How to find the angle bisectors of 2 lines

Question

I tried gooogling how to do it but there were none that could relate to the problem I have.

Find the angle bisectors of the lines: $$g: r= \begin{pmatrix} 2 \\ 5 \\ -9 \\ \end{pmatrix}+u \begin{pmatrix} 8 \\ 4 \\ 1 \\ \end{pmatrix} $$

$$f: r= \begin{pmatrix} 2 \\ 5 \\ -9 \\ \end{pmatrix}+v \begin{pmatrix} 12 \\ 4 \\ 3 \\ \end{pmatrix} $$

Could someone give me an idea on how to start solving this question? I know that in order to determine the direction of the angle bisectors of the lines need to have the same length. So if

$u$ or $v$ =$0$ then they would have the same length no? But idk how to go from there.

Answer 1

$\|(8,4,1)\| = \sqrt{64+16+1} = 9\\ \|(12,4,3)\| = \sqrt{144+16+9} = 13$

If we scale our vectors by $\frac {1}9,\frac {1}{13}$ respectively we will have two unit vectors pointing in different directions. When we sum them, the sum will bisect the angle between them.

$r = \pmatrix{2\\5\\-9} + s\left(\frac 19 \pmatrix{8\\4\\1} + \frac {1}{13}\pmatrix{12\\4\\3}\right)$

Answer 2

First normalise the two directing vectors. A directing vector of the angle bisectors of the two lines are simply the sum and the difference of the two normalised directing vectors, by a property of rhombuses.