I have two lines $l_{1}$ and $l_{2}$ with respective equations $\textbf{r} = \lambda(\textbf{i+j+k})$ and $\textbf{r} = (2\textbf{i}+\textbf{j}) + \mu(3\textbf{i} - \textbf{j} +4\textbf{k})$. Here $\lambda, \mu$ are parameters. I'd like to find the shortest distance between two lines.
My approach:
Let point $A$ be on $l_{1}$, point $B$ on $l_{2}$ such that length$|\vec{AB}|$ gives us shortest distance. Then $\vec{AB}$ is normal to both $(\bf{i+j+k})$ and $(3\textbf{i} - \textbf{j} +4\textbf{k})$. I can use that $\vec{AB} \cdot (\textbf{i+j+k}) = 0$ and $ \vec{AB} \cdot (3\textbf{i} - \textbf{j} +4\textbf{k}) = 0$ to form two simultaneous equations and solve for $\lambda$ and $\mu$ to get coordinates of $A$ and $B$. Then I can simply compute $|\vec{AB}|$.
But this does not give me the same value for distance as the traditional formula for the shortest distance between two lines (using the vector product approach between their directions). I don't understand why? Please help.
Doesn't my distance $|\vec{AB}|$ correspond to the formula?
I thought my approach has the advantage of additionally giving the coordinates of points on the lines that correspond to shortest distance, too.
My calculation:
$\vec{AB} = (2 + 3\mu - \lambda)\textbf{i} + (1 - \mu - \lambda)\textbf{j} + (4\mu - \lambda)\textbf{k}.$
Now
$ \vec{AB} \cdot (\textbf{i} + \textbf{j} + \textbf{k}) =(2 + 3\mu - \lambda) + (1-\mu - \lambda) + (4\mu - \lambda) = 0,$
and
$ \vec{AB} \cdot (3\textbf{i} - \textbf{j} +4\textbf{k}) = 3(2 + 3\mu - \lambda) - (1-\mu - \lambda) + 4(4\mu - \lambda) = 0.$
This two equations yield $\mu = -\frac{5}{26}$ and $\lambda = \frac{8}{13}$, yielding $|\vec{AB}| = \sqrt{\frac{981}{338}}$. But using the traditional formula gives distance $\frac{9}{\sqrt{42}}$.
Checking using MATHEMATICA gives $ \mu = \frac{1}{14}$ and $\lambda = \frac{8}{7}$. In which case $|AB|$ simply vanishes, and I have no idea where the mistake is...
You seem to have made a mistake in simplyfing and/or solving the 2 equations.
The 2 equations boil down to
$$ 0 = 3+ 6\mu - 3\lambda,$$
and
$$ 0 = 5+26\mu -6\lambda.$$
The solution of this is
$$\mu=\frac1{14}, \lambda = \frac{16}{14}.$$
This leads to
$$\overrightarrow{AB}=\left(\frac{15}{14}, \frac{-3}{14}, \frac{-12}{14}\right)$$
and finally to
$$|\overrightarrow{AB}|=\frac{\sqrt{15^2 + (-3)^2 + (-12)^2}}{14} =\frac{\sqrt{378}}{14}=\frac{\sqrt{2\times 3^3 \times 7}}{14}=\frac9{\sqrt{42}}, $$
which is the result you got with the other method.