Algebra Question: vector equations

[Accurim]

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

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

\frac{4a+3b}{5}=\frac{6a+b}{6}
a=\frac{13}{6}b

I'm not exactly sure what the question is asking, is there anything else I have to do?

 

[Dick]

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

 

[andrevdh]

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

\vec{r_b} = (1,2) + S(x,y)

The direction vector of r_b will then be midway between the direction vectors of r1 and r2.