Algebra Question: vector equations

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SUMMARY

The discussion focuses on finding the equation for the bisector of the angles between two lines, L1 and L2, represented by vector equations. The lines are defined as L1: r1 = (1,2) + T(6,1) and L2: r2 = (1,2) + K(4,3). The solution involves calculating the cosine of the angle between the direction vectors and establishing a relationship between the variables a and b, ultimately leading to the vector equation of the bisector line in the form of r_b = (1,2) + S(x,y), where the direction vector is the average of the direction vectors of L1 and L2.

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