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Algebra Question: vector equations

  1. Feb 14, 2007 #1
    1. The problem statement, all variables and given/known data

    The problem -
    Consider lines L1 and L2:
    L1: r1 = (X,Y) = (1,2) + T (6,1)
    L2: r2 = (X,Y) = (1,2) + K (4,3)

    Find an equation for either of the bisectors of the angles between L1 and L2


    2. Relevant equations



    3. The attempt at a solution

    [tex]cos\alpha = \frac{(a,b)\bullet(4,3)}{\|(a,b)\|\|(4,3)\|}=\frac{(a,b)\bullet(6,1)}{\|(a,b)\|\|(6,1)\|}[/tex]

    [tex]\frac{4a+3b}{5}=\frac{6a+b}{6}[/tex]
    [tex]a=\frac{13}{6}b[/tex]

    I'm not exactly sure what the question is asking, is there anything else I have to do?
     
  2. jcsd
  3. Feb 14, 2007 #2

    Dick

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    Homework Helper

    Well, ||(6,1)|| is not 6. So what is (a,b)? You don't have an equation for a line yet.
     
  4. Feb 15, 2007 #3

    andrevdh

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    Homework Helper

    Notice that both vectors go through the point P(1,2) so the position vector will also be [tex]\vec{p}(1,2)[/tex]. The vector equation of the bisector line will then be of the form

    [tex]\vec{r_b} = (1,2) + S(x,y)[/tex]

    The direction vector of [tex]r_b[/tex] will then be midway between the direction vectors of r1 and r2.
     
    Last edited: Feb 15, 2007
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