How do I prove the lines are skew lines and the distance between them?

Question

What's the approach to prove two lines are skew lines aka, not parallel and non-intersecting AND the distance between the two?

Line 1: $\frac {x-3}2 = 4-y = z = \frac{z-1}3 $

Line 2: $\frac {x-1}4 = \frac{3-y}2 = \frac{z - 4}5$

Answer 1

These are just basic vector calculations:

To find if they are skew:

1. Convert to parametric equations

2. Equate the $x$, $y$ and $z$ components,

3. Solve the $x$ and $y$ components simultaneously

4. Substitute the solved values into the $z$ equation

5. If you get a simultaneous solution the lines intersect and are therefore not skew, no simultaneous solution shows that the lines are do not intersect.

6. To determine the angle suppose one line was translated to intersect the other, then evaluate whether one direction vector is parallel to the other i.e. $v_a=k*v_b$ $\forall k\in \Bbb R$

7. If the lines are not parallel and not intersecting, then they are skew :)

Answer 2

OP (opponent) has some problems in line 1.

If $l_1: P=\vec{a_1}+\vec{u_1}t$ and $l_2: P=\vec{a_2}+\vec{u_2}t$ then $$d=\frac{|(\vec{u_1}\times \vec{u_2})\bullet (\vec{a_1} - \vec{a_2})|}{||\vec{u_1}\times \vec{u_2}||}$$ is the disince between any two lines in $\Bbb{R}^3$. Is this formula correct? Maybe except paralell lines... I always forget it and then google about it.

The two lines above are parallel if $\vec{u_1}$ is a multiple of $\vec{u_2}$.