平面向量数量积的求解三角形ABC中,AC=2,BC=6,已知点O是三角形ABC内一点,且满足OA+3OB+4OC=0（都是向量）,求OC与（BA+2BC）的数量积

Question

平面向量数量积的求解
三角形ABC中,AC=2,BC=6,已知点O是三角形ABC内一点,且满足OA+3OB+4OC=0（都是向量）,求OC与（BA+2BC）的数量积

Answer 1

|AC|=|OC-OA|=2,
∴OC^2-2OA*OC+OA^2=4,
OA*OC=(OA^2+OC^2-4)/2,
同理,OC^2-2OB*OC+OB^2=36,
OB*OC=(OB^2+OC^2-36)/2,
由OA+3OB+4OC=0得
OA=-3OB-4OC,
平方得OA^2=9OB^2+24OB*OC+16OC^2
=9OB^2+12(OB^2+OC^2-36)+16OC^2
=21OB^2+28OC^2-432,①
OA*OC+3OB*OC+4OC^2=0,
∴(OA^2+OC^2-4)/2+3（OB^2+OC^2-36)/2+4OC^2=0,
∴OA^2+3OB^2+12OC^2=112,
24OB^2+40OC^2-432=112,
3OB^2=68-5OC^2,②
把②代入①,OA^2=7(68-5OC^2)+28OC^2-432=44-7OC^2,
OC*(BA+2BC)
=OC*(OA-OB+2OC-2OB)
=OC*(OA-3OB+2OC)
=OA*OC-3OB*OC+2OC^2
=(OA^2+OC^2-4)/2-3(OB^2+OC^2-36)/2+2OC^2
=(1/2)(OA^2-3OB^2+2OC^2-40)
=(1/2)(44-7OC^2+5OC^2-68+2OC^2-40)
=-32.