## Derivation of Formula For Distance Between Parallel Lines in Three Dimensions

If two lines in three dimensions are parallel, the coefficients in the Cartesian equations are the same, since the coefficients of the Cartesian equation are the components of the tangent, which is the same for parallel lines.
Suppose we want to find the distance between the parallel lines
$\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}=t_1$
(1)
$\frac{x-x_2}{a}=\frac{y-y_2}{b}=\frac{z-z_2}{c}=t_2$
(2)
/> Rearrange (1) into vector form to give
$\mathbf{r}_1(t_1)= \begin{pmatrix}x_1\\y_1\\z_1\end{pmatrix}+ \begin{pmatrix}a\\b\\c\end{pmatrix}t_1$

Rearrange (2) into vector form to give
$\mathbf{r}_2(t_2)= \begin{pmatrix}x_2\\y_2\\z_2\end{pmatrix}+ \begin{pmatrix}a\\b\\c\end{pmatrix}t_2$

(2)-(1) gives
$\mathbf{r}_{12}= \begin{pmatrix}x_2-x_1\\y_2-y_1\\z_2-z_1\end{pmatrix}+ \begin{pmatrix}a\\b\\c\end{pmatrix}(t_2-t_1)$

We can take this vector as perpendicular to the tangent vector
$\begin{pmatrix}a\\b\\c\end{pmatrix}$
for suitable values
$t'_1, \; t'_2$
of
$t_1, \; t_2$
. Hence
\begin{aligned} &( \begin{pmatrix}x_2-x_1\\y_2-y_1\\z_2-z_1\end{pmatrix}+ \begin{pmatrix}a\\b\\c\end{pmatrix}(t'_2-t'_1)) \cdot \begin{pmatrix}a\\b\\c\end{pmatrix} \\ &= \begin{pmatrix}x_2-x_1+a(t'_2-t'_1)\\y_2-y_1++b(t'_2-t'_1)\\z_2-z_1+c(t'_2-t'_1)\end{pmatrix} \cdot \begin{pmatrix}a\\b\\c\end{pmatrix}=0 \end{aligned}

This gives the equation
\begin{aligned} & a(x_2-x_1)+a^2(t'_2-t'_1)+b(y_2-y_1)+b^2(t'_2-t'_1)+c(z_2-z_1)+c^2(t'_2-t'_1)=0 \\ & \rightarrow t'_2-t'_1=- \frac{a(x_2-x_1)+b(y_2-y_1)+c(z_2-z_1)}{a^2+b^2+c^2} \end{aligned}

\begin{aligned} \left| \mathbf{r}_{12} \right| &= \sqrt{(x_2-x_1+a(t'_2-t'_1))^2+(y_2-y_1+b(t'_2-t'_1))^2+(z_2-z_1+c(t'_2-t'_1))^2} \\ &= \sqrt{(x_2-x_1-a( \frac{a(x_2-x_1)+b(y_2-y_1)+c(z_2-z_1)}{a^2+b^2+c^2}))^2+(y_2-y_1-b( \frac{a(x_2-x_1)+b(y_2-y_1)+c(z_2-z_1)}{a^2+b^2+c^2}))^2+(z_2-z_1-c( \frac{a(x_2-x_1)+b(y_2-y_1)+c(z_2-z_1)}{a^2+b^2+c^2}))^2} \\ &= \sqrt{\frac{(a(x_2-x_1)+b(y_2-y_1)+c(z_2-z_1))^2}{a^2+b^2+c^2}} \end{aligned}