Exercise 11.2 · Q13: Shortest distance between two lines in Cartesian form

Basics you need

Solution

1. Points and directions as vectors

   $\frac{x - (-1)}{7} = \frac{y - (-1)}{-6} = \frac{z - (-1)}{1}$ passes through $S(-1, -1, -1)$⁠; $\frac{x - 3}{1} = \frac{y - 5}{-2} = \frac{z - 7}{1}$ passes through $T(3, 5, 7)$⁠. So

   $$\begin{aligned} \vec a_1 &= -\hat i - \hat j - \hat k, & \vec b_1 &= 7\hat i - 6\hat j + \hat k, \\ \vec a_2 &= 3\hat i + 5\hat j + 7\hat k, & \vec b_2 &= \hat i - 2\hat j + \hat k. \end{aligned}$$
   Each numerator $x + 1$ is $x - (-1)$⁠, so every coordinate of $S$ is $-1$⁠. The denominators, with their signs, are the direction ratios, and direction ratios $a, b, c$ give the direction vector $a\hat i + b\hat j + c\hat k$⁠.
   Cartesian equation (3) and the Remark of Section 11.3.1. Basics: point plus direction

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

   $$\begin{aligned} \vec a_2 - \vec a_1 &= (3 + 1)\hat i + (5 + 1)\hat j \\ &\quad + (7 + 1)\hat k \\ &= 4\hat i + 6\hat j + 8\hat k. \end{aligned}$$
   This is $\overrightarrow{ST}$⁠. Subtracting $-1$ adds 1 to each coordinate of $T$⁠.
   Vector joining two points (Section 10.5.2). Basics: points in space

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

   $$\begin{aligned} \vec b_1\times\vec b_2 &= \begin{vmatrix} \hat i & \hat j & \hat k \\ 7 & -6 & 1 \\ 1 & -2 & 1 \end{vmatrix} \\ &= \big((-6)(1) - (1)(-2)\big)\hat i \\ &\quad - \big((7)(1) - (1)(1)\big)\hat j \\ &\quad + \big((7)(-2) - (-6)(1)\big)\hat k \\ &= -4\hat i - 6\hat j - 8\hat k. \end{aligned}$$
   Expand along the top row, with a minus sign on the $\hat j$ term. Check: $(-4)(7) + (-6)(-6) + (-8)(1)$ $= -28 + 36 - 8 = 0$ and $(-4)(1) + (-6)(-2) + (-8)(1)$ $= -4 + 12 - 8 = 0$⁠, 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

   $$\begin{aligned} |\vec b_1\times\vec b_2| &= \sqrt{16 + 36 + 64} \\ &= \sqrt{116} = 2\sqrt{29}. \end{aligned}$$
   $116 = 4 \times 29$ and $\sqrt4 = 2$⁠; 29 is prime, so $\sqrt{29}$ does not simplify further.
   Length of a vector. Basics: Pythagoras in 3D

5. The dot product

   $$\begin{aligned} &(\vec b_1\times\vec b_2)\cdot(\vec a_2 - \vec a_1) \\ &= (-4)(4) + (-6)(6) + (-8)(8) \\ &= -16 - 36 - 64 = -116. \end{aligned}$$
   Multiply matching components and add. Notice that $\vec a_2 - \vec a_1 = 4\hat i + 6\hat j + 8\hat k$ is exactly $-(\vec b_1\times\vec b_2)$⁠; Step 8 explains what that means.
   $\vec a\cdot\vec b = a_1b_1 + a_2b_2 + a_3b_3$⁠. Basics: the dot product

6. Put it into the formula

   $$\begin{aligned} d &= \left|\frac{(\vec b_1\times\vec b_2)\cdot(\vec a_2 - \vec a_1)}{|\vec b_1\times\vec b_2|}\right| \\ &= \left|\frac{-116}{2\sqrt{29}}\right| = \frac{58}{\sqrt{29}} \\ & = \frac{58\sqrt{29}}{29} = 2\sqrt{29}. \end{aligned}$$
   Cancel 2, then rationalise: $\frac{58}{\sqrt{29}} \times \frac{\sqrt{29}}{\sqrt{29}} = \frac{58\sqrt{29}}{29}$⁠, and $58 = 2 \times 29$⁠. The modulus makes the distance positive. As in the other shortest-distance questions, note that 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.
   Section 11.5.1, shortest distance between skew lines. Lesson: the shortest distance

7. Check with the Cartesian formula

   Section 11.5.1 also gives the distance directly from the Cartesian data, as a determinant over a square root. The determinant has rows $x_2 - x_1, y_2 - y_1, z_2 - z_1$⁠, then $a_1, b_1, c_1$⁠, then $a_2, b_2, c_2$⁠:

   $$\begin{vmatrix} 4 & 6 & 8 \\ 7 & -6 & 1 \\ 1 & -2 & 1 \end{vmatrix} = -116,$$

   and the square root is $\sqrt{(-6 + 2)^2 + (1 - 7)^2 + (-14 + 6)^2} = \sqrt{116}$⁠. So $d = \frac{116}{\sqrt{116}} = \sqrt{116} = 2\sqrt{29}$ again.

   The determinant is the same number as the dot product of Step 5 (a scalar triple product), and the three brackets under the root are the components of $\vec b_1\times\vec b_2$ from Step 3. So the Cartesian formula is the vector formula written out in coordinates. Expanding along the first row: $4(-6 + 2) - 6(7 - 1) + 8(-14 + 6)$ $= -16 - 36 - 64$⁠.
   Cartesian form of the shortest distance, Section 11.5.1. Lesson: the shortest distance

8. Why the given points are the closest points

   $\overrightarrow{ST} = 4\hat i + 6\hat j + 8\hat k$ is perpendicular to both lines:

   $$\begin{aligned} \overrightarrow{ST}\cdot\vec b_1 &= 28 - 36 + 8 = 0, \\ \overrightarrow{ST}\cdot\vec b_2 &= 4 - 12 + 8 = 0. \end{aligned}$$

   So $ST$ itself is the common perpendicular, and $d = ST = \sqrt{16 + 36 + 64} = 2\sqrt{29}$⁠.

   In general the feet $P$⁠, $Q$ of the common perpendicular are found from $\overrightarrow{PQ}\cdot\vec b_1 = \overrightarrow{PQ}\cdot\vec b_2 = 0$⁠; here those equations are solved by $\lambda = \mu = 0$⁠, that is $P = S$ and $Q = T$⁠. That is why the formula gave $|\overrightarrow{ST}|$ exactly: the projection of $\overrightarrow{ST}$ on its own direction is its full length. The book chose the points so that this happens.
   The line of shortest distance is perpendicular to both lines (Section 11.5). Lesson: skew lines

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