Miscellaneous · Q4: Shortest distance between lines through $6\hat i + 2\hat j + 2\hat k$ and $-4\hat i - \hat k$

Basics you need

Solution

1. Read off the points and directions

   $$\begin{aligned} \vec a_1 &= 6\hat i + 2\hat j + 2\hat k, & \vec b_1 &= \hat i - 2\hat j + 2\hat k, \\ \vec a_2 &= -4\hat i - \hat k, & \vec b_2 &= 3\hat i - 2\hat j - 2\hat k. \end{aligned}$$
   The vector multiplied by the parameter is the direction, the rest is a point. $\vec a_2 = -4\hat i - \hat k$ has no $\hat j$ term, so its $\hat j$-component is 0. So $l_1$ passes through $S(6, 2, 2)$ and $l_2$ through $T(-4, 0, -1)$⁠.
   Vector equation of a line, Section 11.3.1. Basics: a⃗ + λb⃗

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

   $$\begin{aligned} \vec a_2 - \vec a_1 &= (-4 - 6)\hat i + (0 - 2)\hat j \\ &\quad + (-1 - 2)\hat k \\ &= -10\hat i - 2\hat j - 3\hat k. \end{aligned}$$
   This is $\overrightarrow{ST}$⁠, from $S$ on $l_1$ to $T$ on $l_2$⁠: terminal minus initial. Writing the missing $\hat j$-component of $\vec a_2$ as 0 avoids a slip.
   Vector joining two points (Section 10.5.2). Basics: points in space

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

   $$\begin{aligned} \vec b_1\times\vec b_2 &= \begin{vmatrix} \hat i & \hat j & \hat k \\ 1 & -2 & 2 \\ 3 & -2 & -2 \end{vmatrix} \\ &= \big((-2)(-2) - (2)(-2)\big)\hat i \\ &\quad - \big((1)(-2) - (2)(3)\big)\hat j \\ &\quad + \big((1)(-2) - (-2)(3)\big)\hat k \\ &= 8\hat i + 8\hat j + 4\hat k. \end{aligned}$$
   The $\hat j$ term is $-(-2 - 6) = +8$⁠: the minus sign of the middle term and the negative bracket cancel. Check: $8(1) + 8(-2) + 4(2) = 0$ and $8(3) + 8(-2) + 4(-2) = 0$⁠, 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

   $$\begin{aligned} |\vec b_1\times\vec b_2| &= \sqrt{8^2 + 8^2 + 4^2} \\ &= \sqrt{64 + 64 + 16} \\ &= \sqrt{144} = 12. \end{aligned}$$
   144 is a perfect square, $12^2$⁠, so the length is a whole number.
   Length of a vector. Basics: Pythagoras in 3D

5. The dot product

   $$\begin{aligned} &(\vec b_1\times\vec b_2)\cdot(\vec a_2 - \vec a_1) \\ &= (8)(-10) + (8)(-2) + (4)(-3) \\ &= -80 - 16 - 12 = -108. \end{aligned}$$
   Multiply matching components and add. The result is negative only because $\vec b_1\times\vec b_2$ happens to point against $\overrightarrow{ST}$⁠; the modulus in the formula removes the sign.
   $\vec a\cdot\vec b = a_1b_1 + a_2b_2 + a_3b_3$⁠. Basics: the dot product

6. Put it into the formula

   $$\begin{aligned} d &= \left|\frac{(\vec b_1\times\vec b_2)\cdot(\vec a_2 - \vec a_1)}{|\vec b_1\times\vec b_2|}\right| \\ &= \left|\frac{-108}{12}\right| = |-9| = 9. \end{aligned}$$
   This is the length of the projection of $\overrightarrow{ST}$ on the unit vector along the common perpendicular. A distance is never negative, so we take the modulus. 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

   $$\begin{aligned} \overrightarrow{PQ} &= (-10 + 3\mu - \lambda)\hat i \\ &\quad + (-2 - 2\mu + 2\lambda)\hat j \\ &\quad + (-3 - 2\mu - 2\lambda)\hat k. \end{aligned}$$

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

   $$\begin{aligned} -12 - 9\lambda + 3\mu &= 0, \\ -20 - 3\lambda + 17\mu &= 0, \end{aligned}$$

   so $\lambda = -1$ and $\mu = 1$⁠. Then $P(5, 4, 0)$⁠, $Q(-1, -2, -3)$ and

   $$\begin{aligned} \overrightarrow{PQ} &= -6\hat i - 6\hat j - 3\hat k, \\ PQ &= \sqrt{36 + 36 + 9} = \sqrt{81} = 9. \end{aligned}$$
   The first condition gives $\mu = 4 + 3\lambda$⁠; putting it into the second gives $-20 - 3\lambda + 68 + 51\lambda = 0$⁠, so $48\lambda = -48$⁠, $\lambda = -1$ and $\mu = 1$⁠. Then $P = (6 - 1, 2 + 2, 2 - 2)$ and $Q = (-4 + 3, 0 - 2, -1 - 2)$⁠. $\overrightarrow{PQ} = -\frac34(8\hat i + 8\hat j + 4\hat k)$ lies along the common perpendicular, and its length 9 agrees with the formula. 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

Every fact used, and where it comes from