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Vector - Section formula

  1. Jul 25, 2011 #1
    Does anyone know why if
    (i) -1 < λ/μ < 0 then P is outside AB and nearer to A;
    (ii) λ/μ < -1 then P is outside AB and nearer to B?

    Can anybody give me the proof? Thanks in advance!

    Additional information: AP:PB = λ:μ

    dBXcF.jpg
     
  2. jcsd
  3. Jul 28, 2011 #2
    AP:PB = λ:μ is equivalent to AP/PB=λ/μ

    (i) -1 < λ/μ < 0 then P is outside AB and nearer to A;


    P is outside AB: |AP|>|AB|
    P is nearer to A: |AP| < |BP|

    -1 < λ/μ < 0
    -1 < AP/PB < 0
    1 > AP/BP > 0

    Let AP be +ve therefore BP is +ve (since AP/BP > 0)
    1 > AP/BP > 0
    BP > AP > 0
    BP > AP
    P is nearer to A
    BP > AP > 0
    AB+BP > AB+AP > AB
    AP > AB
    P is outside AB

    I can't be bothered with (ii),
     
  4. Aug 1, 2011 #3
    Thank you brother~
    But I think the condition that "P is outside AB: |AP|>|AB|" is not necessary.
     
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