Browse Questions

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

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

Toolbox:
• If $l,m,n$are direction cosines and $a,b,c$ are direction ration of a line, then
• $\large\frac{a}{\lambda l},\frac{b}{\lambda m},\frac{c}{\lambda n}$ for any non zero $\lambda \in R$
• If $l,m,n$ are direction cosines are $\large\frac{l}{\sqrt {l^2+m^2+n^2}},\frac{m}{\sqrt {l^2+m^2+n^2}},\frac{n}{\sqrt {l^2+m^2+n^2}}$
Step 1:
Let $A(3,5,-4),B(-1,1,2)$ and $(-5,-5,-2)$
Direction ratios of the line joining A and B are
$(-1-3),(1-5),(2-(-4))$
$=(-4,-4,6)$
Therefore $AB=\sqrt {(-4)^2+(-4)^2+6^2}=\sqrt {16+16+36}$
$AB=\sqrt {68}$
Therefore the direction cosines are $\large\frac{-4}{\sqrt {68}},\frac{-4}{\sqrt {68}},\frac{6}{\sqrt {68}}$
$= \large\frac{-2}{\sqrt {17}},\frac{-2}{\sqrt {17}},\frac{3}{\sqrt {17}}$
Step 2:
Direction ratios of the line joining B and C are
$(-5(-1)),(-5-1),(-2-(-2))$
$=(-4,-6,4)$
Therefore $BC=\sqrt {(-4)^2+(-6)^2+(-4)^2}=\sqrt {16+36+16}$
$BC=\sqrt {68}=2 \sqrt {17}$
Therefore the direction cosines are $\large\frac{-4}{2\sqrt {17}},\frac{-6}{2\sqrt {17}},\frac{-4}{2\sqrt {17}}$
$= \large\frac{-2}{\sqrt {17}},\frac{-3}{\sqrt {17}},\frac{-2}{\sqrt {17}}$
Step 3:
Direction ratios of the line joining C and A are
$(3-(-5)),(5-(-5)),(-4-(-2))$
$=(8,10,-2)$
$\sqrt {(8)^2+(10)^2+(-2)^2}=\sqrt {64+100+4}$
$=\sqrt {168}$
$=2 \sqrt {42}$
Therefore the direction cosines are $\large\frac{8}{2\sqrt {42}},\frac{10}{2\sqrt {42}},\frac{-2}{2\sqrt {42}}$
$= \large\frac{4}{\sqrt {42}},\frac{5}{\sqrt {42}},\frac{-1}{\sqrt {42}}$