已知A、B、C是平面上不共线三点,动点P满足向量OP=1/3[(1-λ)向量OA+(1-λ)向量OB+(1+2λ)向量已知A、B、C是平面上不共线三点,动点P满足向量OP=1/3[(1-λ)向量OA+(1-λ)向量OB+(1+2λ)向量OC](λ∈R且λ≠0),O为
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/29 22:02:05
![已知A、B、C是平面上不共线三点,动点P满足向量OP=1/3[(1-λ)向量OA+(1-λ)向量OB+(1+2λ)向量已知A、B、C是平面上不共线三点,动点P满足向量OP=1/3[(1-λ)向量OA+(1-λ)向量OB+(1+2λ)向量OC](λ∈R且λ≠0),O为](/uploads/image/z/3941474-50-4.jpg?t=%E5%B7%B2%E7%9F%A5A%E3%80%81B%E3%80%81C%E6%98%AF%E5%B9%B3%E9%9D%A2%E4%B8%8A%E4%B8%8D%E5%85%B1%E7%BA%BF%E4%B8%89%E7%82%B9%2C%E5%8A%A8%E7%82%B9P%E6%BB%A1%E8%B6%B3%E5%90%91%E9%87%8FOP%3D1%2F3%5B%281-%CE%BB%29%E5%90%91%E9%87%8FOA%2B%281-%CE%BB%29%E5%90%91%E9%87%8FOB%2B%281%2B2%CE%BB%29%E5%90%91%E9%87%8F%E5%B7%B2%E7%9F%A5A%E3%80%81B%E3%80%81C%E6%98%AF%E5%B9%B3%E9%9D%A2%E4%B8%8A%E4%B8%8D%E5%85%B1%E7%BA%BF%E4%B8%89%E7%82%B9%2C%E5%8A%A8%E7%82%B9P%E6%BB%A1%E8%B6%B3%E5%90%91%E9%87%8FOP%3D1%2F3%5B%281-%CE%BB%29%E5%90%91%E9%87%8FOA%2B%281-%CE%BB%29%E5%90%91%E9%87%8FOB%2B%281%2B2%CE%BB%29%E5%90%91%E9%87%8FOC%5D%28%CE%BB%E2%88%88R%E4%B8%94%CE%BB%E2%89%A00%29%2CO%E4%B8%BA)
已知A、B、C是平面上不共线三点,动点P满足向量OP=1/3[(1-λ)向量OA+(1-λ)向量OB+(1+2λ)向量已知A、B、C是平面上不共线三点,动点P满足向量OP=1/3[(1-λ)向量OA+(1-λ)向量OB+(1+2λ)向量OC](λ∈R且λ≠0),O为
已知A、B、C是平面上不共线三点,动点P满足向量OP=1/3[(1-λ)向量OA+(1-λ)向量OB+(1+2λ)向量
已知A、B、C是平面上不共线三点,动点P满足向量OP=1/3[(1-λ)向量OA+(1-λ)向量OB+(1+2λ)向量OC](λ∈R且λ≠0),O为坐标原点,则P的轨迹一定通过△ABC的().
A.内心 B.垂心 C.重心 D.AB边的中点
“O是坐标原点”没有这句话- -
已知A、B、C是平面上不共线三点,动点P满足向量OP=1/3[(1-λ)向量OA+(1-λ)向量OB+(1+2λ)向量已知A、B、C是平面上不共线三点,动点P满足向量OP=1/3[(1-λ)向量OA+(1-λ)向量OB+(1+2λ)向量OC](λ∈R且λ≠0),O为
好吧,我来帮你做:
OP=OA+AP,OP=OB+BP,OP=OC+CP
故:3OP=(OA+OB+OC)-(PA+PB+PC)
而:3OP=(1-λ)OA+(1-λ)OB+(1+2λ)OC
=(OA+OB+OC)-λ(OA+OB-2OC)
故:PA+PB+PC=λ(OA+OB-2OC)
取线段AB的中点为D
OA+OB-2OC=2OD-2OC=2CD
而:PA+PB=2PD,即:2PD+PC=2λCD=2λ(PD-PC)
故:2(λ-1)PD=(1+2λ)PC
λ=1时,OP=OC,即:P点与C点重合
λ=-1/2时,2OP=OA+OB,即:P点与D点重合
λ≠1和-1/2时,PD与PC共线,即:C、P、D共线
CD为△ABC的一条中线,故P点定过△ABC的重心
设坐标分别为:P(x,y);A(x1,y1);B(x2,y2);C(x3,y3)
则有:
x=1/3[(1-λ)x1+(1-λ)x2+(1+2λ)x3=(x1+x2+x3)/3-(x1+x2-2x3)λ/3
y=1/3[(1-λ)y1+(1-λ)y2+(1+2λ)y3=(y1+y2+y3)/3-(y1+y2-2y3)λ/3
将λ消去可得:
y-(y1+y2+...
全部展开
设坐标分别为:P(x,y);A(x1,y1);B(x2,y2);C(x3,y3)
则有:
x=1/3[(1-λ)x1+(1-λ)x2+(1+2λ)x3=(x1+x2+x3)/3-(x1+x2-2x3)λ/3
y=1/3[(1-λ)y1+(1-λ)y2+(1+2λ)y3=(y1+y2+y3)/3-(y1+y2-2y3)λ/3
将λ消去可得:
y-(y1+y2+y3)/3=(y1+y2-2y3)/(x1+x2-2x3)*[x-(x1+x2+x3)/3]
所以,当x=(x1+x2+x3)/3时,y=(y1+y2+y3)/3。
因此,过重心。
收起