已知G是△ABC的重心,若PQ过△ABC的重心G,且OA=a,OB=b,OP=ma,OQ=nb求证1/m+1/n=3http://zhidao.baidu.com/q?ct=17&pn=0&tn=ikask&rn=12&word=%B8%DF%D2%BB%CA%FD%D1%A7%A3%A1%A3%A1%B8%DF%CA%D6%CB%D9%C0%B4%21%21%21%21%21&cm=1&lm=394496这里全都
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![已知G是△ABC的重心,若PQ过△ABC的重心G,且OA=a,OB=b,OP=ma,OQ=nb求证1/m+1/n=3http://zhidao.baidu.com/q?ct=17&pn=0&tn=ikask&rn=12&word=%B8%DF%D2%BB%CA%FD%D1%A7%A3%A1%A3%A1%B8%DF%CA%D6%CB%D9%C0%B4%21%21%21%21%21&cm=1&lm=394496这里全都](/uploads/image/z/7123923-27-3.jpg?t=%E5%B7%B2%E7%9F%A5G%E6%98%AF%E2%96%B3ABC%E7%9A%84%E9%87%8D%E5%BF%83%2C%E8%8B%A5PQ%E8%BF%87%E2%96%B3ABC%E7%9A%84%E9%87%8D%E5%BF%83G%2C%E4%B8%94OA%3Da%2COB%3Db%2COP%3Dma%2COQ%3Dnb%E6%B1%82%E8%AF%811%2Fm%2B1%2Fn%3D3http%3A%2F%2Fzhidao.baidu.com%2Fq%3Fct%3D17%26pn%3D0%26tn%3Dikask%26rn%3D12%26word%3D%25B8%25DF%25D2%25BB%25CA%25FD%25D1%25A7%25A3%25A1%25A3%25A1%25B8%25DF%25CA%25D6%25CB%25D9%25C0%25B4%2521%2521%2521%2521%2521%26cm%3D1%26lm%3D394496%E8%BF%99%E9%87%8C%E5%85%A8%E9%83%BD)
已知G是△ABC的重心,若PQ过△ABC的重心G,且OA=a,OB=b,OP=ma,OQ=nb求证1/m+1/n=3http://zhidao.baidu.com/q?ct=17&pn=0&tn=ikask&rn=12&word=%B8%DF%D2%BB%CA%FD%D1%A7%A3%A1%A3%A1%B8%DF%CA%D6%CB%D9%C0%B4%21%21%21%21%21&cm=1&lm=394496这里全都
已知G是△ABC的重心,若PQ过△ABC的重心G,且OA=a,OB=b,OP=ma,OQ=nb
求证1/m+1/n=3
http://zhidao.baidu.com/q?ct=17&pn=0&tn=ikask&rn=12&word=%B8%DF%D2%BB%CA%FD%D1%A7%A3%A1%A3%A1%B8%DF%CA%D6%CB%D9%C0%B4%21%21%21%21%21&cm=1&lm=394496
这里全都是向量!我打的△ABC是△ABO,请见谅!
已知G是△ABC的重心,若PQ过△ABC的重心G,且OA=a,OB=b,OP=ma,OQ=nb求证1/m+1/n=3http://zhidao.baidu.com/q?ct=17&pn=0&tn=ikask&rn=12&word=%B8%DF%D2%BB%CA%FD%D1%A7%A3%A1%A3%A1%B8%DF%CA%D6%CB%D9%C0%B4%21%21%21%21%21&cm=1&lm=394496这里全都
这道题应该根据PG和PQ共线来解
PG=PA+AG
=OA-OP+AC
=-am+1/3a+1/3b
PQ=OQ-OP
=nb-ma
∴PG=μPQ
μ·(nb-ma)=-am+1/3a+1/3b
kn=1/3.①
km=m-1/3.②
∴m=3mn-n
即1/m+1/n=3
应该是△ABO吧。以下字母均带箭头:
首先求出OG=1/3a+1/3b(这个总会吧),然后求出PG=PO+OG=-ma+1/3(a+b)=(1/3-m)a+1/3b 然后求出QG=QO+OG=1/3b+(1/3-n)b
因为PG与QG共线,所以PG=xQG 其中x为常数,带入可得x=1-3m;x=1/(1-3n)
解方程可得1/m+1/n=3。
CG=1/3(a+b)
PQ=nb-ma
PG=CG-CP=1/3(a+b)-ma
PQ,PG共线
1/3(a+b)-ma=k(nb-ma)=knb-kma
kn=1/3。。。。。。。。。①
km=m-1/3。。。。。。。。②
解得
m=3mn-n
即1/m+1/n=3