%99%7Ca%2Bb%2Bc%7C%3D%3F)
若平面向量a,b,c两两所成的角为120度,|a|=|b|=2|c|=2,则|a+b+c|=?
若平面向量a,b,c两两所成的角为120度,|a|=|b|=2|c|=2,则|a+b+c|=?
若平面向量a,b,c两两所成的角为120度,|a|=|b|=2|c|=2,则|a+b+c|=?
|a+b+c|²
=|a|²+|b|²+|c|²+2a*b+2b*c+2c*a
=4+4+1-4-2-2
=1
得|a+b+c|=1
|a+b+c|^2
=|a|+|b|^2+|c|^2 + 2(a.b+b.c+c.a)
=4+4+1+2(4+2+2)(-1/2)
=1
|a+b+c|=1
|a+b+c|²=a²+b²+c²+2ab+2bc+2ca=1²+2²+3²+2IaIIbIcosθ1+2IbIIcIcosθ2+2IcIIaIcosθ3
=1²+2²+3²...
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|a+b+c|²=a²+b²+c²+2ab+2bc+2ca=1²+2²+3²+2IaIIbIcosθ1+2IbIIcIcosθ2+2IcIIaIcosθ3
=1²+2²+3²+2x1x2·cosθ1+2x2x3·cosθ2+2x3x1·cosθ3
当两两所成角相等:θ1=θ2=θ3=0°时,|a+b+c|²=36,∴|a+b+c|=6
当两两所成角相等:θ1=θ2=θ3=120°时,|a+b+c|²=3,,∴|a+b+c|=√3
故|a+b+c|的长度为√3或6.
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