← 上一题 返回专题 下一题 → 平面向量 a → = ( cos α , sin α ) , b → = ( cos ( α + β ) , sin ( α + β ) ) , c → = ( cos ( α + 2 β ) , sin ( α + 2 β ) ) \overrightarrow{a} = \left( {\cos \alpha ,\sin \alpha }\right) ,\overrightarrow{b} = \left( {\cos \left( {\alpha + \beta }\right) ,\sin \left( {\alpha + \beta }\right) }\right) ,\overrightarrow{c} = \left( {\cos \left( {\alpha + {2\beta }}\right) ,\sin \left( {\alpha + {2\beta }}\right) }\right) a = ( cos α , sin α ) , b = ( cos ( α + β ) , sin ( α + β ) ) , c = ( cos ( α + 2 β ) , sin ( α + 2 β ) ) 0 ∘ < β < 180 ∘ {0}^{\circ} < \beta < {180}^{\circ} 0 ∘ < β < 180 ∘ ∣ a → − b → ∣ = ∣ b → − c → ∣ \left| {\overrightarrow{a} - \overrightarrow{b}}\right| = \left| {\overrightarrow{b} - \overrightarrow{c}}\right| a − b = b − c ( a → + c → ) / / b → \left( {\overrightarrow{a} + \overrightarrow{c}}\right) //\overrightarrow{b} ( a + c ) // b ∣ a → + c → ∣ = ∣ b → ∣ \left| {\overrightarrow{a} + \overrightarrow{c}}\right| = \left| \overrightarrow{b}\right| a + c = b β = 30 ∘ \beta = {30}^{\circ} β = 30 ∘ a → + b → + c → = 0 → \overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c} = \overrightarrow{0} a + b + c = 0 β = 120 ∘ \beta = {120}^{\circ} β = 120 ∘
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