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

Question

$\begin{array}{1 1} \large(\frac{2}{5},\frac{2}{5},\frac{3}{5}), \large(\frac{2}{5},\frac{3}{5},\frac{2}{5})\; and \;\large(\frac{4}{\sqrt{42}},\frac{5}{\sqrt{42}},\frac{1}{\sqrt{42}}) \\\large(\frac{2}{5},\frac{2}{5},-\frac{3}{5}), \large(\frac{2}{5},\frac{3}{5},\frac{2}{5})\; and \;\large(\frac{4}{\sqrt{42}},\frac{5}{\sqrt{42}},\frac{1}{\sqrt{42}}) \\\large(\frac{2}{5},\frac{2}{5},\frac{3}{5}), \large(\frac{2}{5},\frac{3}{5},\frac{2}{5}) and \large(\frac{4}{\sqrt{42}},\frac{5}{\sqrt{42}},-\frac{1}{\sqrt{42}}) \\ \large(\frac{-2}{\sqrt {17}},\frac{2}{\sqrt {17}},\frac{3}{\sqrt {17}}), \large(\frac{-2}{\sqrt {17}},\frac{-3}{\sqrt {17}},\frac{-2}{\sqrt {17}})\; and\; \large(\frac{4}{\sqrt{42}},\frac{5}{\sqrt{42}},-\frac{1}{\sqrt{42}}) \end{array}$