My teacher gave me a list with the following exercise:
Let $r$ be the line that passes through $P = (1, 0, 5)$ and has directional vector $\vec{u} = (0, 1, -3)$ and $s$ the line that intersects the planes $\pi_{1} = \{(x, y, z); x = 1\}$ and $\pi_{2} = \{(x, y, z); y = 2\}$.
Determine if $r$ and $s$ are concurrent, and find their intersection point. Also, calculate the angle between those lines.
I managed to find the point where they intersect. This is what I did:
$$P = (1, 0, 5) + t(0, 1, -3) = (1, t, 5 - 3t)$$
As it was given to us, $x = 1$ and $y = 2$. That means that $t = 2$ and $z = -1$.
Therefore, the point that intersects the lines is $I = (1, 2, -1)$.
However, I'm having problems to calculate the angle between them.
Find the direction vector for the second line. It is trivial to observe the line will be parallel to $z$-axis, the direction vercor can be $(0, 0, 1)$. Now simply calculate the angle between the two vectors. For example, from $\vec{a} \cdot \vec{b} = |a||b|\;\text{cos}(\alpha)$