CM=(1/6)CB+(2/3)CA
MA=CA-CM
MB=CB-CM
MA*MB=(CA-CM)(CB-CM)=CA*CB-[(1/6)CB+(2/3)CA]*[CA+CB-(1/6)CB-(2/3)CA]=CA*CB-[(2/3)CA+(1/6)CB]*[(1/3)CA+(5/6)CB]=(7/18)CA*CB-(2/9)CA*CA-(5/36)*CB*CB
三角形ABC是等边三角形,个边模=2根号3,设向量CA=2根号3（cos&+isin&）,向量CB是向量CA逆时针旋转60°所得向量,则CB=2根号3[cos(&+60°）+isin（&+60°）]
所以CA*CB=12*[cos（2&+60°）+isin(2&+60°）]
CA*CA=12[cos(2&)+isin(2*)]
CB*CB=12[cos(2&+120°）+isin（2&+120°）]
将这三个关系式代入上式的
MA*MB=7/18)CA*CB-(2/9)CA*CA-(5/36)*CB*CB