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的角平分线同方向的单位向量.要有过程啦,谢谢
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/21 11:04:24
![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的角平分线同方向的单位向量.要有过程啦,谢谢](/uploads/image/z/3747217-49-7.jpg?t=%E6%95%B0%E5%AD%A6%EF%BC%9A%E5%B7%B2%E7%9F%A5%E4%B8%A4%E5%90%91%E9%87%8Fa%2Cb%E6%BB%A1%E8%B6%B3a%2Bb%3D%280%2C5%29%2Ca-b%3D%284%2C-3%29.%E8%8B%A5%E5%90%91%E9%87%8FOA%3D2a%2COB%3D3b%2C%E8%AF%95%E5%86%99%E5%87%BA%E4%B8%8E%E2%88%A0AOB%E7%9A%84%E8%A7%92%E5%B9%B3%E5%88%86%E7%BA%BF%E5%90%8C%E6%96%B9%E5%90%91%E7%9A%84%E5%8D%95%E4%BD%8D%E5%90%91%E9%87%8F.%E8%A6%81%E6%9C%89%E8%BF%87%E7%A8%8B%E5%95%A6%EF%BC%8C%E8%B0%A2%E8%B0%A2)
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的角平分线同方向的单位向量.要有过程啦,谢谢
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的角平分线同方向的单位向量.
要有过程啦,谢谢
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的角平分线同方向的单位向量.要有过程啦,谢谢
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'|...
全部展开
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)
我就拿个2分……
收起