Exercise 11.2 · Q12: Shortest distance between lines through $\hat i + 2\hat j + \hat k$ and $2\hat i - \hat j - \hat k$

Basics you need

Solution

1. Read off the points and directions

   $$\begin{aligned} \vec a_1 &= \hat i + 2\hat j + \hat k, & \vec b_1 &= \hat i - \hat j + \hat k, \\ \vec a_2 &= 2\hat i - \hat j - \hat k, & \vec b_2 &= 2\hat i + \hat j + 2\hat k. \end{aligned}$$
   The vector multiplied by the parameter is the direction, the rest is a point. So $l_1$ passes through $S(1, 2, 1)$ and $l_2$ through $T(2, -1, -1)$⁠. The brackets around $\hat i + 2\hat j + \hat k$ in the first equation only group the point; they change nothing.
   Vector equation of a line, Section 11.3.1. Basics: a⃗ + λb⃗

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

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

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

   $$\begin{aligned} \vec b_1\times\vec b_2 &= \begin{vmatrix} \hat i & \hat j & \hat k \\ 1 & -1 & 1 \\ 2 & 1 & 2 \end{vmatrix} \\ &= \big((-1)(2) - (1)(1)\big)\hat i \\ &\quad - \big((1)(2) - (1)(2)\big)\hat j \\ &\quad + \big((1)(1) - (-1)(2)\big)\hat k \\ &= -3\hat i + 0\hat j + 3\hat k. \end{aligned}$$
   Expand along the top row; the $\hat j$ term carries a minus sign, and here its bracket is $2 - 2 = 0$⁠. The result is perpendicular to both directions: $(-3)(1) + 0(-1) + 3(1) = 0$ and $(-3)(2) + 0(1) + 3(2) = 0$⁠. It is not the zero vector, so the lines are not parallel and the skew-lines formula applies.
   Determinant form of the cross product (Section 10.6.3). Basics: a⃗ × b⃗ stands up from both

4. Its length

   $$\begin{aligned} |\vec b_1\times\vec b_2| &= \sqrt{(-3)^2 + 0^2 + 3^2} \\ &= \sqrt{18} = 3\sqrt2. \end{aligned}$$
   $18 = 9 \times 2$ and $\sqrt9 = 3$⁠. Dividing by this length turns $\vec b_1\times\vec b_2$ into the unit vector $\hat n$ along the common perpendicular (equation (3) of Section 11.5.1).
   Length of a vector. Basics: Pythagoras in 3D

5. The dot product

   $$\begin{aligned} &(\vec b_1\times\vec b_2)\cdot(\vec a_2 - \vec a_1) \\ &= (-3)(1) + (0)(-3) + (3)(-2) \\ &= -3 + 0 - 6 = -9. \end{aligned}$$
   Multiply matching components of $-3\hat i + 0\hat j + 3\hat k$ and $\hat i - 3\hat j - 2\hat k$ and add. The result is negative; that only means $\overrightarrow{ST}$ points against $\vec b_1\times\vec b_2$⁠, and the modulus in the formula takes care of it.
   $\vec a\cdot\vec b = a_1b_1 + a_2b_2 + a_3b_3$⁠. Basics: the dot product

6. Put it into the formula

   $$\begin{aligned} d &= \left|\frac{(\vec b_1\times\vec b_2)\cdot(\vec a_2 - \vec a_1)}{|\vec b_1\times\vec b_2|}\right| \\ &= \left|\frac{-9}{3\sqrt2}\right| = \frac{3}{\sqrt2} = \frac{3\sqrt2}{2}. \end{aligned}$$
   This is the length of the projection of $\overrightarrow{ST}$ on the unit vector $\hat n$ along the common perpendicular, $d = ST\,|\cos\theta|$⁠. Cancel 3, then multiply top and bottom by $\sqrt2$ to rationalise. One misprint to watch: 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

   $$\begin{aligned} \overrightarrow{PQ} &= (1 + 2\mu - \lambda)\hat i + (-3 + \mu + \lambda)\hat j \\ &\quad + (-2 + 2\mu - \lambda)\hat k. \end{aligned}$$

   $PQ$ is shortest when it is perpendicular to both lines, $\overrightarrow{PQ}\cdot\vec b_1 = 0$ and $\overrightarrow{PQ}\cdot\vec b_2 = 0$⁠:

   $$\begin{aligned} 2 + 3\mu - 3\lambda &= 0, \\ -5 + 9\mu - 3\lambda &= 0, \end{aligned}$$

   so $\mu = \frac76$ and $\lambda = \frac{11}{6}$⁠. Then $P\left(\frac{17}{6}, \frac16, \frac{17}{6}\right)$⁠, $Q\left(\frac{13}{3}, \frac16, \frac43\right)$ and

   $$\begin{aligned} \overrightarrow{PQ} &= \frac32\hat i - \frac32\hat k, \\ PQ &= \frac32\sqrt2 = \frac{3}{\sqrt2}. \end{aligned}$$
   This works the problem from scratch, without the formula, and gets the same distance. Subtracting the first condition from the second gives $-7 + 6\mu = 0$⁠. $\overrightarrow{PQ} = -\frac12(-3\hat i + 3\hat k)$ is a multiple of $\vec b_1\times\vec b_2$⁠, as the formula assumes. Since $d \ne 0$ the lines do not meet, and Step 3 showed they are not parallel, so they are skew.
   The line of shortest distance is perpendicular to both lines (Section 11.5). Lesson: skew lines

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