Chapter 11 · Three Dimensional Geometry

Example 10: Distance between two parallel lines along $2\hat i + 3\hat j + 6\hat k$

ExamplesDistance between parallel lines · Section 11.5.2

The question

Find the distance between the lines $l_1$ and $l_2$ given by

\begin{aligned} \vec r = {}&\hat i + 2\hat j - 4\hat k \\ &+ \lambda(2\hat i + 3\hat j + 6\hat k) \end{aligned}

and

\begin{aligned} \vec r = {}&3\hat i + 3\hat j - 5\hat k \\ &+ \mu(2\hat i + 3\hat j + 6\hat k) \end{aligned}

Exam answer

Both lines are parallel to $\vec b = 2\hat i + 3\hat j + 6\hat k$ ⁠, so they are parallel.

Here $\vec a_1 = \hat i + 2\hat j - 4\hat k$ and $\vec a_2 = 3\hat i + 3\hat j - 5\hat k$ ⁠.

\vec a_2 - \vec a_1 = 2\hat i + \hat j - \hat k

\begin{aligned} &\vec b\times(\vec a_2 - \vec a_1) \\ &= \begin{vmatrix} \hat i & \hat j & \hat k \\ 2 & 3 & 6 \\ 2 & 1 & -1 \end{vmatrix} \\ &= (-3 - 6)\hat i - (-2 - 12)\hat j \\ &\quad + (2 - 6)\hat k \\ &= -9\hat i + 14\hat j - 4\hat k \end{aligned}

\begin{aligned} &|\vec b\times(\vec a_2 - \vec a_1)| \\ &= \sqrt{81 + 196 + 16} = \sqrt{293} \end{aligned}

|\vec b| = \sqrt{4 + 9 + 36} = \sqrt{49} = 7

The distance between the parallel lines is

d = \frac{|\vec b\times(\vec a_2 - \vec a_1)|}{|\vec b|}

$\therefore\ d = \dfrac{\sqrt{293}}{7}$ units.

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

The one idea this problem is built on

Parallel lines are a constant distance apart. Take a point $S$ on $l_1$ and a point $T$ on $l_2$ ⁠: the parallelogram on $\vec b$ and $\overrightarrow{ST}$ has area $|\vec b\times\overrightarrow{ST}|$ ⁠, and also area = base $\times$ height = $|\vec b| \times d$ ⁠. So $d = \frac{|\vec b\times(\vec a_2 - \vec a_1)|}{|\vec b|}$ ⁠.

Given, to find, and the plan

Given: $l_1$ ⁠: $\vec r = \hat i + 2\hat j - 4\hat k$ $+\ \lambda(2\hat i + 3\hat j + 6\hat k)$ ⁠; $l_2$ ⁠: $\vec r = 3\hat i + 3\hat j - 5\hat k$ $+\ \mu(2\hat i + 3\hat j + 6\hat k)$ ⁠.
To find: The distance $d$ between $l_1$ and $l_2$ ⁠.
Plan: (1) Show the lines are parallel (the book's "Why?"). (2) Find $\vec a_2 - \vec a_1$ ⁠. (3) Find $\vec b\times(\vec a_2 - \vec a_1)$ ⁠. (4) Find the two lengths. (5) Divide. (6) Check with the foot of the perpendicular.

Basics you need

1Distance between parallel lines

For $\vec r = \vec a_1 + \lambda\vec b$ and $\vec r = \vec a_2 + \mu\vec b$ (Section 11.5.2):

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

Lesson: the shortest distance

2The cross product and area

$|\vec a\times\vec b| = |\vec a||\vec b|\sin\theta$ ⁠, the area of the parallelogram on $\vec a$ and $\vec b$ (Section 10.6.3). Basics: the cross product

3Parallel lines

Lines with the same (or proportional) direction vectors are parallel (Section 11.4). Basics: a line is a point and a direction

4Length of a vector

$|x\hat i + y\hat j + z\hat k| = \sqrt{x^2 + y^2 + z^2}$ ⁠. Basics: Pythagoras in 3D

Solution

Why the lines are parallel

Both equations have the same direction vector $\vec b = 2\hat i + 3\hat j + 6\hat k$ ⁠, so $l_1 \parallel l_2$ ⁠. We take
$\begin{aligned} \vec a_1 &= \hat i + 2\hat j - 4\hat k, \\ \vec a_2 &= 3\hat i + 3\hat j - 5\hat k. \end{aligned}$
This answers the book's "(Why?)": the vector multiplied by the parameter is the direction of each line, and here it is the same vector $2\hat i + 3\hat j + 6\hat k$ in both. The lines are also different lines: the point $T(3, 3, -5)$ of $l_2$ is not on $l_1$ ⁠, because $\overrightarrow{ST} = 2\hat i + \hat j - \hat k$ (from $S(1, 2, -4)$ on $l_1$ ⁠) is not a multiple of $\vec b$ ( $\frac22 \ne \frac13$ ⁠). So the distance is not 0.

Parallel lines have proportional direction ratios (Section 11.4). Basics: a⃗ + λb⃗
The joining vector
$\begin{aligned} \vec a_2 - \vec a_1 &= (3 - 1)\hat i + (3 - 2)\hat j \\ &\quad + (-5 + 4)\hat k \\ &= 2\hat i + \hat j - \hat k. \end{aligned}$
This is $\overrightarrow{ST}$ ⁠, from the point $S$ of $l_1$ to the point $T$ of $l_2$ ⁠. The last component is $-5 - (-4) = -1$ ⁠.

Vector joining two points (Section 10.5.2). Basics: points in space
The cross product $\vec b\times(\vec a_2 - \vec a_1)$
$\begin{aligned} \vec b\times(\vec a_2 - \vec a_1) &= \begin{vmatrix} \hat i & \hat j & \hat k \\ 2 & 3 & 6 \\ 2 & 1 & -1 \end{vmatrix} \\ &= \big((3)(-1) - (6)(1)\big)\hat i \\ &\quad - \big((2)(-1) - (6)(2)\big)\hat j \\ &\quad + \big((2)(1) - (3)(2)\big)\hat k \\ &= -9\hat i + 14\hat j - 4\hat k. \end{aligned}$
The top row is $\hat i, \hat j, \hat k$ ⁠, the second row $\vec b$ ⁠, the third $\vec a_2 - \vec a_1$ ⁠, in that order. The $\hat j$ term is $-(-2 - 12) = +14$ ⁠: the minus sign of the middle term and the negative bracket cancel.

Determinant form of the cross product (Section 10.6.3). Basics: the cross product
The two lengths
$\begin{aligned} &|\vec b\times(\vec a_2 - \vec a_1)| \\ &\quad = \sqrt{81 + 196 + 16} = \sqrt{293}, \\ &|\vec b| = \sqrt{4 + 9 + 36} = \sqrt{49} = 7. \end{aligned}$
$(-9)^2 = 81$ ⁠, $14^2 = 196$ ⁠, $(-4)^2 = 16$ ⁠, and $81 + 196 + 16 = 293$ ⁠. 293 is a prime number, so $\sqrt{293}$ does not simplify.

Length of a vector. Basics: Pythagoras in 3D
Divide
$d = \frac{|\vec b\times(\vec a_2 - \vec a_1)|}{|\vec b|} = \frac{\sqrt{293}}{7}.$
In the book's Fig 11.7, with $P$ the foot of the perpendicular from $T$ to $l_1$ and $\theta$ the angle between $\overrightarrow{ST}$ and $\vec b$ ⁠, the distance is $PT = ST\sin\theta$ ⁠. Since $|\vec b\times\overrightarrow{ST}| = |\vec b|\,ST\sin\theta = |\vec b|\,PT$ ⁠, dividing by $|\vec b|$ leaves $PT$ ⁠. In words: parallelogram area divided by base gives the height.

Section 11.5.2, distance between parallel lines. Lesson: Examples 9–10
Check with the foot of the perpendicular

The foot $P$ of the perpendicular from $T$ to $l_1$ is the point $S + \lambda\vec b$ with $\overrightarrow{PT}\cdot\vec b = 0$ ⁠:
$\lambda = \frac{\overrightarrow{ST}\cdot\vec b}{|\vec b|^2} = \frac{4 + 3 - 6}{49} = \frac{1}{49}.$
So $P = \left(\frac{51}{49}, \frac{101}{49}, -\frac{190}{49}\right)$ and
$\overrightarrow{PT} = \frac{1}{49}(96\hat i + 46\hat j - 55\hat k),$ $\begin{aligned} PT &= \frac{\sqrt{9216 + 2116 + 3025}}{49} \\ &= \frac{\sqrt{14357}}{49} \\ &= \frac{7\sqrt{293}}{49} = \frac{\sqrt{293}}{7}. \end{aligned}$

Oblique projection, not to scale (lengths and right angles are not shown true). The parallel lines $l_1$ and $l_2$ ⁠, the given points $S$ and $T$ ⁠, and the perpendicular $TP$ of length $\frac{\sqrt{293}}{7}$ ⁠. $P$ is very close to $S$ ⁠, because $\lambda = \frac{1}{49}$ there.

$\overrightarrow{PT}\cdot\vec b = \frac{1}{49}(192 + 138 - 330) = 0$ ⁠, so $TP$ really is perpendicular to the lines, and its length agrees with the formula ( $14357 = 49 \times 293$ ⁠). $P$ is so close to $S$ because $\overrightarrow{ST}$ is almost perpendicular to $\vec b$ ⁠: $\overrightarrow{ST}\cdot\vec b = 1$ is small compared with $|\overrightarrow{ST}|\,|\vec b| = 7\sqrt6$ ⁠.

The distance between parallel lines is the length of the perpendicular from a point of one to the other (Section 11.5). Basics: projection

Result

The lines are parallel, and the distance between them is $\frac{\sqrt{293}}{7}$ units.

(Textbook answer: $\frac{\sqrt{293}}{7}$ ⁠.)

Check it with numbers

$\sqrt{293} \approx 17.117243$ ⁠, so $d \approx 2.445320$ ⁠. The joining segment $ST = \sqrt{4 + 1 + 1} = \sqrt6 \approx 2.449490$ is only slightly longer than $d$ ⁠, as it should be, since $S$ is close to the foot $P$ ⁠.

Common mistakes

Using the skew-lines formula. For parallel lines $\vec b_1\times\vec b_2 = \vec 0$ ⁠, so that formula divides by 0; use $\frac{|\vec b\times(\vec a_2 - \vec a_1)|}{|\vec b|}$ ⁠.
Dividing by $|\vec b|^2 = 49$ instead of $|\vec b| = 7$ ⁠.
Sign slips in the determinant, especially the middle term: $-\big((2)(-1) - (6)(2)\big) = +14$ ⁠.
Taking $|\vec a_2 - \vec a_1| = \sqrt6$ as the distance. $ST$ is a slanting segment; the distance is measured perpendicular to the lines.

What to take away

Parallel lines: $d = \frac{|\vec b\times(\vec a_2 - \vec a_1)|}{|\vec b|}$ ⁠, area of the parallelogram divided by its base.
Same direction vector in both equations means parallel lines; check a point to be sure they are not the same line.
Compare with skew lines (Example 9), where the cross product of the two directions gives the common perpendicular.

Every fact used, and where it comes from

Fact	Used in	Source
Same direction vector means parallel lines	Step 1	Section 11.4 (Basics)
Vector joining two points	Step 2	Section 10.5.2
Determinant form of the cross product	Step 3	Section 10.6.3 (Basics)
$\|\vec b\times\overrightarrow{ST}\| = \|\vec b\|\,ST\sin\theta$ ⁠, $PT = ST\sin\theta$	Step 5	Section 11.5.2, equation (3) and Fig 11.7
$d = \left\|\frac{\vec b\times(\vec a_2 - \vec a_1)}{\|\vec b\|}\right\|$	Step 5	Section 11.5.2 (Lesson)
Distance between parallel lines = perpendicular from a point of one to the other	Step 6	Section 11.5 (Basics)

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Example 10: Distance between two parallel lines along 2i^+3j^+6k^2\hat i + 3\hat j + 6\hat k2i^+3j^​+6k^

Basics you need

Solution

Why the lines are parallel

The joining vector

The cross product b⃗×(a⃗2−a⃗1)\vec b\times(\vec a_2 - \vec a_1)b×(a2​−a1​)

The two lengths

Divide

Check with the foot of the perpendicular

Every fact used, and where it comes from

Example 10: Distance between two parallel lines along $2\hat i + 3\hat j + 6\hat k$

The cross product $\vec b\times(\vec a_2 - \vec a_1)$