作业帮平面向量题目一道..在三角形ABC中 D为BC上一点,P为AD上一点 且满足(向量)AP=5/13(向量)AB+4/13(向量)AC,试求(模长)BD:CD 4 思路.
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平面向量题目一道..在三角形ABC中 D为BC上一点,P为AD上一点 且满足(向量)AP=5/13(向量)AB+4/13(向量)AC,试求(模长)BD:CD 4 思路.
平面向量题目一道..
在三角形ABC中 D为BC上一点,P为AD上一点 且满足(向量)AP=5/13(向量)AB+4/13(向量)AC,试求(模长)BD:CD
4 思路.
平面向量题目一道..在三角形ABC中 D为BC上一点,P为AD上一点 且满足(向量)AP=5/13(向量)AB+4/13(向量)AC,试求(模长)BD:CD 4 思路.
设BD:CD =m:n,则
向量AD=n/(m+n)向量AB+m/(m+n)向量AC
A,P,D三点共线,
向量AD=λ向量AP=(5/13)λ向量AB+(4/13)λ向量AC
所以,m/(m+n):n/(m+n)=(4/13)λ:(5/13)λ
所以,m:n=4:5
BD:CD =m:n=4:5.
AP=(5/13)AB+(4/13)AC
D is on BC
let: BD = kBC ( k is a constant )
P is on AD
let |AP|/|PD| = m
BP = (BA+ mBD)/(1+m)
BA+AP = (BA+ mBD)/(1+m)
AP = [m/(1+m)]AB + (m/(...
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AP=(5/13)AB+(4/13)AC
D is on BC
let: BD = kBC ( k is a constant )
P is on AD
let |AP|/|PD| = m
BP = (BA+ mBD)/(1+m)
BA+AP = (BA+ mBD)/(1+m)
AP = [m/(1+m)]AB + (m/(1+m)) kBC
=[m/(1+m)]AB + (m/(1+m)) k(-AB+AC)
= [m(1-k)/(1+m)]AB + (km/(1+m)) AC
(5/13)AB+(4/13)AC= [m(1-k)/(1+m)]AB + (km/(1+m)) AC
=>
m(1-k)/(1+m) =5/13 (1) and
(km/(1+m)) = 4/13 (2)
(1)/(2)
(1-k)/k = 5/4
5k= 4-4k
k = 4/9
ie
BD = (4/9)BC
BD+DC = BC
=> DC = (5/9)BC
|BD|/|DC| = 4/5
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