Example 21: Collinear points $A(-2\hat i+3\hat j+5\hat k)$⁠, $B(\hat i+2\hat j+3\hat k)$⁠, $C(7\hat i-\hat k)$

Basics you need

Solution

1. The three vectors between the points

   $$\begin{aligned} \overrightarrow{AB} &= (1 + 2)\hat i + (2 - 3)\hat j + (3 - 5)\hat k \\ &= 3\hat i - \hat j - 2\hat k, \\ \overrightarrow{BC} &= (7 - 1)\hat i + (0 - 2)\hat j + (-1 - 3)\hat k \\ &= 6\hat i - 2\hat j - 4\hat k, \\ \overrightarrow{AC} &= (7 + 2)\hat i + (0 - 3)\hat j + (-1 - 5)\hat k \\ &= 9\hat i - 3\hat j - 6\hat k. \end{aligned}$$
   Each vector is terminal minus initial, component by component. For $\overrightarrow{AB}$ the $\hat i$ part is $1 - (-2) = 1 + 2 = 3$⁠, which is how the book writes it. For C the missing $\hat j$ term is a 0, so the $\hat j$ parts of $\overrightarrow{BC}$ and $\overrightarrow{AC}$ are $0 - 2$ and $0 - 3$⁠.
   $\overrightarrow{P_1P_2} = \overrightarrow{OP_2} - \overrightarrow{OP_1}$ (Section 10.5.2). Basics: three numbers fix a point

2. Their lengths, with a common root

   $$\begin{aligned} |\overrightarrow{AB}| &= \sqrt{9 + 1 + 4} = \sqrt{14}, \\ |\overrightarrow{BC}| &= \sqrt{36 + 4 + 16} = \sqrt{56}, \\ |\overrightarrow{AC}| &= \sqrt{81 + 9 + 36} = \sqrt{126}. \end{aligned}$$

   Since $56 = 4 \times 14$ and $126 = 9 \times 14$⁠, $\sqrt{56} = 2\sqrt{14}$ and $\sqrt{126} = 3\sqrt{14}$⁠.

   Each square comes from the components: $3^2 + (-1)^2 + (-2)^2 = 9 + 1 + 4$⁠, and so on. Writing every length as a multiple of the same root $\sqrt{14}$ is what makes them comparable by eye; $\sqrt{14} + \sqrt{56}$ and $\sqrt{126}$ are hard to compare until they are simplified. The rule used is $\sqrt{k^2m} = k\sqrt m$ for $k \ge 0$⁠.
   $|\vec r| = \sqrt{x^2 + y^2 + z^2}$ (Section 10.5.1). Basics: use Pythagoras twice

3. The two shorter lengths add up to the longest

   $$\begin{aligned} |\overrightarrow{AB}| + |\overrightarrow{BC}| &= \sqrt{14} + 2\sqrt{14} \\ &= 3\sqrt{14} = |\overrightarrow{AC}|. \end{aligned}$$

   So $\text{AC} = \text{AB} + \text{BC}$⁠: going from A to C through B is no longer than going straight from A to C. That can only happen when B lies on the segment AC, so A, B and C are on one line.

   If B were off the line AC, the points would form a genuine triangle, and in a triangle each side is strictly shorter than the sum of the other two: $\text{AC} < \text{AB} + \text{BC}$⁠. Here the two are equal, so there is no triangle, and B must be on the segment between A and C. This is exactly the Remark after Example 20: equality in the triangle inequality means collinear points.
   $|\vec a + \vec b| \le |\vec a| + |\vec b|$ with $\vec a = \overrightarrow{AB}$⁠, $\vec b = \overrightarrow{BC}$ (Example 20 and its Remark). Lesson: Examples 13–21

4. A second view: the vectors are multiples of each other

   $$\begin{aligned} \overrightarrow{BC} &= 6\hat i - 2\hat j - 4\hat k \\ &= 2(3\hat i - \hat j - 2\hat k) = 2\,\overrightarrow{AB}, \\ \overrightarrow{AC} &= 9\hat i - 3\hat j - 6\hat k = 3\,\overrightarrow{AB}. \end{aligned}$$

   $\overrightarrow{AB}$ and $\overrightarrow{BC}$ are parallel and point the same way, and they share the point B, so the line through A and B and the line through B and C are the same line. Also $\text{AB} : \text{BC} = 1 : 2$⁠: B divides AC internally in the ratio $1 : 2$⁠.

   Parallel vectors alone would allow two separate parallel lines; it is the shared point B that forces one line. The book's Note warns about the opposite trap: $\overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CA} = \vec 0$ is true here, but it is true for any three points, so it shows nothing about a triangle or a line.
   $\vec b = \lambda\vec a$ means $\vec a$⁠, $\vec b$ are collinear (Section 10.5.1, Remarks (i)). Lesson: the section formula

Every fact used, and where it comes from