Elementary geometry question: How to calculate distance between two skew lines? I am helping someone with highschool maths but I got stacked in a elementary geometry problem.
I am given the equation of two straigh lines in the space $r\equiv \begin{cases} x=1 \\ y=1 \\z=\lambda -2 \end{cases}$ and $s\equiv\begin{cases} x=\mu \\ y=\mu -1 \\ z=-1\end{cases}$ and asked for some calculations. First I am asked the relative position of them so I get they are skew lines. After that I am asked for the distance between the two lines. In order to get the distance I have to calculate the line that is perpendicular to both of them in the "skewing" point, check the points where it touches the other two lines (sorry, not sure about the word in English) and calculate the module of this vector.
Im having trouble calculating the perpendicular line. I know I can get the director vector using vectorial product, but  I'm not sure how to find a point so that I can build the line.

$\def\v{\mathbf v}\def\d{\mathbf d}$
The general expression can be computed as follows. Let the lines be given as:
$$
x=x_i+a_it,\quad y=y_i+b_it,\quad z=z_i+c_it
$$
with $i=1,2$.
Then the vector perpendicular to both given lines can be found from:
$$
\v_3=\v_1\times\v_2=\begin{vmatrix}
i&j&k\\
a_1&b_1&c_1\\
a_2&b_2&c_2\\
\end{vmatrix}
=(b_1c_2-c_1b_2,c_1a_2-a_1c_2,a_1b_2-b_1c_2).
$$
Now the distance $D$ between the lines can be found as the projection of the vector
$$
\d=(x_2-x_1,y_2-y_1,z_2-z_1)
$$
onto $\v_3$:
$$
D=\frac{|\v_3\cdot\d|}{|\v_3|}=\frac{|(b_1c_2-c_1b_2)(x_2-x_1)
+(c_1a_2-a_1c_2)(y_2-y_1)+(a_1b_2-b_1c_2)(z_2-z_1)|}{\sqrt{(b_1c_2-c_1b_2)^2+(c_1a_2-a_1c_2)^2+(a_1b_2-b_1c_2)^2}}.
$$
