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Algebra Question: vector equations

  • Thread starter Accurim
  • Start date
  • #1
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Homework Statement



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


Homework Equations





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?
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
26,258
618
Well, ||(6,1)|| is not 6. So what is (a,b)? You don't have an equation for a line yet.
 
  • #3
andrevdh
Homework Helper
2,128
116
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.
 
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