The shortest distance between two lines

What you will be able to do

• find the shortest distance between two skew lines in vector and in Cartesian form;
• find the distance between two parallel lines;
• find the direction of the common perpendicular.

The closest two cables come

Take two skew cables. Pick a point on each and measure how far apart they are. Different pairs of points give different distances. Is there a smallest distance? Yes, and it is the length of the segment that is perpendicular to both cables. This segment is the common perpendicular, and the shortest distance is its length.

The formula for skew lines

The direction of the common perpendicular must be at right angles to $\vec{b}_1$ and to $\vec{b}_2$. The cross product $\vec{n} = \vec{b}_1 \times \vec{b}_2$ has exactly this property. The vector $\vec{a}_2 - \vec{a}_1$ joins a point of line 1 to a point of line 2. The shortest distance is the length of the projection of $\vec{a}_2 - \vec{a}_1$ along $\vec{n}$:

The numerator is the triple product from the last lesson. So $d = 0$ exactly when the lines are coplanar and not parallel, which means that the lines meet.

In Cartesian form, with the lines of the last lesson, the numerator is the determinant with rows $(x_2 - x_1,\ y_2 - y_1,\ z_2 - z_1)$, $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$. The denominator is the length of the cross product $\vec{b}_1 \times \vec{b}_2$. Its components are $P = b_1c_2 - b_2c_1$, $Q = c_1a_2 - c_2a_1$ and $R = a_1b_2 - a_2b_1$, so the length is $\sqrt{P^2 + Q^2 + R^2}$.

A-Level and IB books reach the same number in another way. They write a general vector $\overrightarrow{PQ}$ between a point $P$ of one line and a point $Q$ of the other, and make it perpendicular to both directions. The two methods give the same distance.

The formula for parallel lines

If the lines are parallel, then $\vec{b}_1 \times \vec{b}_2 = \vec{0}$ and the formula above fails. Now both lines have the direction $\vec{b}$, and the distance between them is the distance of a point of one line from the other line:

A worked example

Find the shortest distance between $\vec{r} = (\hat{i} - \hat{j} + 2\hat{k}) + \lambda(\hat{i} + 2\hat{j} + 2\hat{k})$ and $\vec{r} = (-\hat{i} + 3\hat{j} + 2\hat{k}) + \mu(2\hat{i} + \hat{j} - 2\hat{k})$.

1. The directions and the points. $\vec{b}_1 = (1, 2, 2)$, $\vec{b}_2 = (2, 1, -2)$. Then $\vec{a}_2 - \vec{a}_1 = (-2, 4, 0)$.
2. The cross product. $\vec{b}_1 \times \vec{b}_2 = (2 \times (-2) - 2 \times 1,\ 2 \times 2 - 1 \times (-2),\ 1 \times 1 - 2 \times 2) = (-6, 6, -3)$.
3. Its length. $\sqrt{36 + 36 + 9} = 9$.
4. The dot product. $(-2, 4, 0) \cdot (-6, 6, -3) = 12 + 24 + 0 = 36$.
5. The distance. $d = \dfrac{36}{9} = 4$ units.

A parallel pair. The lines $\dfrac{x - 1}{2} = \dfrac{y - 2}{3} = \dfrac{z - 3}{6}$ and $\dfrac{x - 6}{2} = \dfrac{y + 1}{3} = \dfrac{z - 11}{6}$ have $\vec{b} = (2, 3, 6)$, $\lvert \vec{b} \rvert = 7$ and $\vec{a}_2 - \vec{a}_1 = (5, -3, 8)$. The cross product $(5, -3, 8) \times (2, 3, 6)$ is $(-42, -14, 21)$, whose length is $49$. So $d = \dfrac{49}{7} = 7$ units.

Practise with new numbers

Try it in Python

The program finds the shortest distance between two lines. It uses the formula for skew lines, and it switches to the formula for parallel lines when the cross product is zero.

Common mistakes

• Forgetting the length of $\vec{n}$. The triple product alone is a volume, not a distance.
• Using the skew formula for parallel lines. The denominator becomes $0$. Use the parallel formula.
• Dropping the modulus. A distance is never negative.
• Mixing up the two cross products. In the skew formula the cross product is $\vec{b}_1 \times \vec{b}_2$. In the parallel formula it is $(\vec{a}_2 - \vec{a}_1) \times \vec{b}$.

Remember these

• Skew lines: $d = \dfrac{\lvert (\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) \rvert}{\lvert \vec{b}_1 \times \vec{b}_2 \rvert}$.
• Parallel lines: $d = \dfrac{\lvert (\vec{a}_2 - \vec{a}_1) \times \vec{b} \rvert}{\lvert \vec{b} \rvert}$.
• The direction of the common perpendicular is $\vec{b}_1 \times \vec{b}_2$.
• $d = 0$ for lines that are not parallel means that they meet.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Find the shortest distance between the $x$-axis and the line through $(2, 0, 5)$ parallel to the $y$-axis.

2. Find the distance between the $x$-axis and the line through $(0, 3, 4)$ parallel to the $x$-axis.

3. Find the shortest distance between the line through the origin with direction $(1, 1, 0)$ and the line through $(2, 0, 1)$ with direction $(0, 1, 1)$.

4. Why does the cross product $\vec{b}_1 \times \vec{b}_2$ give the direction of the common perpendicular?

5. Two lines that are not parallel have shortest distance $0$. What can you say about them?

References

• OpenStax, Calculus Volume 3, Chapter 2, section on equations of lines and planes in space (distance between skew lines).
• NCERT, Mathematics Part II for Class XII, Chapter 11, section on the shortest distance between two lines.