△ABC内接于以O为圆心,1为半径的圆,且向量3OA+4OB+5OC=O,①求向量OA·OB,OB·OC,OC·OA.②求△ABC的面积.HELP!
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![△ABC内接于以O为圆心,1为半径的圆,且向量3OA+4OB+5OC=O,①求向量OA·OB,OB·OC,OC·OA.②求△ABC的面积.HELP!](/uploads/image/z/3818727-63-7.jpg?t=%E2%96%B3ABC%E5%86%85%E6%8E%A5%E4%BA%8E%E4%BB%A5O%E4%B8%BA%E5%9C%86%E5%BF%83%2C1%E4%B8%BA%E5%8D%8A%E5%BE%84%E7%9A%84%E5%9C%86%2C%E4%B8%94%E5%90%91%E9%87%8F3OA%2B4OB%2B5OC%3DO%2C%E2%91%A0%E6%B1%82%E5%90%91%E9%87%8FOA%C2%B7OB%2COB%C2%B7OC%2COC%C2%B7OA.%E2%91%A1%E6%B1%82%E2%96%B3ABC%E7%9A%84%E9%9D%A2%E7%A7%AF.HELP%21)
△ABC内接于以O为圆心,1为半径的圆,且向量3OA+4OB+5OC=O,①求向量OA·OB,OB·OC,OC·OA.②求△ABC的面积.HELP!
△ABC内接于以O为圆心,1为半径的圆,且向量3OA+4OB+5OC=O,①求向量OA·OB,OB·OC,OC·OA.②求△ABC的面积.
HELP!
△ABC内接于以O为圆心,1为半径的圆,且向量3OA+4OB+5OC=O,①求向量OA·OB,OB·OC,OC·OA.②求△ABC的面积.HELP!
(1).∵A,B,C在单位圆上,∴|OA|=|OB|=|OC|=1
取OC与X轴的负向重合,于是OC=icos180?+jsin180?=-i,
5oc=-5i.
∵3OA+4OB=-5OC=5i,故可在x轴的正向上取一点D,使|OD|=5,
并以OD为斜边,以3|OA|=3,4|OB|=4作直角三角形,便有:
3OA+4OB=OD=5i.故OA⊥OB,
OA与X轴正向的夹角α=arccos(3/5),(OA在第一象限)
OB与X轴正向的夹角β=arccos(4/5).(OB在第四象限)
于是∠AOC=180°-α=180°-arccos(3/5)
∠COB=180°-β=180°-arccos(4/5)
故OA•OB=|OA||OB|cos90°=0
OB•OC=|OB||OC|cos∠COB=cos[180°-arccos(4/5)]
=-cosarccos(4/5)=-4/5
OC•OA=|OC||OA|cos∠AOC=cos[180°-arccos(3/5)]
=-cosarccos(3/5)=-3/5.
(2).又因OA,OB,OC已知,可得cosA,COSB,COSC,可得sinA,SINB,SINC,可得,oab,oac,obc面积
3OA+4OB=-5OC,两边平方后可得OA.OB,同理得OB.OC,OC.OA
又因OA,OB,OC已知,可得cosA,COSB,COSC,可得sinA,SINB,SINC,之后可得,oab,oac,obc面积,之后.......