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

Let the vertices of the triangle be A, B and C respectively.

Here, $|AB| = \sqrt {{{( - 1 - 3)}^2} + {{(1 - 5)}^2} + {{(2 + 4)}^2}} = \sqrt {68} = 2\sqrt {17}$

$|BC| = \sqrt {{{( - 5 + 1)}^2} + {{( - 1 - 5)}^2} + {{( - 2 - 2)}^2}} = \sqrt {68}$ $= 2\sqrt {17}$

$|CA| = \sqrt {{{(3 + 5)}^2} + {{(5 + 5)}^2} + {{( - 4 + 2)}^2}} = \sqrt {168} = 2\sqrt {42}$

$\therefore$ Direction cosines of AB are $< \cfrac{{( - 1 - 3)}}{{|AB|}},\cfrac{{(1 - 5)}}{{|AB|}},\cfrac{{(2 + 4)}}{{|AB|}} >$

or $< \cfrac{{ - 2}}{{\sqrt {17} }},\cfrac{{ - 2}}{{\sqrt {17} }},\cfrac{3}{{\sqrt {17} }} >$

$\therefore$ Direction cosines of BC are $< \cfrac{{ - 5 + 1}}{{|BC|}},\cfrac{{ - 5 - 1}}{{|BC|}},\cfrac{{ - 2 - 2}}{{|BC|}} >$

or $< \cfrac{{ - 2}}{{\sqrt {17} }},\cfrac{{ - 3}}{{\sqrt {17} }},\cfrac{{ - 2}}{{\sqrt {17} }} >$

Direction cosines of CA are $< \cfrac{{3 + 5}}{{|CA|}},\cfrac{{5 + 5}}{{|CA|}},\cfrac{{ - 4 + 2}}{{|CA|}} >$

or $< \cfrac{4}{{\sqrt {42} }},\cfrac{5}{{\sqrt {42} }},\cfrac{{ - 1}}{{\sqrt {42} }} >$

Exercise - 11.2