已知三点A(cosa,sina)B(cosb,sinb)C(cosc,sinc)若向量OA+k向量OB+(2—k)向量OC=0,(k为常数,且0
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![已知三点A(cosa,sina)B(cosb,sinb)C(cosc,sinc)若向量OA+k向量OB+(2—k)向量OC=0,(k为常数,且0](/uploads/image/z/3775699-19-9.jpg?t=%E5%B7%B2%E7%9F%A5%E4%B8%89%E7%82%B9A%EF%BC%88cosa%2Csina%29B%28cosb%2Csinb%29C%28cosc%2Csinc%29%E8%8B%A5%E5%90%91%E9%87%8FOA%2Bk%E5%90%91%E9%87%8FOB%2B%EF%BC%882%E2%80%94k%EF%BC%89%E5%90%91%E9%87%8FOC%3D0%2C%EF%BC%88k%E4%B8%BA%E5%B8%B8%E6%95%B0%2C%E4%B8%940)
已知三点A(cosa,sina)B(cosb,sinb)C(cosc,sinc)若向量OA+k向量OB+(2—k)向量OC=0,(k为常数,且0
已知三点A(cosa,sina)B(cosb,sinb)C(cosc,sinc)若向量OA+k向量OB+(2—k)向量OC=0,(k为常数,且0
已知三点A(cosa,sina)B(cosb,sinb)C(cosc,sinc)若向量OA+k向量OB+(2—k)向量OC=0,(k为常数,且0
题目看不明白,我为你可以解答,数学问题请Q463863369!
(2—k)这是减吗?
∵向量a+b+c=0,
∴cosA+kcosB+(2-k)cosC=0
sinA+ksinB+(2-k)sinC=0
(sinA)^2+(cosA)^2
=[kcosB+(2-k)cosC]^2+[ksinB+(2-k)sinC]^2
=k^2+(2-k)^2+2k(2-k)(cosBcosC+sinBsinC)
=k^2+(2-...
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∵向量a+b+c=0,
∴cosA+kcosB+(2-k)cosC=0
sinA+ksinB+(2-k)sinC=0
(sinA)^2+(cosA)^2
=[kcosB+(2-k)cosC]^2+[ksinB+(2-k)sinC]^2
=k^2+(2-k)^2+2k(2-k)(cosBcosC+sinBsinC)
=k^2+(2-k)^2+2k(2-k)cos(B-C)=1
即cos(B-C)=[k^2+(2-k)^2-1]/2k(k-2)
=(2k^2-4k+3)/(2k^2-4k)
=1+3/(2k^2-4k)
∵0<k<2∴-2<(2k^2-4k<0
1/(2k^2-4k<-1/2
cos(B-C)<-1/2
又-1≤cos(B-C)≤1
∴-1≤cos(B-C)<-1/2
当cos(B-C)=-1时,1+3/(2k^2-4k)=-1
解得k=3/2,k=1/2
收起