Chapter 11 · Three Dimensional Geometry

Exercise 11.2 · Q15: Shortest distance between two lines with parameters $t$ and $s$

Exercise 11.2Shortest distance between skew lines · Section 11.5.1

The question

Find the shortest distance between the lines whose vector equations are

$\vec r = (1 - t)\hat i + (t - 2)\hat j + (3 - 2t)\hat k$ and

$\vec r = (s + 1)\hat i + (2s - 1)\hat j - (2s + 1)\hat k$

Exam answer

Separating the terms with the parameters:

\begin{aligned} \vec r &= (\hat i - 2\hat j + 3\hat k) \\ &\quad + t(-\hat i + \hat j - 2\hat k) \end{aligned}

\begin{aligned} \vec r &= (\hat i - \hat j - \hat k) \\ &\quad + s(\hat i + 2\hat j - 2\hat k) \end{aligned}

Comparing with $\vec r = \vec a_1 + t\vec b_1$ and $\vec r = \vec a_2 + s\vec b_2$ ⁠:

$\vec a_1 = \hat i - 2\hat j + 3\hat k$ ⁠, $\vec b_1 = -\hat i + \hat j - 2\hat k$

$\vec a_2 = \hat i - \hat j - \hat k$ ⁠, $\vec b_2 = \hat i + 2\hat j - 2\hat k$

\vec a_2 - \vec a_1 = \hat j - 4\hat k

\begin{aligned} \vec b_1\times\vec b_2 &= \begin{vmatrix} \hat i & \hat j & \hat k \\ -1 & 1 & -2 \\ 1 & 2 & -2 \end{vmatrix} \\ &= (-2 + 4)\hat i - (2 + 2)\hat j \\ &\quad + (-2 - 1)\hat k \\ &= 2\hat i - 4\hat j - 3\hat k \end{aligned}

|\vec b_1\times\vec b_2| = \sqrt{4 + 16 + 9} = \sqrt{29}

\begin{aligned} &(\vec b_1\times\vec b_2)\cdot(\vec a_2 - \vec a_1) \\ &= 2(0) + (-4)(1) + (-3)(-4) \\ &= 0 - 4 + 12 = 8 \end{aligned}

The shortest distance is

\begin{aligned} d &= \left|\frac{(\vec b_1\times\vec b_2)\cdot(\vec a_2 - \vec a_1)}{|\vec b_1\times\vec b_2|}\right| \\ &= \left|\frac{8}{\sqrt{29}}\right| \end{aligned}

$\therefore$ The shortest distance is $\dfrac{8}{\sqrt{29}}$ units.

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

These equations are vector equations of lines written component by component. Collecting the terms without the parameter and the terms with it puts each line in the form $\vec r = \vec a + t\vec b$ ⁠: a point plus a multiple of a direction. Then the skew-lines formula applies as usual.

Given, to find, and the plan

Given: $l_1$ ⁠: $\vec r = (1 - t)\hat i + (t - 2)\hat j + (3 - 2t)\hat k$ ⁠; $l_2$ ⁠: $\vec r = (s + 1)\hat i + (2s - 1)\hat j - (2s + 1)\hat k$ ⁠.
To find: The shortest distance $d$ between $l_1$ and $l_2$ ⁠.
Plan: (1) Rewrite each line as point + parameter $\times$ direction. (2) Find $\vec a_2 - \vec a_1$ ⁠. (3) Find $\vec b_1\times\vec b_2$ ⁠. (4) Its length. (5) The dot product. (6) The formula. (7) Check by finding the closest points.

Basics you need

1Vector form of a line

$\vec r = \vec a + \lambda\vec b$ ⁠: the vector multiplied by the parameter is the direction; the rest is a point (Section 11.3.1). The parameter may be called $t$ ⁠, $s$ ⁠, $\lambda$ or $\mu$ ⁠. Basics: go to A, then walk λ copies of b

2Shortest distance between skew lines

For $\vec r = \vec a_1 + \lambda\vec b_1$ and $\vec r = \vec a_2 + \mu\vec b_2$ (Section 11.5.1):

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

Lesson: project onto the common perpendicular

3The cross product

$\vec b_1\times\vec b_2$ is perpendicular to both $\vec b_1$ and $\vec b_2$ (Section 10.6.3). Basics: the cross product

4Projection

The projection of $\vec v$ on the direction of $\vec n$ has length $\frac{|\vec v\cdot\vec n|}{|\vec n|}$ (Section 10.6.2). Basics: the shadow on the gap direction

Solution

Separate the point and the direction
$\begin{aligned} \vec r &= (1 - t)\hat i + (t - 2)\hat j + (3 - 2t)\hat k \\ &= (\hat i - 2\hat j + 3\hat k) + t(-\hat i + \hat j - 2\hat k), \\ \vec r &= (s + 1)\hat i + (2s - 1)\hat j - (2s + 1)\hat k \\ &= (\hat i - \hat j - \hat k) + s(\hat i + 2\hat j - 2\hat k). \end{aligned}$
So $\vec a_1 = \hat i - 2\hat j + 3\hat k$ ⁠, $\vec b_1 = -\hat i + \hat j - 2\hat k$ ⁠, $\vec a_2 = \hat i - \hat j - \hat k$ ⁠, $\vec b_2 = \hat i + 2\hat j - 2\hat k$ ⁠.

Split each bracket into its constant part and its parameter part: $1 - t$ gives $1$ and $-t$ ⁠, so $\hat i$ and $-t\hat i$ ⁠. Watch the last term of the second line: $-(2s + 1)\hat k = -\hat k - 2s\hat k$ ⁠, so the constant part is $-\hat k$ and the direction has $-2\hat k$ ⁠. Putting $t = 0$ and $s = 0$ gives the points $S(1, -2, 3)$ and $T(1, -1, -1)$ ⁠.

Vector equation of a line, Section 11.3.1. Basics: a⃗ + λb⃗
The joining vector $\vec a_2 - \vec a_1$
$\begin{aligned} \vec a_2 - \vec a_1 &= (1 - 1)\hat i + (-1 + 2)\hat j \\ &\quad + (-1 - 3)\hat k \\ &= 0\hat i + \hat j - 4\hat k. \end{aligned}$
This is $\overrightarrow{ST}$ ⁠. Its $\hat i$ component is 0, which must be kept in mind for the dot product.

Vector joining two points (Section 10.5.2). Basics: points in space
The cross product $\vec b_1\times\vec b_2$
$\begin{aligned} \vec b_1\times\vec b_2 &= \begin{vmatrix} \hat i & \hat j & \hat k \\ -1 & 1 & -2 \\ 1 & 2 & -2 \end{vmatrix} \\ &= \big((1)(-2) - (-2)(2)\big)\hat i \\ &\quad - \big((-1)(-2) - (-2)(1)\big)\hat j \\ &\quad + \big((-1)(2) - (1)(1)\big)\hat k \\ &= 2\hat i - 4\hat j - 3\hat k. \end{aligned}$
Each bracket has two signed products, so go slowly: $-2 + 4 = 2$ ⁠; $2 + 2 = 4$ ⁠, with the middle minus sign giving $-4$ ⁠; $-2 - 1 = -3$ ⁠. Check: $2(-1) + (-4)(1) + (-3)(-2) = 0$ and $2(1) + (-4)(2) + (-3)(-2) = 0$ ⁠. It is not $\vec 0$ ⁠, so the lines are not parallel.

Determinant form of the cross product (Section 10.6.3). Basics: a⃗ × b⃗ stands up from both
Its length
$\begin{aligned} |\vec b_1\times\vec b_2| &= \sqrt{2^2 + (-4)^2 + (-3)^2} \\ &= \sqrt{4 + 16 + 9} = \sqrt{29}. \end{aligned}$
29 is prime, so $\sqrt{29}$ does not simplify.

Length of a vector. Basics: Pythagoras in 3D
The dot product
$\begin{aligned} &(\vec b_1\times\vec b_2)\cdot(\vec a_2 - \vec a_1) \\ &= (2)(0) + (-4)(1) + (-3)(-4) \\ &= 0 - 4 + 12 = 8. \end{aligned}$
Multiply matching components of $2\hat i - 4\hat j - 3\hat k$ and $0\hat i + \hat j - 4\hat k$ and add; $(-3)(-4) = +12$ ⁠.

$\vec a\cdot\vec b = a_1b_1 + a_2b_2 + a_3b_3$ ⁠. Basics: the dot product
Put it into the formula
$\begin{aligned} d &= \left|\frac{(\vec b_1\times\vec b_2)\cdot(\vec a_2 - \vec a_1)}{|\vec b_1\times\vec b_2|}\right| \\ &= \left|\frac{8}{\sqrt{29}}\right| = \frac{8}{\sqrt{29}}. \end{aligned}$
This is the length of the projection of $\overrightarrow{ST}$ on the unit vector along the common perpendicular. Rationalised, it is $\frac{8\sqrt{29}}{29}$ ⁠; the answer key keeps $\frac{8}{\sqrt{29}}$ ⁠. The final formula for $d$ on page 387 of the book (set in bold) prints $(\vec a_2 \times \vec a_1)$ where $(\vec a_2 - \vec a_1)$ is meant; the derivation just before it uses the difference.

Section 11.5.1, shortest distance between skew lines. Lesson: the shortest distance
Check: find the closest points

Let $P = \vec a_1 + t\vec b_1$ on $l_1$ and $Q = \vec a_2 + s\vec b_2$ on $l_2$ ⁠. Then
$\begin{aligned} \overrightarrow{PQ} &= (s + t)\hat i + (1 + 2s - t)\hat j \\ &\quad + (-4 - 2s + 2t)\hat k. \end{aligned}$
$\overrightarrow{PQ}\cdot\vec b_1 = 0$ and $\overrightarrow{PQ}\cdot\vec b_2 = 0$ give
$\begin{aligned} 9 + 5s - 6t &= 0, \\ 10 + 9s - 5t &= 0, \end{aligned}$
so $t = \frac{31}{29}$ and $s = -\frac{15}{29}$ ⁠. Then $P\left(-\frac{2}{29}, -\frac{27}{29}, \frac{25}{29}\right)$ ⁠, $Q\left(\frac{14}{29}, -\frac{59}{29}, \frac{1}{29}\right)$ and
$\begin{aligned} \overrightarrow{PQ} &= \frac{8}{29}(2\hat i - 4\hat j - 3\hat k), \\ PQ &= \frac{8\sqrt{29}}{29} = \frac{8}{\sqrt{29}}. \end{aligned}$

Oblique projection, not to scale (lengths and right angles are not shown true). $l_1$ passes through $S(1, -2, 3)$ ( $t = 0$ ⁠) and $l_2$ through $T(1, -1, -1)$ ( $s = 0$ ⁠). The common perpendicular $PQ$ has length $\frac{8}{\sqrt{29}}$ ⁠; $ST$ (dashed) is $\sqrt{17}$ ⁠.

Solving the two linear equations: $6t - 5s = 9$ and $5t - 9s = 10$ give $29t = 31$ (multiply the first by 9, the second by 5, and subtract). $\overrightarrow{PQ}$ is a multiple of $\vec b_1\times\vec b_2$ ⁠, as the formula assumes, and its length is the same $d$ ⁠. Since $d \ne 0$ and the lines are not parallel, they are skew.

The line of shortest distance is perpendicular to both lines (Section 11.5). Lesson: skew lines

Result

The shortest distance between the lines is $\frac{8}{\sqrt{29}}$ units $\left(= \frac{8\sqrt{29}}{29}\right)$ ⁠. The lines are skew.

(Textbook answer: $\frac{8}{\sqrt{29}}$ ⁠.)

Check it with numbers

$\sqrt{29} \approx 5.385165$ ⁠, so $d \approx 1.485563$ ⁠. The closest points are $P \approx$ $(-0.068966,\ -0.931034,\ 0.862069)$ at $t \approx 1.068966$ ⁠, and $Q \approx$ $(0.482759,\ -2.034483,\ 0.034483)$ at $s \approx -0.517241$ ⁠; the distance between them is $1.485563$ again. For comparison, $ST = \sqrt{0 + 1 + 16} = \sqrt{17} \approx 4.123106$ ⁠.

Common mistakes

Mishandling $-(2s + 1)\hat k$ : it is $-\hat k - 2s\hat k$ ⁠, so $\vec a_2$ has $-\hat k$ and $\vec b_2$ has $-2\hat k$ ⁠. Taking $+\hat k$ or $+2\hat k$ changes the answer.
Reading $\vec b_1$ as $\hat i + \hat j + 2\hat k$ : the parameter terms are $-t$ ⁠, $+t$ and $-2t$ ⁠, so $\vec b_1 = -\hat i + \hat j - 2\hat k$ ⁠.
Using $\vec a_2\times\vec a_1$ from the misprinted final formula for $d$ on page 387 (set in bold) instead of $\vec a_2 - \vec a_1$ ⁠.

What to take away

A vector equation written component by component can always be split into point + parameter $\times$ direction.
Then $d = \frac{|(\vec b_1\times\vec b_2)\cdot(\vec a_2 - \vec a_1)|}{|\vec b_1\times\vec b_2|}$ ⁠, whatever the parameters are called.
Putting the parameter equal to 0 gives the point $\vec a$ ⁠; the coefficients of the parameter give $\vec b$ ⁠.

Every fact used, and where it comes from

Fact	Used in	Source
Vector equation $\vec r = \vec a + \lambda\vec b$ ⁠: point and direction	Step 1	Section 11.3.1 (Basics)
Vector joining two points	Step 2	Section 10.5.2
$\vec b_1\times\vec b_2$ as a determinant; it is perpendicular to both vectors	Step 3	Section 10.6.3 (Basics)
$d = \left\|\frac{(\vec b_1\times\vec b_2)\cdot(\vec a_2 - \vec a_1)}{\|\vec b_1\times\vec b_2\|}\right\|$ ⁠; the final formula for $d$ on p. 387 (set in bold) misprints $\vec a_2 - \vec a_1$ as $\vec a_2\times\vec a_1$	Steps 4 to 6	Section 11.5.1 (Lesson)
Dot product in components	Steps 3, 5, 7	Section 10.6.1 (Basics)
Shortest segment is perpendicular to both skew lines	Step 7	Section 11.5 (Lesson)

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Exercise 11.2 · Q15: Shortest distance between two lines with parameters ttt and sss

Basics you need

Solution

Separate the point and the direction

The joining vector a⃗2−a⃗1\vec a_2 - \vec a_1a2​−a1​

The cross product b⃗1×b⃗2\vec b_1\times\vec b_2b1​×b2​

Its length

The dot product

Put it into the formula

Check: find the closest points

Every fact used, and where it comes from

Exercise 11.2 · Q15: Shortest distance between two lines with parameters $t$ and $s$

The joining vector $\vec a_2 - \vec a_1$

The cross product $\vec b_1\times\vec b_2$