Question 5: Find the direction cosines of the sides of the triangle whose vertices are (3, 5, – 4), (– 1, 1, 2) and (– 5, – 5, – 2).
Answer:
Given vertices of the triangle $\triangle ABC$ (3, 5, – 4), (– 1, 1, 2) and (– 5, – 5, – 2).
Finding each side direction ratios;
$\Rightarrow$ Direction ratios of side AB are $(-1-3), (1-5),\ and\ (2-(-4))$ i.e.,
$-4,-4,\ and\ 6.$
Therefore its direction cosines values are;
$\frac{-4}{\sqrt{(-4)^2+(-4)^2+(6)^2}},\ \frac{-4}{\sqrt{(-4)^2+(-4)^2+(6)^2}},\ \frac{6}{\sqrt{(-4)^2+(-4)^2+(6)^2}}$ $or\ \frac{-4}{2\sqrt{17}},\frac{-4}{2\sqrt{17}},\frac{6}{2\sqrt{17}}\ or\ \frac{-2}{\sqrt{17}},\frac{-2}{\sqrt{17}},\frac{3}{\sqrt{17}}$
SImilarly for side BC;
$\Rightarrow$ Direction ratios of side BC are $(-5-(-1)), (-5-1),\ and\ (-2-2)$ i.e.,
$-4,-6,\ and\ -4.$
Therefore its direction cosines values are;
$\frac{-4}{\sqrt{(-4)^2+(-6)^2+(-4)^2}},\ \frac{-6}{\sqrt{(-4)^2+(-6)^2+(-4)^2}},\ \frac{-4}{\sqrt{(-4)^2+(-6)^2+(-4)^2}}$ $or\ \frac{-4}{2\sqrt{17}},\frac{-6}{2\sqrt{17}},\frac{-4}{2\sqrt{17}}\ or\ \frac{-2}{\sqrt{17}},\frac{-3}{\sqrt{17}},\frac{-2}{\sqrt{17}}$
$\Rightarrow$ Direction ratios of side CA are $(-5-3), (-5-5),\ and\ (-2-(-4))$ i.e.,
$-8,-10,\ and\ 2.$
Therefore its direction cosines values are;
$\frac{-8}{\sqrt{(-8)^2+(10)^2+(2)^2}},\ \frac{-5}{\sqrt{(-8)^2+(10)^2+(2)^2}},\ \frac{2}{\sqrt{(-8)^2+(10)^2+(2)^2}}$ $or\ \frac{-8}{2\sqrt{42}},\frac{-10}{2\sqrt{42}},\frac{2}{2\sqrt{42}}\ or\ \frac{-4}{\sqrt{42}},\frac{-5}{\sqrt{42}},\frac{1}{\sqrt{42}}$
