Find the shortest distance between the line $x$ $=$ $2$ $-$ $t$ $y$ $=$ $-1$ $+$ $2t$ and $z$ $=$ $-1$ $+$ $t$ and the line $x$ $=$ $5$ $+$ $3s$ $y$ $=$ $0$ and $z$ $=$ $2$ $-$ $s$. Find the points on these two lines that give this shortest distance.
Using the cross product, I got the normal vector to be:
\begin{pmatrix}-2\\ -2\\ -6\end{pmatrix}
The I selected two points on the lines with $t$ $=$ $-1$:
\begin{pmatrix}3\\ -3\\ -2\end{pmatrix}
and with $s$ $=$ $1$:
\begin{pmatrix}8\\ 0\\ 1\end{pmatrix}
Therefore the distance between the these two points gave the vector:
\begin{pmatrix}-5\\ \:-3\\ \:-3\end{pmatrix}
So then I projected this vector:
$\left(\frac{\left(-5,\:-3,\:-3\right)\cdot \left(-5,\:-3,\:-3\right)}{\left(-2,\:-2,\:-6\right)\cdot \left(-2,\:-2,\:-6\right)}\right)\cdot$ $\begin{pmatrix}-2\\ -2\\ -6\end{pmatrix}$
This left me with:
$\frac{43}{\:44}\cdot$ $\begin{pmatrix}-2\\ -2\\ -6\end{pmatrix}$
Which left me with a final distance of:
$\frac{43}{44}\sqrt{44}$
But this answer isn't correct. I think I have made a mistake in my projection calculation but I am not sure where. Also, how would I find the two points that are closest together. I am also a little confused about that.
Any help would be highly appreciated!