3D Geometry: Shortest Distance Between Parallel and Skew Lines

This lesson works through three sure questions on lines in 3D geometry: forming the equation of a line, and finding the shortest distance between skew lines and between parallel lines.

Equation of a line through a point in a given direction

We want the line parallel to the vector $\vec b = 2\hat\imath - \hat\jmath + 3\hat k$ that passes through the point $(5, -2, 4)$.

The direction ratios are $a = 2,\ b = -1,\ c = 3$, and the position vector of the given point is

$\vec a = 5\hat\imath - 2\hat\jmath + 4\hat k.$

Vector form. The line is $\vec r = \vec a + \lambda\,\vec b$, that is

$\vec r = (5\hat\imath - 2\hat\jmath + 4\hat k) + \lambda\,(2\hat\imath - \hat\jmath + 3\hat k).$

Cartesian form. Using $\dfrac{x - x_1}{a} = \dfrac{y - y_1}{b} = \dfrac{z - z_1}{c}$,

$\frac{x - 5}{2} = \frac{y + 2}{-1} = \frac{z - 4}{3}.$

Shortest distance between two skew lines

The two lines are given as

$\vec r = (1 - t)\hat\imath + (t - 2)\hat\jmath + (3 - 2t)\hat k,$ $\vec r = (s + 1)\hat\imath + (2s - 1)\hat\jmath - (2s + 1)\hat k.$

Writing them in $\vec r = \vec a + (\text{parameter})\,\vec b$ form. Grouping the constant terms and the terms in $t$ and $s$:

$\vec r = (\hat\imath - 2\hat\jmath + 3\hat k) + t\,(-\hat\imath + \hat\jmath - 2\hat k),$ $\vec r = (\hat\imath - \hat\jmath - \hat k) + s\,(\hat\imath + 2\hat\jmath - 2\hat k).$

The direction vectors are not parallel, so the lines are skew. We read off

$\vec a_1 = \hat\imath - 2\hat\jmath + 3\hat k,\quad \vec b_1 = -\hat\imath + \hat\jmath - 2\hat k,$ $\vec a_2 = \hat\imath - \hat\jmath - \hat k,\quad \vec b_2 = \hat\imath + 2\hat\jmath - 2\hat k.$

The distance formula for skew lines is

$d = \frac{\left|(\vec b_1 \times \vec b_2)\cdot(\vec a_2 - \vec a_1)\right|}{\left|\vec b_1 \times \vec b_2\right|}.$

Cross product.

$\vec b_1 \times \vec b_2 = \begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ -1 & 1 & -2 \\ 1 & 2 & -2 \end{vmatrix} = 2\hat\imath - 4\hat\jmath - 3\hat k.$

So $\left|\vec b_1 \times \vec b_2\right| = \sqrt{2^2 + (-4)^2 + (-3)^2} = \sqrt{4 + 16 + 9} = \sqrt{29}.$

The connecting vector.

$\vec a_2 - \vec a_1 = (\hat\imath - \hat\jmath - \hat k) - (\hat\imath - 2\hat\jmath + 3\hat k) = 0\hat\imath + \hat\jmath - 4\hat k.$

Dot product.

$(\vec b_1 \times \vec b_2)\cdot(\vec a_2 - \vec a_1) = (2)(0) + (-4)(1) + (-3)(-4) = 0 - 4 + 12 = 8.$

Result. Since distance is always positive,

$d = \frac{8}{\sqrt{29}}\ \text{units}.$

Distance between two parallel lines

The lines are

$L_1:\ \vec r = (\hat\imath + 2\hat\jmath - 4\hat k) + \lambda\,(2\hat\imath + 3\hat\jmath + 6\hat k),$ $L_2:\ \vec r = (3\hat\imath + 3\hat\jmath - 5\hat k) + \mu\,(2\hat\imath + 3\hat\jmath + 6\hat k).$

Both have the same direction vector, so the lines are parallel. Here

$\vec a_1 = \hat\imath + 2\hat\jmath - 4\hat k,\quad \vec a_2 = 3\hat\imath + 3\hat\jmath - 5\hat k,\quad \vec b = 2\hat\imath + 3\hat\jmath + 6\hat k.$

The distance formula for parallel lines is

$d = \frac{\left|\vec b \times (\vec a_2 - \vec a_1)\right|}{\left|\vec b\right|}.$

The connecting vector.

$\vec a_2 - \vec a_1 = 2\hat\imath + \hat\jmath - \hat k.$

Cross product.

$\vec b \times (\vec a_2 - \vec a_1) = \begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ 2 & 3 & 6 \\ 2 & 1 & -1 \end{vmatrix} = -9\hat\imath + 14\hat\jmath - 4\hat k.$

Its magnitude is $\sqrt{(-9)^2 + 14^2 + (-4)^2} = \sqrt{81 + 196 + 16} = \sqrt{293}.$

Also $\left|\vec b\right| = \sqrt{2^2 + 3^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7.$

Result.

$d = \frac{\sqrt{293}}{7}\ \text{units}.$

Key takeaways

• A line through a point with position vector $\vec a$ and direction $\vec b$ is $\vec r = \vec a + \lambda\,\vec b$, with Cartesian form $\dfrac{x - x_1}{a} = \dfrac{y - y_1}{b} = \dfrac{z - z_1}{c}$.
• Two lines are parallel when their direction vectors are proportional; otherwise (and when they do not meet) they are skew.
• For skew lines, $d = \dfrac{\left|(\vec b_1 \times \vec b_2)\cdot(\vec a_2 - \vec a_1)\right|}{\left|\vec b_1 \times \vec b_2\right|}$; for parallel lines, $d = \dfrac{\left|\vec b \times (\vec a_2 - \vec a_1)\right|}{\left|\vec b\right|}$.