Exercise 11.1 · Q5: Direction cosines of the sides of the triangle $(3, 5, -4)$⁠, $(-1, 1, 2)$⁠, $(-5, -5, -2)$

Basics you need

Solution

1. Name the vertices and the order

   Let $A(3, 5, -4)$⁠, $B(-1, 1, 2)$⁠, $C(-5, -5, -2)$⁠, and take the sides in the order $A \to B$⁠, $B \to C$⁠, $C \to A$⁠.

   A side such as $AB$ is a line, so it has two sets of direction cosines, one for each direction. Travelling round the triangle in one order is a natural choice that fixes all three; it is also the choice the answer key makes. Taking a side the other way, say $B \to A$⁠, just reverses all three signs for that side.
   Direction cosines of a directed line (Section 11.2). Basics: three numbers fix a point

2. Side $AB$

   Direction ratios: $-1 - 3,\ 1 - 5,\ 2 - (-4) = -4, -4, 6$⁠. Length:

   $$AB = \sqrt{16 + 16 + 36} = \sqrt{68} = 2\sqrt{17}.$$

   Direction cosines:

   $$\begin{aligned} &\frac{-4}{2\sqrt{17}}, \frac{-4}{2\sqrt{17}}, \frac{6}{2\sqrt{17}} \\ &= -\frac{2}{\sqrt{17}}, -\frac{2}{\sqrt{17}}, \frac{3}{\sqrt{17}}. \end{aligned}$$
   Second point minus first point gives how far you move along each axis from $A$ to $B$⁠. Dividing these moves by the length $AB$ gives the cosines of the angles $AB$ makes with the axes (the right triangle $PNQ$ of Fig 11.2). Each fraction is cancelled by 2; in particular $\frac{6}{2\sqrt{17}} = \frac{3}{\sqrt{17}}$⁠.
   $\frac{x_2 - x_1}{PQ}, \frac{y_2 - y_1}{PQ}, \frac{z_2 - z_1}{PQ}$ (Section 11.2.1). Example 3: the same method

3. Side $BC$

   Direction ratios: $-5 - (-1),\ -5 - 1,\ -2 - 2 = -4, -6, -4$⁠. Length:

   $$BC = \sqrt{16 + 36 + 16} = \sqrt{68} = 2\sqrt{17}.$$

   Direction cosines:

   $$\begin{aligned} &\frac{-4}{2\sqrt{17}}, \frac{-6}{2\sqrt{17}}, \frac{-4}{2\sqrt{17}} \\ &= -\frac{2}{\sqrt{17}}, -\frac{3}{\sqrt{17}}, -\frac{2}{\sqrt{17}}. \end{aligned}$$
   The same rule with $B$ first and $C$ second. $-5 - (-1) = -5 + 1 = -4$⁠. The sides $AB$ and $BC$ happen to have the same length, so the triangle is isosceles.
   Distance formula. Basics: Pythagoras twice

4. Side $CA$

   Direction ratios: $3 - (-5),\ 5 - (-5),\ -4 - (-2) = 8, 10, -2$⁠. Length:

   $$\begin{aligned} CA &= \sqrt{64 + 100 + 4} \\ &= \sqrt{168} = 2\sqrt{42}. \end{aligned}$$

   Direction cosines:

   $$\begin{aligned} &\frac{8}{2\sqrt{42}}, \frac{10}{2\sqrt{42}}, \frac{-2}{2\sqrt{42}} \\ &= \frac{4}{\sqrt{42}}, \frac{5}{\sqrt{42}}, -\frac{1}{\sqrt{42}}. \end{aligned}$$
   Now $C$ is the first point and $A$ the second, closing the loop. $-4 - (-2) = -2$⁠. Since $168 = 4 \times 42$ and 42 has no square factor, $\sqrt{168} = 2\sqrt{42}$ is fully simplified.
   Direction ratios of $CA$⁠: $x_A - x_C$⁠, and so on. Exercise 11.1 player

5. Check

   Squares: $\frac{4 + 4 + 9}{17} = 1$⁠, $\frac{4 + 9 + 4}{17} = 1$⁠, $\frac{16 + 25 + 1}{42} = 1$⁠.

   Closed loop: $(-4, -4, 6) + (-4, -6, -4)$ $+\ (8, 10, -2) = (0, 0, 0)$⁠.

   A real triangle: $\frac{-4}{-4} = 1$ but $\frac{-4}{-6} = \frac23$⁠, so $AB$ and $BC$ are not parallel.

   Each set of direction cosines must have squares adding up to 1. Going $A \to B \to C \to A$ returns to the start, so the three moves add up to zero, which checks the subtractions. Finally, if the direction ratios of $AB$ and $BC$ were proportional, the three points would be collinear and there would be no triangle; they are not.
   $l^2 + m^2 + n^2 = 1$ (Section 11.2). Basics: direction cosines

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