Bisector of two lines in the euclidean space $\mathbb{E}_3$ Let $$r: \begin{cases} x  +  z = 0 \\ y + z + 1 = 0\end{cases}$$ and $$s: \begin{cases} x  - y - 1  = 0 \\ 2x - z -1 = 0\end{cases}$$ 
be two lines in the euclidean space $\mathbb{E}_3$. It is easily seen that their intersection is one point. 
How do I find the cartesian equation of the bisector of the angle that these lines form?
 A: The unit direction vectors of the first line and the second line are 
$$\overline a=\left(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}}\right)$$
and  
$$\overline b=\left(\frac{1}{\sqrt{6}},\frac{1}{\sqrt{6}},\frac{2}{\sqrt{6}}\right),$$
respectively. (The scalar product of these two vectors is $0$. That is, our lines are perpendicular.)
(Proof: Take $x=0$ and $x=1$. In the case of the first line we have  $(0,-1,0)$ for $x=0$ and $(1,0,-1)$ for $x=1$. Subtracting the first vector from the second one: a direction vector is $(1,1,-1)$; the length of this vector is $\sqrt{3}$. Then, in the case of the second line if $x=0$ then $(0,-1,-1)$ , if $x=1$ then we have $(1,0,1)$ . Subtracting the first vector from the second one we get $(1,1,2)$, the length is then $\sqrt{6}$.)
Summing $\overline a$ and $\overline b $ we get the following vector 
$$\overline c=\left(\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{6}},\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{6}},-\frac{1}{\sqrt{3}}+\frac{2}{\sqrt{6}}\right).$$
I claim that $(\overline c, \overline a, \overline b)$ are coplanar and $\overline c$ is their angle bisector. This follows from the paralelogramma rule but it can be checked by taking the following scalar products $$(\overline a,\overline c)=(\overline b,\overline c)=1.$$ So, these angles equal. (Don't be mistaken, the length of $\overline c$ is not $1$. So, the value $(1)$ of the scalar products does not tell exactly the value of the angle, it tells only that the angles equal. But we know that the angle is $\frac{\pi}{4}.$
The two lines meet at the point  
$$\overline m=\left(\frac{1}{3},-\frac{2}{3},-\frac{1}{3}\right).$$ 
As a result the angle bisector line in parametric form is
$$\overline  l= \overline m+ t\overline c, \ -\infty \le t \le \infty.$$
Or
$$x(t)=\frac{1}{3}+t\left(\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{6}}\right) \tag 1,$$
$$y(t)=-\frac{2}{3}+t\left(\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{6}}\right) \tag 2,$$
$$z(t)=-\frac{1}{3}+t\left(-\frac{1}{\sqrt{3}}+\frac{2}{\sqrt{6}}\right) \tag 3.$$

In order to get the Cartesian equation of the angle bisector line, first, we express $t$ from $(1)$:
$$t=\frac{x-\frac{1}{3}}{\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{6}}} \tag 4$$
Subtracting $(2)$ from $(1)$ we get
$$x-y=1. \tag 5$$
Comparing $(3)$ and $(4)$ gives that
$$z=-\frac{1}{3}+\left(x-\frac{1}{3}\right)\frac{2-\sqrt 2}{1+\sqrt 2}  \tag 6.$$
Finally, $(5)$ and $(6)$ are the Cartesian equations of the angle bisector.
A: You said "easy to see" so I am assuming you know how to find it: you find the parametric equation of the lines in the form:
$x=a_1t+k$, $y=b_1t+l$, $z=c_1t+m$
And then your vectors become $(a_1,b_1,c_1)$ and $(a_2,b_2,c_2)$
And then you use the following formula and find $\theta$

