数学：已知两向量a,b满足a+b=(0,5),a-b=(4,-3).若向量OA=2a,OB=3b,试写出与∠AOB的角平

OA=2a = (a+b) + (a-b) = (0,5) + ( 4,-3) = (4,2)
OB=3b = 1.5[(a+b) - (a-b)] = 1.5 [(0,5)-(4,-3)] = (-6,12)
OA .OB = 0
所以两个向量垂直
我们取OA' = OA/2 = (2,1)
OB'= OB/6 = (-1,2)
显然|OA'| = |OB'|,OA'B'构成等腰直角三角形
OB' + OC' = (1,3)是AOB的角平方线
其单位向量为(1,3)/|(1,3)| = (1/根号10,3/根号10)

OA=2a = (a+b) + (a-b) = (0,5) + ( 4,-3) = (4,2)
OB=3b = 1.5[(a+b) - (a-b)] = 1.5 [(0,5)-(4,-3)] = (-6, 12)
OA . OB = 0
所以两个向量垂直
我们取OA' = OA/2 = (2,1)
OB'= OB/6 = (-1,2)
显然|OA'|...