# Distance between two skew lines.

I am to show that the lines

$$x = y = z$$ and $$x + 1 = \frac{y}{2} = \frac{z}{3}$$

are skew and to find the distance between them.

My attempt: Let $$L_1: x=t \:\:\:\:\:\: y=t \:\:\:\:\:\: z=t$$ and

$$L_2: x=s-1 \:\:\:\:\:\: y = 2s \:\:\:\:\:\: z = 3s.$$

$t=s-1 \Rightarrow s = t+1 \Rightarrow t = 2t+2 \:\: \mbox{and} \:\: t = 3t+3$ which is impossible. So the lines do not have an intersection point and are not parallel because $\langle1,1,1\rangle$ and $\langle1,2,3\rangle$ are not parallel. Therefore the lines are skew.

Then, I have to find the distance between the lines. I sketched the following to do so:

$$\vec{m} = \vec{v_1} \times \vec{v_2} = \langle -1,2,-1 \rangle$$ $$\vec{n} = \vec{m} \times \vec{v_2} = \langle8,2-4\rangle$$

$$\Downarrow$$

$$P: 4x+y-2z+4 = 0$$

$$\Downarrow$$

$$P \cap L_1 : (-\frac{4}{3},-\frac{4}{3},-\frac{4}{3})$$

$$\Downarrow$$

$$L: x = -\frac{4}{3} -t \:\:\:\:\:\: y = -\frac{4}{3} +2t \:\:\:\:\:\: z =-\frac{4}{3} -t$$

$$\Downarrow$$

$L \cap L_2$ does not exist! But it must be.

Where did I go wrong?

You have correctly identified the vector perpendicular to both lines, i.e. $$\left(\begin{matrix}1\\-2\\1\end{matrix}\right)$$

Now find the vector joining any two point on the respective lines, for example, that joining $(0,0,0)$ and $(-1,0,0)$, therefore $$\left(\begin{matrix}1\\0\\0\end{matrix}\right)$$

and calculate the projection of this vector onto the normal.

Therefore the required distance is $$\frac{\left(\begin{matrix}1\\0\\0\end{matrix}\right)\cdot\left(\begin{matrix}1\\-2\\1\end{matrix}\right)}{\sqrt{1+4+1}}=\frac{1}{\sqrt{6}}$$

$L \cap L_2 = \left( - \frac 32 , -1 , - \frac 32 \right)$

From what I have found, using normal geometry extended to three different planes, as per your drawing, was:
L1:
y=x
y=z
z=x
L2:
y=2x+2
y=2/3*z
x=1/3*z-1
Then, I solved for each equation to find the three closest points:
{y=2x+2,y=x} » (-2,-2)
{y=2/3*z,y=z} » (0,0)
{x=1/3*z-1,x=z} » (-3/2,-3/2)
Then, I used the distance formula between each point: in (x,y,z) form
(-2,-2,any),(any,0,0)» distance 2
(-3/2,any,-3/2),(any,0,0)» distance 3/2
(-2,-2,any),(-3/2,any,-3/2)» distance 1/2
The distance is √(2^2+3^2/2^2+1/2^2)≈2.5495097568