设G为三角形ABC重心 则 mAG向量+2nBG向量+3kCG向量=0向量 且m不等于0 求m:n:k= ?
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解决时间 2021-01-04 00:48
- 提问者网友：你给我的爱
- 2021-01-03 07:22
设G为三角形ABC重心 则 mAG向量+2nBG向量+3kCG向量=0向量 且m不等于0 求m:n:k= ?
最佳答案
- 五星知识达人网友：何以畏孤独
- 2021-01-10 05:51
G is CG
(OA+OB+OC)/3 = OG
3OG = OA+OB+OC (1)
mAG+2nBG+3kCG=0
m(OG-OA)+2n(OG-OB)+3k(OG-OC)=0
(m+2n+3k)OG = mOA+2nOB+3kOG
3OG = [3/(m+2n+3k)] [mOA+2nOB+3kOG] (2)
from (1) ,(2), we have
3m/(m+2n+3k) :6n/(m+2n+3k) :9k/(m+2n+3k) =1:1:1
=> 3m:6n:9k
m:n:k = 6:3:2
(OA+OB+OC)/3 = OG
3OG = OA+OB+OC (1)
mAG+2nBG+3kCG=0
m(OG-OA)+2n(OG-OB)+3k(OG-OC)=0
(m+2n+3k)OG = mOA+2nOB+3kOG
3OG = [3/(m+2n+3k)] [mOA+2nOB+3kOG] (2)
from (1) ,(2), we have
3m/(m+2n+3k) :6n/(m+2n+3k) :9k/(m+2n+3k) =1:1:1
=> 3m:6n:9k
m:n:k = 6:3:2
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- 1楼网友：詩光轨車
- 2021-01-10 06:52
搜一下：设G为三角形ABC重心 则 mAG向量+2nBG向量+3kCG向量=0向量 且m不等于0 求m:n:k= ?
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