在平面直角坐标系中,O为坐标原点,A,B,C三点满足A,B,C三点满足向量OC=1/3向量OA+2/3向量OB(1)求证:A、B、C三点共线(2)已知A(1,cosx)、B(1+sinx,cosx),x属于[o,π/2],f(x)=向量OA点乘向量OC-(2m^2+2/3)点乘I向量ABI
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![在平面直角坐标系中,O为坐标原点,A,B,C三点满足A,B,C三点满足向量OC=1/3向量OA+2/3向量OB(1)求证:A、B、C三点共线(2)已知A(1,cosx)、B(1+sinx,cosx),x属于[o,π/2],f(x)=向量OA点乘向量OC-(2m^2+2/3)点乘I向量ABI](/uploads/image/z/2187732-12-2.jpg?t=%E5%9C%A8%E5%B9%B3%E9%9D%A2%E7%9B%B4%E8%A7%92%E5%9D%90%E6%A0%87%E7%B3%BB%E4%B8%AD%2CO%E4%B8%BA%E5%9D%90%E6%A0%87%E5%8E%9F%E7%82%B9%2CA%2CB%2CC%E4%B8%89%E7%82%B9%E6%BB%A1%E8%B6%B3A%2CB%2CC%E4%B8%89%E7%82%B9%E6%BB%A1%E8%B6%B3%E5%90%91%E9%87%8FOC%3D1%2F3%E5%90%91%E9%87%8FOA%2B2%2F3%E5%90%91%E9%87%8FOB%281%29%E6%B1%82%E8%AF%81%3AA%E3%80%81B%E3%80%81C%E4%B8%89%E7%82%B9%E5%85%B1%E7%BA%BF%282%29%E5%B7%B2%E7%9F%A5A%281%2Ccosx%29%E3%80%81B%281%2Bsinx%2Ccosx%29%2Cx%E5%B1%9E%E4%BA%8E%5Bo%2C%CF%80%2F2%5D%2Cf%28x%29%3D%E5%90%91%E9%87%8FOA%E7%82%B9%E4%B9%98%E5%90%91%E9%87%8FOC-%282m%5E2%2B2%2F3%29%E7%82%B9%E4%B9%98I%E5%90%91%E9%87%8FABI)
在平面直角坐标系中,O为坐标原点,A,B,C三点满足A,B,C三点满足向量OC=1/3向量OA+2/3向量OB(1)求证:A、B、C三点共线(2)已知A(1,cosx)、B(1+sinx,cosx),x属于[o,π/2],f(x)=向量OA点乘向量OC-(2m^2+2/3)点乘I向量ABI
在平面直角坐标系中,O为坐标原点,A,B,C三点满足
A,B,C三点满足向量OC=1/3向量OA+2/3向量OB(1)求证:A、B、C三点共线(2)已知A(1,cosx)、B(1+sinx,cosx),x属于[o,π/2],f(x)=向量OA点乘向量OC-(2m^2+2/3)点乘I向量ABI的最小值为1/2,求实数m的值
在平面直角坐标系中,O为坐标原点,A,B,C三点满足A,B,C三点满足向量OC=1/3向量OA+2/3向量OB(1)求证:A、B、C三点共线(2)已知A(1,cosx)、B(1+sinx,cosx),x属于[o,π/2],f(x)=向量OA点乘向量OC-(2m^2+2/3)点乘I向量ABI
∵(1)OC=
13OA+
23OB,∴AC=OC-
OA=-23OA+23OB,AB=OB-
OA,…(1分)
∴AB=23AC …(4分),∴AC∥AB,即A,B,C三点共线. …(5分)
(2)由A(1,cosx),B(1+sinx,cosx),x∈[0,
π2],…(6分)
∵AB=(sinx,0),∴|
AB|=
sin2x=sinx,…(7分)
从而f(x)=
OA•
OC-(2m2+
23)•|
AB|=1+
23sinx+cos2x-(2m2+
23)sinx. …(10分)
又x∈[0,
π2],则sinx∈[0,1],
当0≤m2<
12时,f(x)的最小值f(
π2)=-(1+m2)2+m4+2=
12.∴m2=
14,∴m=±
12. …(12分)
当m2≥
12时,f(x)的最小值f(0)=-(0+m2)2+m4+2=
12.∴m无解,
综上,m=±
12.