Exercise 11.2 · Q14: Shortest distance between lines through $\hat i + 2\hat j + 3\hat k$ and $4\hat i + 5\hat j + 6\hat k$

Basics you need

Solution

1. Read off the points and directions

   $$\begin{aligned} \vec a_1 &= \hat i + 2\hat j + 3\hat k, & \vec b_1 &= \hat i - 3\hat j + 2\hat k, \\ \vec a_2 &= 4\hat i + 5\hat j + 6\hat k, & \vec b_2 &= 2\hat i + 3\hat j + \hat k. \end{aligned}$$
   The vector multiplied by the parameter is the direction, the rest is a point. So $l_1$ passes through $S(1, 2, 3)$ and $l_2$ through $T(4, 5, 6)$⁠.
   Vector equation of a line, Section 11.3.1. Basics: a⃗ + λb⃗

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

   $$\begin{aligned} \vec a_2 - \vec a_1 &= (4 - 1)\hat i + (5 - 2)\hat j \\ &\quad + (6 - 3)\hat k \\ &= 3\hat i + 3\hat j + 3\hat k. \end{aligned}$$
   This is $\overrightarrow{ST}$⁠, from $S$ on $l_1$ to $T$ on $l_2$⁠: terminal minus initial.
   Vector joining two points (Section 10.5.2). Basics: points in space

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

   $$\begin{aligned} \vec b_1\times\vec b_2 &= \begin{vmatrix} \hat i & \hat j & \hat k \\ 1 & -3 & 2 \\ 2 & 3 & 1 \end{vmatrix} \\ &= \big((-3)(1) - (2)(3)\big)\hat i \\ &\quad - \big((1)(1) - (2)(2)\big)\hat j \\ &\quad + \big((1)(3) - (-3)(2)\big)\hat k \\ &= -9\hat i + 3\hat j + 9\hat k. \end{aligned}$$
   The $\hat j$ term is $-(1 - 4) = +3$⁠: the minus sign of the middle term and the negative bracket cancel. Check: $(-9)(1) + 3(-3) + 9(2) = 0$ and $(-9)(2) + 3(3) + 9(1) = 0$⁠, so it is perpendicular to both lines; 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

4. Its length

   $$\begin{aligned} |\vec b_1\times\vec b_2| &= \sqrt{81 + 9 + 81} \\ &= \sqrt{171} = 3\sqrt{19}. \end{aligned}$$
   $171 = 9 \times 19$ and $\sqrt9 = 3$⁠; 19 is prime.
   Length of a vector. Basics: Pythagoras in 3D

5. The dot product

   $$\begin{aligned} &(\vec b_1\times\vec b_2)\cdot(\vec a_2 - \vec a_1) \\ &= (-9)(3) + (3)(3) + (9)(3) \\ &= -27 + 9 + 27 = 9. \end{aligned}$$
   Every component of $\vec a_2 - \vec a_1$ is 3, so the dot product is $3(-9 + 3 + 9) = 3 \times 3 = 9$⁠.
   $\vec a\cdot\vec b = a_1b_1 + a_2b_2 + a_3b_3$⁠. Basics: the dot product

6. Put it into the formula

   $$\begin{aligned} d &= \left|\frac{(\vec b_1\times\vec b_2)\cdot(\vec a_2 - \vec a_1)}{|\vec b_1\times\vec b_2|}\right| \\ &= \left|\frac{9}{3\sqrt{19}}\right| = \frac{3}{\sqrt{19}} = \frac{3\sqrt{19}}{19}. \end{aligned}$$
   This is the length of the projection of $\overrightarrow{ST}$ on the unit vector along the common perpendicular. Cancel 3; the rationalised form is $\frac{3\sqrt{19}}{19}$⁠, and the answer key keeps $\frac{3}{\sqrt{19}}$⁠. 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

7. Check: find the closest points

   Let $P = \vec a_1 + \lambda\vec b_1$ on $l_1$ and $Q = \vec a_2 + \mu\vec b_2$ on $l_2$⁠. Then

   $$\begin{aligned} \overrightarrow{PQ} &= (3 + 2\mu - \lambda)\hat i + (3 + 3\mu + 3\lambda)\hat j \\ &\quad + (3 + \mu - 2\lambda)\hat k. \end{aligned}$$

   $\overrightarrow{PQ}\cdot\vec b_1 = 0$ and $\overrightarrow{PQ}\cdot\vec b_2 = 0$ give

   $$\begin{aligned} -14\lambda - 5\mu &= 0, \\ 18 + 5\lambda + 14\mu &= 0, \end{aligned}$$

   so $\lambda = \frac{10}{19}$ and $\mu = -\frac{28}{19}$⁠. Then $P\left(\frac{29}{19}, \frac{8}{19}, \frac{77}{19}\right)$⁠, $Q\left(\frac{20}{19}, \frac{11}{19}, \frac{86}{19}\right)$ and

   $$\begin{aligned} \overrightarrow{PQ} &= \frac{1}{19}(-9\hat i + 3\hat j + 9\hat k), \\ PQ &= \frac{\sqrt{171}}{19} = \frac{3\sqrt{19}}{19} = \frac{3}{\sqrt{19}}. \end{aligned}$$
   The first condition gives $\lambda = -\frac{5\mu}{14}$⁠; putting it into the second gives $18 + \frac{171\mu}{14} = 0$⁠, so $\mu = -\frac{252}{171} = -\frac{28}{19}$⁠. $\overrightarrow{PQ}$ is $\frac{1}{19}(\vec b_1\times\vec b_2)$⁠, along the common perpendicular, and its length agrees with the formula: $\frac{3\sqrt{19}}{19} = \frac{3}{\sqrt{19}}$⁠. 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

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