Example 9: Shortest distance between two skew lines in vector form

Basics you need

Solution

1. Read off the points and directions

   $$\begin{aligned} \vec a_1 &= \hat i + \hat j, & \vec b_1 &= 2\hat i - \hat j + \hat k, \\ \vec a_2 &= 2\hat i + \hat j - \hat k, & \vec b_2 &= 3\hat i - 5\hat j + 2\hat k. \end{aligned}$$
   Compare (1) with $\vec r = \vec a_1 + \lambda\vec b_1$ and (2) with $\vec r = \vec a_2 + \mu\vec b_2$⁠: the vector multiplied by the parameter is the direction, the rest is a point. So $l_1$ passes through $S(1, 1, 0)$ (there is no $\hat k$ term, so its $z$-coordinate is 0) and $l_2$ through $T(2, 1, -1)$⁠.
   Vector equation of a line, Section 11.3.1. Basics: a⃗ + λb⃗

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

   $$\begin{aligned} \vec a_2 - \vec a_1 &= (2 - 1)\hat i + (1 - 1)\hat j \\ &\quad + (-1 - 0)\hat k \\ &= \hat i - \hat k. \end{aligned}$$
   This is $\overrightarrow{ST}$⁠, from the point $S$ of $l_1$ to the point $T$ of $l_2$⁠. Its $\hat j$ component is 0, which we must remember when taking the dot product later.
   Vector joining two points: terminal minus initial (Section 10.5.2). Basics: points in space

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

   $$\begin{aligned} \vec b_1\times\vec b_2 &= \begin{vmatrix} \hat i & \hat j & \hat k \\ 2 & -1 & 1 \\ 3 & -5 & 2 \end{vmatrix} \\ &= \big((-1)(2) - (1)(-5)\big)\hat i \\ &\quad - \big((2)(2) - (1)(3)\big)\hat j \\ &\quad + \big((2)(-5) - (-1)(3)\big)\hat k \\ &= 3\hat i - \hat j - 7\hat k. \end{aligned}$$
   Expand along the top row; the $\hat j$ term carries a minus sign. The result is perpendicular to both directions: $(3)(2) + (-1)(-1) + (-7)(1) = 0$ and $(3)(3) + (-1)(-5) + (-7)(2) = 0$⁠. So it points along the common perpendicular of the two lines. It is not the zero vector, 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

   $$\begin{aligned} |\vec b_1\times\vec b_2| &= \sqrt{3^2 + (-1)^2 + (-7)^2} \\ &= \sqrt{9 + 1 + 49} = \sqrt{59}. \end{aligned}$$
   We divide by this length so that only the direction of $\vec b_1\times\vec b_2$ counts: $\frac{\vec b_1\times\vec b_2}{|\vec b_1\times\vec b_2|}$ is the unit vector $\hat n$ of equation (3) in Section 11.5.1.
   Length of a vector. Basics: Pythagoras in 3D

5. The dot product

   $$\begin{aligned} &(\vec b_1\times\vec b_2)\cdot(\vec a_2 - \vec a_1) \\ &= (3)(1) + (-1)(0) + (-7)(-1) \\ &= 3 - 0 + 7 = 10. \end{aligned}$$
   Multiply matching components of $3\hat i - \hat j - 7\hat k$ and $\hat i + 0\hat j - \hat k$ and add. The last product is $(-7)(-1) = +7$⁠: two negatives give a positive.
   $\vec a\cdot\vec b = a_1b_1 + a_2b_2 + a_3b_3$⁠. Basics: the dot product

6. Put it into the formula

   $$\begin{aligned} d &= \left|\frac{(\vec b_1\times\vec b_2)\cdot(\vec a_2 - \vec a_1)}{|\vec b_1\times\vec b_2|}\right| \\ &= \left|\frac{10}{\sqrt{59}}\right| = \frac{10}{\sqrt{59}}. \end{aligned}$$
   This is the length of the projection of $\overrightarrow{ST}$ on the unit vector $\hat n$ along the common perpendicular: $d = ST\,|\cos\theta|$⁠, where $\theta$ is the angle between $\overrightarrow{ST}$ and $\hat n$⁠. The modulus makes the distance positive whichever way $\hat n$ points. 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, and this example, both use 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 $l_2$⁠. The segment $PQ$ is shortest when it is perpendicular to both lines:

   $$\overrightarrow{PQ}\cdot\vec b_1 = 0, \qquad \overrightarrow{PQ}\cdot\vec b_2 = 0.$$

   Here $\overrightarrow{PQ} = (\vec a_2 - \vec a_1) + \mu\vec b_2 - \lambda\vec b_1$⁠, that is

   $$\begin{aligned} \overrightarrow{PQ} &= (1 + 3\mu - 2\lambda)\hat i + (\lambda - 5\mu)\hat j \\ &\quad + (-1 + 2\mu - \lambda)\hat k, \end{aligned}$$

   and the two conditions become

   $$\begin{aligned} 1 + 13\mu - 6\lambda &= 0, \\ 1 + 38\mu - 13\lambda &= 0, \end{aligned}$$

   so $\lambda = \frac{25}{59}$ and $\mu = \frac{7}{59}$⁠. Then

   $$\begin{aligned} \overrightarrow{PQ} &= \frac{10}{59}(3\hat i - \hat j - 7\hat k), \\ PQ &= \frac{10}{59}\sqrt{59} = \frac{10}{\sqrt{59}}. \end{aligned}$$
   This works the problem from scratch, without the formula, and gets the same distance. $\overrightarrow{PQ}$ comes out as 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 given points are not the closest ones: $ST = |\hat i - \hat k| = \sqrt2 > \frac{10}{\sqrt{59}}$⁠.
   The line of shortest distance is perpendicular to both lines (Section 11.5). Lesson: skew lines

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