# Algebra Question: vector equations

## 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

## 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?

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

andrevdh
Homework Helper
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.

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