Distance between two straight lines in space

The distance between two straight lines $r$ and $r'$, $\text{d}(r,r')$, is the minimal distance between any point of $r$ and any other point of $r'$.

• If the straight lines coincide or are secant, the distance between them is zero, $\text{d}(r,r')=0$.
• If the straight lines are parallel, the distance between them can be calculated at any point of one of the lines, $P\in r$ or $P'\in r'$, and finding the distance to the other straight line: $$\text{d}(r,r')=\text{d}(P,r')=\text{d}(r,P')$$
• If the straight lines cross, the following general formula is deduced to calculate the distance between them:

We take a point $A$ belonging to $r$ and another point $A'$ belonging to $r'$. Let $\vec{v}$ and $\vec{v}'$ be the governing vectors of $r$ and $r'$. We join the points $A$ and $A'$. The volume of the parallelepiped determined by $\overrightarrow{AA'}$, $\vec{v}$ and $\vec{v}'$, is the absolute value of the mixed product of these vectors: $$v_p=|[\overrightarrow{AA'},\vec{v},\vec{v}']|$$

On the other hand we can also calculate this volume by multiplying the area of the base and the height: $$v_p=|\vec{v}\times\vec{v}'|\text{d}(r,r')|$$

Therefore: $$\text{d}(r,r')=\dfrac{|[\overrightarrow{AA'},\vec{v},\vec{v}']|} {|\vec{v}\times\vec{v}'|}$$

We are going to calculate the distance between the straight lines: $$ r:x-2=\dfrac{y+3}{2}=z \qquad r':x=y=z$$

First we determine its relative position. To do it we must write the implicit equations of the straight line: $$ r:\left\{ \begin{array}{l} 2x-y-7=0 \\ x-z-2=0 \end{array} \right. \qquad r':\left\{ \begin{array}{l} x-y=0 \\ x-z=0 \end{array} \right.$$

And we calculate the rank of the matrix of the resulting systems of equations: $$|M'|=\begin{vmatrix} 2 & -1 & 0 & 7 \\ 1 & 0 & -1 & 2 \\ 1 & -1 & 0 & 0 \\ 1 & 0 & -1 & 0 \end{vmatrix} =2 \neq 0 $$

Therefore $\text{rank}(M')=4$ and the two straight lines intersect. Now we must find a point and the governing vector of each line.

For the straight line $r$: $A=(2,-3,0)$ and $\vec{v}=(1,2,1)$.

For the straight line $r'$: $A'=(0,0,0)$ and $\vec{v}=(1,1,1)$.

So we have: $\overrightarrow{AA'}=(-2,3,0)$

$$\begin{array}{rl} |\vec{v}\times\vec{v}'|=&\left| \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 1 & 2 & 1 \\ 1 & 1 & 1 \end{vmatrix} \right|= |2\vec{i}+\vec{j}+\vec{k}-2\vec{k}-\vec{j}-\vec{i}|= |\vec{i}-\vec{k}| \\ =& |(1,0,-1)| = \sqrt{1^2+0^2+(-1)^2}=\sqrt{2} \end{array}$$

$$[\overrightarrow{AA'},\vec{v},\vec{v}']= \begin{vmatrix} -2 & 3 & 0 \\ 1 & 2 & 1 \\ 1 & 1 & 1 \end{vmatrix} = -4+3+2-3=-2 $$

Finally: $$\text{d}(r,r')=\dfrac{|[\overrightarrow{AA'},\vec{v},\vec{v}']|} {|\vec{v}\times\vec{v}'|}= \dfrac{|-2|}{\sqrt{2}}=\sqrt{2}$$