Acute angle bisector between two 3D lines

Question

If $L_1$ and $L_2$ are two 3D lines represented by the equation ${L_1}:\frac{{x - 1}}{1} = \frac{y}{{ - 1}} = \frac{{z - 1}}{3}$ & ${L_2}:\frac{{x - 1}}{{ - 3}} = \frac{y}{{ - 1}} = \frac{{z - 1}}{1}$. If the line L bisects the acute angle between the lines $L_1$ and $L_2$. Then find the equation of the Line "L".

My approach is as follow the point of intersection is $(1,0,1)$.

${L_1}:\frac{{x - 1}}{1} = \frac{y}{{ - 1}} = \frac{{z - 1}}{3} \Rightarrow \overrightarrow r = \hat i + \hat k + \mu \left( {\hat i - \hat j + 3\hat k} \right) = \overrightarrow a + \mu \overrightarrow c $

${L_2}:\frac{{x - 1}}{{ - 3}} = \frac{y}{{ - 1}} = \frac{{z - 1}}{1} \Rightarrow \overrightarrow r = \hat i + \hat k + \mu \left( { - 3\hat i - \hat j + \hat k} \right) = \overrightarrow b + \lambda \overrightarrow d $

The direction vector of bisector between $L_1$ and $L_2$ is

$\overrightarrow T = \frac{{\overrightarrow c }}{{\left| {\overrightarrow c } \right|}} + \frac{{\overrightarrow d }}{{\left| {\overrightarrow d } \right|}} = \frac{{\hat i - \hat j + 3\hat k}}{{\left| {\hat i - \hat j + 3\hat k} \right|}} + \frac{{ - 3\hat i - \hat j + \hat k}}{{\left| { - 3\hat i - \hat j + \hat k} \right|}} = \frac{{ - 2\hat i - 2\hat j + 4\hat k}}{{\sqrt {11} }} = \left\langle {1,1, - 2} \right\rangle = \left\langle {\ell ,m, - 2} \right\rangle $

$\overrightarrow U = \frac{{\overrightarrow c }}{{\left| {\overrightarrow c } \right|}} - \frac{{\overrightarrow d }}{{\left| {\overrightarrow d } \right|}} = \frac{{\hat i - \hat j + 3\hat k}}{{\left| {\hat i - \hat j + 3\hat k} \right|}} - \frac{{ - 3\hat i - \hat j + \hat k}}{{\left| { - 3\hat i - \hat j + \hat k} \right|}} = \frac{{4\hat i + 2\hat k}}{{\sqrt {11} }} = \left\langle { - 4,0, - 2} \right\rangle = \left\langle {\ell ,m, - 2} \right\rangle $.

From here I am confused, my assumption is that if $\overrightarrow c .\overrightarrow d > 0$, then $\overrightarrow T $ is the direction vector that is acute angle and if $\overrightarrow c .\overrightarrow d < 0$, then $\overrightarrow U $ is the direction vector that is acute angle. Just want to clarify it.

Answer 1

We see that the direction vectors have the same length, and their dot product is positive. Thus, merely adding them will give a vector pointing in the direction of the bisector of the acute angle: $(1,-1,3)+(-3,-1,1)=(-2,-2,4)$. So the equation of $L$ is .$$(1,0,1)+t(1,1,-2)$$

Answer 2

For two given lines, $\vec{a} \cdot \vec{b} = |a|.|b|\cos \theta$ where $\theta$ is the angle between the lines.

As, $\cos \theta \ge 0 \, $ for $(0 \le \theta \le {\pi/2}); \cos \theta \lt 0 \, $ for $({\pi/2} \lt \theta \le {\pi}), $
it is easy to remember that the if the dot product is positive, then you have an acute angle between the vectors. So just add the unit vectors for the angle bisector. If the dot product is negative, you have obtuse angle between them and you need one of the vectors in the opposite direction to find the acute angle so subtract the unit vectors to find the angle bisector of the acute angle between lines.