Shortest distance between two lines

Two lines in 3D may be skew , neither parallel nor intersecting. The shortest distance between them is the length of the common perpendicular: the unique line segment that meets both lines at right angles.

The skew-line formula

Let line 1 be $\vec r = \vec a_1 + t \vec d_1$ and line 2 be $\vec r = \vec a_2 + s \vec d_2$ , with $\vec d_1 \not\parallel \vec d_2$ .

A vector perpendicular to both directions is $\vec d_1 \times \vec d_2$ . Its unit version is $\hat n = \dfrac{\vec d_1\times\vec d_2}{|\vec d_1\times\vec d_2|}$ .

The shortest distance is the absolute value of the projection of the displacement $\vec a_2 - \vec a_1$ onto $\hat n$ : $D = \left|\frac{(\vec a_2 - \vec a_1)\cdot (\vec d_1 \times \vec d_2)}{|\vec d_1 \times \vec d_2|}\right|.$

The numerator is the signed triple product; the absolute value gives the geometric distance.

Why does this work?

The common perpendicular has direction $\vec d_1 \times \vec d_2$ . To find how "far apart" the lines are along this perpendicular direction, project the line-to-line displacement $\vec a_2 - \vec a_1$ onto $\hat n$ . The result is the gap.

Parallel lines

If $\vec d_1 \parallel \vec d_2$ , the formula above breaks down (the cross product is zero). For two parallel lines, every perpendicular to one is perpendicular to the other; the distance is just the perpendicular distance from any point of line 2 to line 1: $D = \frac{|(\vec a_2 - \vec a_1)\times \vec d_1|}{|\vec d_1|}.$

Cartesian version

Lines $\dfrac{x - x_1}{a_1} = \dfrac{y - y_1}{b_1} = \dfrac{z - z_1}{c_1}$ and $\dfrac{x - x_2}{a_2} = \dfrac{y - y_2}{b_2} = \dfrac{z - z_2}{c_2}$ have shortest distance $D = \frac{\left|\det\begin{pmatrix} x_2 - x_1 & y_2 - y_1 & z_2 - z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2\end{pmatrix}\right|}{\sqrt{(b_1 c_2 - b_2 c_1)^2 + (c_1 a_2 - c_2 a_1)^2 + (a_1 b_2 - a_2 b_1)^2}}.$

(The numerator is the triple product written componentwise; the denominator is $|\vec d_1 \times \vec d_2|$ .)

Intersecting check

Distance $= 0 \iff$ the two lines intersect. So computing this number doubles as a test: zero means they meet, non-zero means they don't.

Worked examples

Example 1. Find the shortest distance between $\vec r = \hat i + 2\hat j + 3\hat k + t(\hat i + \hat j + \hat k)$ and $\vec r = 2\hat i + 3\hat j + 4\hat k + s(2\hat i - \hat j + \hat k)$ .

$\vec a_1 = (1, 2, 3)$ , $\vec d_1 = (1, 1, 1)$ . $\vec a_2 = (2, 3, 4)$ , $\vec d_2 = (2, -1, 1)$ .

$\vec d_1\times\vec d_2 = \begin{vmatrix}\hat i & \hat j & \hat k\\1 & 1 & 1\\2 & -1 & 1\end{vmatrix} = \hat i(1 + 1) - \hat j(1 - 2) + \hat k(-1 - 2) = 2\hat i + \hat j - 3\hat k$ . Magnitude $\sqrt{4 + 1 + 9} = \sqrt{14}$ .

$\vec a_2 - \vec a_1 = (1, 1, 1)$ . Dot with cross: $2 + 1 - 3 = 0$ .

Distance $= 0/\sqrt{14} = 0$ . Lines intersect.

Example 2. Find the shortest distance between lines $\dfrac{x - 1}{2} = \dfrac{y - 2}{3} = \dfrac{z - 3}{4}$ and $\dfrac{x - 2}{3} = \dfrac{y - 3}{4} = \dfrac{z - 5}{5}$ .

$\vec a_1 = (1, 2, 3)$ , $\vec d_1 = (2, 3, 4)$ . $\vec a_2 = (2, 3, 5)$ , $\vec d_2 = (3, 4, 5)$ . $\vec a_2 - \vec a_1 = (1, 1, 2)$ .

$\vec d_1\times\vec d_2 = \begin{vmatrix}\hat i & \hat j & \hat k\\2 & 3 & 4\\3 & 4 & 5\end{vmatrix} = \hat i(15 - 16) - \hat j(10 - 12) + \hat k(8 - 9) = -\hat i + 2\hat j - \hat k$ . Magnitude $\sqrt{1 + 4 + 1} = \sqrt 6$ .

$(\vec a_2 - \vec a_1)\cdot(\vec d_1\times\vec d_2) = (1)(-1) + (1)(2) + (2)(-1) = -1$ . Absolute value: $1$ .

Distance $= 1/\sqrt 6$ .

Example 3. Show that the lines $\dfrac{x + 1}{3} = \dfrac{y + 3}{5} = \dfrac{z + 5}{7}$ and $\dfrac{x - 2}{1} = \dfrac{y - 4}{3} = \dfrac{z - 6}{5}$ are coplanar.

$\vec a_2 - \vec a_1 = (3, 7, 11)$ . $\vec d_1 = (3, 5, 7)$ , $\vec d_2 = (1, 3, 5)$ .

Triple product: $\det \begin{pmatrix} 3 & 7 & 11 \\ 3 & 5 & 7 \\ 1 & 3 & 5\end{pmatrix} = 3(25 - 21) - 7(15 - 7) + 11(9 - 5) = 12 - 56 + 44 = 0$ .

Triple product zero, so coplanar.

Example 4. Find the distance between the two parallel lines $\vec r = \hat i + \hat j + t(2\hat i - \hat j + \hat k)$ and $\vec r = 2\hat i - \hat j + s(2\hat i - \hat j + \hat k)$ .

Parallel direction $\vec d = (2, -1, 1)$ , $|\vec d| = \sqrt 6$ . $\vec a_2 - \vec a_1 = (1, -2, 0)$ . $(\vec a_2 - \vec a_1)\times\vec d = \begin{vmatrix}\hat i & \hat j & \hat k\\1 & -2 & 0\\2 & -1 & 1\end{vmatrix} = \hat i(-2 - 0) - \hat j(1 - 0) + \hat k(-1 + 4) = -2\hat i - \hat j + 3\hat k$ . Magnitude $\sqrt{4 + 1 + 9} = \sqrt{14}$ .

Distance $= \sqrt{14}/\sqrt 6 = \sqrt{14/6} = \sqrt{7/3}$ .

Example 5. Find the shortest distance between $\vec r = (1, 2, 3) + t(1, 0, 0)$ and $\vec r = (0, 0, 0) + s(0, 0, 1)$ .

$\vec d_1 = (1, 0, 0)$ , $\vec d_2 = (0, 0, 1)$ . $\vec d_1\times\vec d_2 = (0, -1, 0)$ , magnitude $1$ . $\vec a_2 - \vec a_1 = (-1, -2, -3)$ . Dot with cross: $(-1)(0) + (-2)(-1) + (-3)(0) = 2$ . Distance $= 2$ . (Note: this is the $y$ -coordinate difference, since the lines are the $x$ -axis-parallel line at $y = 2, z = 3$ and the $z$ -axis. Closest approach has $y = 2$ in line 1 and $y = 0$ in line 2, so the gap in $y$ alone is 2.)

Example 6. Determine whether the lines $\dfrac{x}{1} = \dfrac{y}{2} = \dfrac{z}{3}$ and $\dfrac{x - 2}{1} = \dfrac{y - 4}{2} = \dfrac{z - 6}{3}$ are coincident, parallel, or skew.

Same direction $(1, 2, 3)$ . Try the point $(2, 4, 6)$ on the second line: does it lie on the first? $\dfrac{2}{1} = \dfrac{4}{2} = \dfrac{6}{3} = 2$ . Yes , coincident.

Try it yourself

Shortest distance between $\vec r = (1, 0, 0) + t(0, 0, 1)$ and $\vec r = (0, 1, 0) + s(1, 0, 0)$ .
Shortest distance between $\dfrac{x - 1}{1} = \dfrac{y - 2}{2} = \dfrac{z - 3}{3}$ and $\dfrac{x - 2}{1} = \dfrac{y - 3}{2} = \dfrac{z - 4}{3}$ , note they may be parallel or coincident.
Find the shortest distance between $\vec r = (3, 5, 7) + t(1, -2, 1)$ and $\vec r = (-1, -1, -1) + s(7, -6, 1)$ .
Show that lines $\dfrac{x + 2}{1} = \dfrac{y - 1}{2} = \dfrac{z - 5}{3}$ and $\dfrac{x + 1}{2} = \dfrac{y}{4} = \dfrac{z + 1}{6}$ are coplanar.
Find the shortest distance between two parallel lines $\dfrac{x - 1}{2} = \dfrac{y + 1}{3} = \dfrac{z}{1}$ and $\dfrac{x - 4}{2} = \dfrac{y}{3} = \dfrac{z - 1}{1}$ .
Find the line that is the common perpendicular of $\vec r = (0, 0, 0) + t(1, 0, 0)$ and $\vec r = (0, 0, 1) + s(0, 1, 0)$ .
For what values of $\lambda$ are the lines $\dfrac{x - 1}{2} = \dfrac{y + 1}{3} = \dfrac{z - 1}{4}$ and $\dfrac{x - 3}{1} = \dfrac{y - \lambda}{2} = \dfrac{z}{1}$ coplanar?
Determine if $\vec r = (1, 2, 3) + t(2, 0, 1)$ and $\vec r = (2, 4, 6) + s(1, 1, 0)$ intersect.
Find the perpendicular distance from origin to $\dfrac{x - 1}{1} = \dfrac{y}{1} = \dfrac{z}{1}$ .
Lines $L_1: \vec r = t\vec d_1$ and $L_2: \vec r = \vec b + s\vec d_2$ both pass through origin (if $\vec b = \vec 0$ ) , find shortest distance for $\vec b = (1, 1, 1)$ , $\vec d_1 = (1, 0, 0)$ , $\vec d_2 = (0, 1, 0)$ .
Find shortest distance between $x$ -axis and $\vec r = (1, 1, 1) + t(0, 0, 1)$ .
The two diagonals of opposite faces of a unit cube , find the distance between them.
Find the foot of the common perpendicular from $(0, 0, 0)$ on the line through $(1, 1, 1)$ with direction $(1, 1, 1)$ .
Two skew edges of a regular tetrahedron with side $1$ , find the shortest distance between them.

Pitfalls and tricks

Test parallelism first before applying the skew formula , division by zero awaits.
Triple product zero $\Leftrightarrow$ coplanar lines (intersect or parallel).
Absolute value outside the entire expression , the unsigned distance is what you want.
For parallel lines use the cross-product formula $|(\vec a_2 - \vec a_1)\times\vec d|/|\vec d|$ .
Sketch when possible. Even a sketch of one line plus a perpendicular helps verify the geometry.