What is the angle between two skew lines?

[Kernul]

Homework Statement

The problem asks me to evaluate the angle between these two lines:
$r : \begin{cases} x - 2y - 3 = 0 \\ 3y + z = 0 \end{cases} s : \begin{cases} x = 1 + 4t \\ y = 2 - 3t \\ z = 3 \end{cases}$
both oriented to the decreasing $y$.

Homework Equations

The Attempt at a Solution

Having found $\vec v_r = (-2, -1, 3)$, $\vec v_s = (4, -3, 0)$, $P_r (1, -1, 3)$, and $P_s (1, 2, 3)$
I already know that the lines are askew. I then found out that in order to find the angle between the two lines, I have to first find a plane containing one of the two lines(for example $r$) that is at the same time parallel to the other one(in this example $s$). In a few words I have to find a line parallel to $s$ that meets the line $r$ in a point that belongs to $r$.
The thing is that I don't know how to find that parallel line to $s$ that at the same time passes into a point $P_r$ belonging to the line $r$.
Should I take one of the Cartesian equations of $r$ and see the projection of $s$ on it so to have the parallel line? And then see the interjection between this parallel line and $r$? Or I should proceed in another way?
By the way, this is the Cartesian form of the $s$ line I found:
$s : \begin{cases} \frac{3}{4}x + y - \frac{11}{4} = 0 \\ z - 3 = 0 \end{cases}$

 

[mfb]

Do you know the dot product = scalar product? You don't have to construct new planes and whatever, once you have v_r and v_s the angle can be found in a single line on paper.

 

[Kernul]

Do you know the dot product = scalar product? You don't have to construct new planes and whatever, once you have v_r and v_s the angle can be found in a single line on paper.

Ohw... I was so concentrated on the parallel line I didn't thought of that...
Thank you and sorry.