在直角坐标系xoy中,已知点A(0,1)和点B（-3,4）,若点C在角AOB的角平分线上,且|向量O

在直角坐标系xoy中,已知点A(0,1)和点B（-3,4）,若点C在角AOB的角平分线上,且|向量O

5向量OA=（0,5）向量OB=（-3,4）则|5向量OA|=|向量OB|=5所以 5向量OA+向量OB对应的向量平分∠AOB5向量OA+向量OB=(-3,9)模是3√5将5向量OA+向量OB单位化即可为(1/3√5)*(-3,9)=(-√5/5,2√5/5)即向量 OC=(-√5/5,2√5/5) C(-√5/5,2√5/5)======以下答案可供参考======供参考答案1:OC=x(5 OA+ OB)，x>0OC=x(-3,9)|OC|=2，则x=sqrt(10)/15,OC=（-sqrt(10)/5,3*sqrt(10)/5）sqrt 开方供参考答案2:|OC| =2OA=(0,1), OB=(-3,4)let D be a point on AB such that O,C,D is on a straight linelet|AD|/|DB| = |OA|/|OB| = 1/5OD = (OA + (1/5)OB)/(1+1/5) = (5/6)[(0,1) + (1/5)(-3,4)] = (5/6) (-3/5,9/5) = (-1/2, 3/2)OC = 2(OD/|OD|) = 2(-1/2, 3/2) / √(1/4+9/4) = 2(-1, 3)/√10 = (√10/5)(-1,3)

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