△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

2个回答

  • (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