Example 3: Direction cosines of the line through $(-2, 4, -5)$ and $(1, 2, 3)$

Basics you need

Solution

1. Name the points

   Take $P(x_1, y_1, z_1) = (-2, 4, -5)$ and $Q(x_2, y_2, z_2) = (1, 2, 3)$⁠.

   The formula is written for a first point $P$ and a second point $Q$⁠; the order decides the direction (from $P$ towards $Q$⁠). The book makes the same choice.
   Coordinates of a point in space. Basics: three numbers fix a point

2. Subtract: the direction ratios

   $$\begin{aligned} x_2 - x_1 &= 1 - (-2) = 3, \\ y_2 - y_1 &= 2 - 4 = -2, \\ z_2 - z_1 &= 3 - (-5) = 8. \end{aligned}$$
   Subtracting a negative number adds it: $1 - (-2) = 3$ and $3 - (-5) = 8$⁠. The three differences are how far you move along each axis from $P$ to $Q$⁠, so they point along the line: they are direction ratios of $PQ$ (the Note in Section 11.2.1).
   Direction ratios of $PQ$⁠: $x_2 - x_1, y_2 - y_1, z_2 - z_1$⁠. Lesson: Examples 1–5

3. Find the length $PQ$

   $$\begin{aligned} PQ &= \sqrt{3^2 + (-2)^2 + 8^2} \\ &= \sqrt{9 + 4 + 64} = \sqrt{77}. \end{aligned}$$
   The distance formula squares the three differences just found, so $(-2)^2 = 4$ is positive. 77 is $7 \times 11$⁠, which has no square factor, so $\sqrt{77}$ does not simplify.
   Distance formula in space. Basics: Pythagoras twice

4. Divide by $PQ$

   $$l = \frac{3}{\sqrt{77}}, \quad m = \frac{-2}{\sqrt{77}}, \quad n = \frac{8}{\sqrt{77}}.$$
   In the book's Fig 11.2, the right triangle $PNQ$ has hypotenuse $PQ$ and the side $NQ = z_2 - z_1$ next to the angle $\gamma$⁠, so $\cos\gamma = \frac{z_2 - z_1}{PQ}$⁠. The same argument with the other two axes gives $\cos\alpha$ and $\cos\beta$⁠. In other words, dividing the direction ratios by $PQ$ makes the squares add up to 1.
   Section 11.2.1: $\cos\alpha = \frac{x_2 - x_1}{PQ}$⁠, and so on. Lesson: direction cosines

5. Check, and the other direction

   $$\frac{9}{77} + \frac{4}{77} + \frac{64}{77} = \frac{77}{77} = 1.$$

   Taking the line from $Q$ to $P$ instead gives $\frac{-3}{\sqrt{77}}, \frac{2}{\sqrt{77}}, \frac{-8}{\sqrt{77}}$⁠.

   The squares adding up to 1 confirms that these are direction cosines. Swapping the two points reverses every difference, which is the same line in the opposite direction (Note in Section 11.2.1: either set of differences may be used). The book gives the direction from the first point to the second. If you prefer rationalised denominators, $\frac{3}{\sqrt{77}} = \frac{3\sqrt{77}}{77}$⁠, and so on.
   $l^2 + m^2 + n^2 = 1$⁠. Basics: direction cosines

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