Q5:

Find the direction cosines of the sides of the triangle whose vertices are $(3, 5, – 4)$, $(– 1, 1, 2)$ and $(– 5, – 5, – 2)$.

Step 1: Coordinate Definition and Vector Representation

Let the vertices of the triangle be defined as:

• $A = (3, 5, -4)$
• $B = (-1, 1, 2)$
• $C = (-5, -5, -2)$

The direction cosines of a line segment connecting two points $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$ are proportional to the direction ratios $(a, b, c) = (x_2-x_1, y_2-y_1, z_2-z_1)$. If $L$ is the length of the segment, where $L = \sqrt{a^2 + b^2 + c^2}$, the direction cosines $(l, m, n)$ are given by $\left( \frac{a}{L}, \frac{b}{L}, \frac{c}{L} \right)$ [Per the definition of direction cosines in $\mathbb{R}^3$].

Step 2: Analysis of Side AB

Direction ratios for $\vec{AB}$: $a_1 = -1 - 3 = -4$, $b_1 = 1 - 5 = -4$, $c_1 = 2 - (-4) = 6$.

Length $L_{AB} = \sqrt{(-4)^2 + (-4)^2 + 6^2} = \sqrt{16 + 16 + 36} = \sqrt{68} = 2\sqrt{17}$.

Direction cosines $(l_1, m_1, n_1) = \left( \frac{-4}{2\sqrt{17}}, \frac{-4}{2\sqrt{17}}, \frac{6}{2\sqrt{17}} \right) = \mathbf{\left( -\frac{2}{\sqrt{17}}, -\frac{2}{\sqrt{17}}, \frac{3}{\sqrt{17}} \right)}$.

Step 3: Analysis of Side BC

Direction ratios for $\vec{BC}$: $a_2 = -5 - (-1) = -4$, $b_2 = -5 - 1 = -6$, $c_2 = -2 - 2 = -4$.

Length $L_{BC} = \sqrt{(-4)^2 + (-6)^2 + (-4)^2} = \sqrt{16 + 36 + 16} = \sqrt{68} = 2\sqrt{17}$.

Direction cosines $(l_2, m_2, n_2) = \left( \frac{-4}{2\sqrt{17}}, \frac{-6}{2\sqrt{17}}, \frac{-4}{2\sqrt{17}} \right) = \mathbf{\left( -\frac{2}{\sqrt{17}}, -\frac{3}{\sqrt{17}}, -\frac{2}{\sqrt{17}} \right)}$.

Step 4: Analysis of Side CA

Direction ratios for $\vec{CA}$: $a_3 = 3 - (-5) = 8$, $b_3 = 5 - (-5) = 10$, $c_3 = -4 - (-2) = -2$.

Length $L_{CA} = \sqrt{8^2 + 10^2 + (-2)^2} = \sqrt{64 + 100 + 4} = \sqrt{168} = 2\sqrt{42}$.

Direction cosines $(l_3, m_3, n_3) = \left( \frac{8}{2\sqrt{42}}, \frac{10}{2\sqrt{42}}, \frac{-2}{2\sqrt{42}} \right) = \mathbf{\left( \frac{4}{\sqrt{42}}, \frac{5}{\sqrt{42}}, -\frac{1}{\sqrt{42}} \right)}$.

Summary Table of Direction Cosines

Final Solution: The direction cosines for the sides are $AB: (-\frac{2}{\sqrt{17}}, -\frac{2}{\sqrt{17}}, \frac{3}{\sqrt{17}})$, $BC: (-\frac{2}{\sqrt{17}}, -\frac{3}{\sqrt{17}}, -\frac{2}{\sqrt{17}})$, and $CA: (\frac{4}{\sqrt{42}}, \frac{5}{\sqrt{42}}, -\frac{1}{\sqrt{42}})$.