# Find the shortest distance between the parallel lines $\large\frac{x-1}{-1}=\frac{y}{3}=\frac{z+3}{2}$ and $\large\frac{x-3}{-1}=\frac{y+1}{3}=\frac{z-1}{2}$

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• Distance between two lines
• (i) Shortest = 0 if they are intersecting.
• (ii) Parallel lines $\overrightarrow r = \overrightarrow a_1+t\overrightarrow u_1, \: \overrightarrow s=\overrightarrow a_2+s\overrightarrow u$
• (iii) Skew line $\overrightarrow r = \overrightarrow a_1+t\overrightarrow u, \: \overrightarrow r = \overrightarrow a_2+s\overrightarrow v$
• $d = \large\frac{[(\overrightarrow a_2-\overrightarrow a_1)\overrightarrow u\: \overrightarrow v]}{|\overrightarrow u \times \overrightarrow v|}$
$\large\frac{x-1}{-1} = \large\frac{y}{3}=\large\frac{z+3}{2} ; \large\frac{x-3}{-1}=\large\frac{y+1}{3} = \large\frac{z-1}{2}$
The lines are parallel to the vector with D.r.s $-1, 3, 2,$ i.e., $\overrightarrow u = -\overrightarrow i+3\overrightarrow j+2\overrightarrow k$
The first line passes through the point $A(1, 0, -3)$ i.e., $\overrightarrow a_1 = \overrightarrow i-3\overrightarrow k$
The second line passes through the point (3, -1, 1) i.e., $\overrightarrow a_2 = 3\overrightarrow i-\overrightarrow j+\overrightarrow k$
$\overrightarrow a_2-\overrightarrow a_1 = (3-1)\overrightarrow i+(-1-0)\overrightarrow j+(1+3)\overrightarrow k = 2\overrightarrow i-\overrightarrow j+4\overrightarrow k$
The distance between them $d = \large\frac{|\overrightarrow u \times (\overrightarrow a_2-\overrightarrow a_1)|}{|\overrightarrow u|}$
$|\overrightarrow u| = \sqrt{1+9+4} = \sqrt{14}$
$\overrightarrow u \times (\overrightarrow a_2-\overrightarrow a_1) = \begin{vmatrix} \overrightarrow i & \overrightarrow j & \overrightarrow k \\ -1 & 3 & 2 \\ 2 & -1 & 4 \end{vmatrix} = 14\overrightarrow i+8\overrightarrow j-5\overrightarrow k$
$|\overrightarrow u \times (\overrightarrow a_2 - \overrightarrow a_1) | =\sqrt{196+64+25} = \sqrt{\large\frac{285}{14}}$ units

edited Jun 24, 2013