Why this mat­ters

In a flat plane, two lines either meet or run par­al­lel. Space gives you a third option: lines that never meet and are not par­al­lel either. We call these skew lines. Pic­ture a fly­over cross­ing a road below it. The short­est gap between two such lines runs along the one direc­tion per­pen­dic­u­lar to both. Here we find that dis­tance for skew lines and for par­al­lel lines, and then wrap up the chap­ter with mixed prac­tice.

Short­est dis­tance between skew lines

Take the lines r⃗=a⃗1+λb⃗1\vec r = \vec a_1 + \lambda\vec b_1 and r⃗=a⃗2+μb⃗2\vec r = \vec a_2 + \mu\vec b_2. The short­est dis­tance is

d=∣(b⃗1×b⃗2)⋅(a⃗2−a⃗1)∣b⃗1×b⃗2∣∣.\displaystyle d = \left\lvert\frac{(\vec b_1 \times \vec b_2) \cdot (\vec a_2 - \vec a_1)}{\lvert\vec b_1 \times \vec b_2\rvert}\right\rvert.

Where does this come from? It is just the pro­jec­tion of a⃗2−a⃗1\vec a_2 - \vec a_1 on the com­mon per­pen­dic­u­lar b⃗1×b⃗2\vec b_1 \times \vec b_2. And if you get d=0d = 0, the lines actu­ally inter­sect.

Exam­ple 1. r⃗=i^+j^+λ(2i^−j^+k^)\vec r = \hat i + \hat j + \lambda(2\hat i - \hat j + \hat k) and r⃗=2i^+j^−k^+μ(3i^−5j^+2k^)\vec r = 2\hat i + \hat j - \hat k + \mu(3\hat i - 5\hat j + 2\hat k). First, b⃗1×b⃗2=(−2+5)i^−(4−3)j^+(−10+3)k^=3i^−j^−7k^\vec b_1 \times \vec b_2 = (-2 + 5)\hat i - (4 - 3)\hat j + (-10 + 3)\hat k = 3\hat i - \hat j - 7\hat k, with length 59\sqrt{59}. Next, a⃗2−a⃗1=i^−k^\vec a_2 - \vec a_1 = \hat i - \hat k, and the dot prod­uct is 3+7=103 + 7 = 10. So d=1059\displaystyle d = \tfrac{10}{\sqrt{59}}.

Schematic of the skew lines of Example 1 lying in two parallel planes, neither parallel nor meeting, joined by their common perpendicular, a red segment of length 10/root 59.
The short­est dis­tance between skew lines runs along their com­mon per­pen­dic­u­lar.

Exam­ple 2 (Carte­sian form). Con­sider x−12=y−23=z−34\displaystyle \tfrac{x - 1}{2} = \tfrac{y - 2}{3} = \tfrac{z - 3}{4} and x−23=y−44=z−55\displaystyle \tfrac{x - 2}{3} = \tfrac{y - 4}{4} = \tfrac{z - 5}{5}. Read the points and direc­tions off the equa­tions. Then b⃗1×b⃗2=(15−16)i^−(10−12)j^+(8−9)k^=−i^+2j^−k^\vec b_1 \times \vec b_2 = (15 - 16)\hat i - (10 - 12)\hat j + (8 - 9)\hat k = -\hat i + 2\hat j - \hat k; a⃗2−a⃗1=i^+2j^+2k^\vec a_2 - \vec a_1 = \hat i + 2\hat j + 2\hat k; the dot prod­uct is −1+4−2=1-1 + 4 - 2 = 1; and so d=16\displaystyle d = \tfrac{1}{\sqrt{6}}.

Par­al­lel lines

When the lines share a direc­tion, their cross prod­uct is zero and the for­mula above breaks down. For r⃗=a⃗1+λb⃗\vec r = \vec a_1 + \lambda\vec b and r⃗=a⃗2+μb⃗\vec r = \vec a_2 + \mu\vec b use this instead:

d=∣b⃗×(a⃗2−a⃗1)∣∣b⃗∣.\displaystyle d = \frac{\lvert\vec b \times (\vec a_2 - \vec a_1)\rvert}{\lvert\vec b\rvert}.

Exam­ple 3. r⃗=i^+2j^+λ(2i^+j^+2k^)\vec r = \hat i + 2\hat j + \lambda(2\hat i + \hat j + 2\hat k) and r⃗=3i^−j^+k^+μ(2i^+j^+2k^)\vec r = 3\hat i - \hat j + \hat k + \mu(2\hat i + \hat j + 2\hat k). Here a⃗2−a⃗1=2i^−3j^+k^\vec a_2 - \vec a_1 = 2\hat i - 3\hat j + \hat k; b⃗×(a⃗2−a⃗1)=(1+6)i^−(2−4)j^+(−6−2)k^=7i^+2j^−8k^\vec b \times (\vec a_2 - \vec a_1) = (1 + 6)\hat i - (2 - 4)\hat j + (-6 - 2)\hat k = 7\hat i + 2\hat j - 8\hat k, which has length 117\sqrt{117}. So d=1173=13\displaystyle d = \tfrac{\sqrt{117}}{3} = \sqrt{13}.

Two parallel lines, both with direction 2i + j + 2k as in Example 3, with a red segment perpendicular to both marking the distance root 13 between them.
Par­al­lel lines stay the same dis­tance apart: here root 13.

Mixed prac­tice for the chap­ter

  1. Find the direc­tion cosines of a line per­pen­dic­u­lar to the lines with direc­tion ratios 1,2,31, 2, 3 and −2,1,4-2, 1, 4.
  2. Find the equa­tion of the line through (1,2,−4)(1, 2, -4) per­pen­dic­u­lar to both x−83=y+19−16=z−107\displaystyle \tfrac{x - 8}{3} = \tfrac{y + 19}{-16} = \tfrac{z - 10}{7} and x−153=y−298=z−5−5\displaystyle \tfrac{x - 15}{3} = \tfrac{y - 29}{8} = \tfrac{z - 5}{-5}.
  3. Find the angle between the lines x2=y2=z1\displaystyle \tfrac{x}{2} = \tfrac{y}{2} = \tfrac{z}{1} and x−54=y−21=z−38\displaystyle \tfrac{x - 5}{4} = \tfrac{y - 2}{1} = \tfrac{z - 3}{8}.
  4. Find the short­est dis­tance between r⃗=i^+2j^+3k^+λ(i^−3j^+2k^)\vec r = \hat i + 2\hat j + 3\hat k + \lambda(\hat i - 3\hat j + 2\hat k) and r⃗=4i^+5j^+6k^+μ(2i^+3j^+k^)\vec r = 4\hat i + 5\hat j + 6\hat k + \mu(2\hat i + 3\hat j + \hat k).
  5. Find the short­est dis­tance between x+17=y+1−6=z+11\displaystyle \tfrac{x + 1}{7} = \tfrac{y + 1}{-6} = \tfrac{z + 1}{1} and x−31=y−5−2=z−71\displaystyle \tfrac{x - 3}{1} = \tfrac{y - 5}{-2} = \tfrac{z - 7}{1}.
  6. Show that the lines x−12=y−23=z−34\displaystyle \tfrac{x - 1}{2} = \tfrac{y - 2}{3} = \tfrac{z - 3}{4} and x−45=y−12=z\displaystyle \tfrac{x - 4}{5} = \tfrac{y - 1}{2} = z inter­sect.
  7. Find the dis­tance between the par­al­lel lines r⃗=λ(i^+2j^+2k^)\vec r = \lambda(\hat i + 2\hat j + 2\hat k) and r⃗=3i^+μ(i^+2j^+2k^)\vec r = 3\hat i + \mu(\hat i + 2\hat j + 2\hat k).
  8. Find the foot of the per­pen­dic­u­lar from (1,6,3)(1, 6, 3) to the line x1=y−12=z−23\displaystyle \tfrac{x}{1} = \tfrac{y - 1}{2} = \tfrac{z - 2}{3} and also the length of that per­pen­dic­u­lar.

Answers

Show answers
  1. The cross prod­uct of (1,2,3)(1, 2, 3) and (−2,1,4)(-2, 1, 4) is (5,−10,5)(5, -10, 5), so the ratios are 1,−2,11, -2, 1 and the cosines are 16,−26,16\displaystyle \tfrac{1}{\sqrt{6}}, -\tfrac{2}{\sqrt{6}}, \tfrac{1}{\sqrt{6}}.
  2. The direc­tion is (3,−16,7)×(3,8,−5)=(24,36,72)(3, -16, 7) \times (3, 8, -5) = (24, 36, 72), giv­ing ratios 2,3,62, 3, 6 and the line x−12=y−23=z+46\displaystyle \tfrac{x - 1}{2} = \tfrac{y - 2}{3} = \tfrac{z + 4}{6}.
  3. cos⁡θ=8+2+83⋅9=23\displaystyle \cos\theta = \tfrac{8 + 2 + 8}{3 \cdot 9} = \tfrac{2}{3}.
  4. b⃗1×b⃗2=−9i^+3j^+9k^\vec b_1 \times \vec b_2 = -9\hat i + 3\hat j + 9\hat k, length 3193\sqrt{19}; a⃗2−a⃗1=3i^+3j^+3k^\vec a_2 - \vec a_1 = 3\hat i + 3\hat j + 3\hat k; the dot prod­uct is 99; so d=319\displaystyle d = \tfrac{3}{\sqrt{19}}.
  5. b⃗1×b⃗2=(−6+2)i^−(7−1)j^+(−14+6)k^=−4i^−6j^−8k^\vec b_1 \times \vec b_2 = (-6 + 2)\hat i - (7 - 1)\hat j + (-14 + 6)\hat k = -4\hat i - 6\hat j - 8\hat k, length 2292\sqrt{29}; a⃗2−a⃗1=4i^+6j^+8k^\vec a_2 - \vec a_1 = 4\hat i + 6\hat j + 8\hat k; d=116229=229\displaystyle d = \tfrac{116}{2\sqrt{29}} = 2\sqrt{29}.
  6. The short­est dis­tance comes out as 00, so they meet, at (−1,−1,−1)(-1, -1, -1).
  7. b⃗×3i^=(0,6,−6)\vec b \times 3\hat i = (0, 6, -6), length 626\sqrt{2}; d=623=22\displaystyle d = \tfrac{6\sqrt{2}}{3} = 2\sqrt{2}.
  8. Take a gen­eral point (λ,2λ+1,3λ+2)(\lambda, 2\lambda + 1, 3\lambda + 2) on the line. The per­pen­dic­u­lar con­di­tion gives λ=1\lambda = 1, so the foot is (1,3,5)(1, 3, 5) and the length is 13\sqrt{13}.