Question

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

Answer 1

Let the vertices of the triangle be A(3,5,-4), B(-1,1,2) and C(-5,-5,-2) respectively.

The direction ratios of side AB are (-1,-3,1-5,2-(-4)) (-4,-4,2+4), i.e., (-4,-4,6).

[$\therefore$ If $A(x_1,y_1,z_1)$ and $B(x_2,y_2,z_2)$ then dr's

of $AB=(x_2-x_1,y_2-y_1,z_2-z_1)]$
Then, magnitude of AB,

$|AB|=\sqrt{(-4{)}^2+(-4{)}^2+(6{)}^2}=\sqrt{16+16+36}=\sqrt{68}=2\sqrt{17}$
Therefore, the direction cosines of AB are

$\frac{-4}{2\sqrt{17}},\frac{-4}{2\sqrt{17}},\frac{6}{2\sqrt{17}}\Rightarrow \frac{-2}{\sqrt{17}},\frac{-2}{\sqrt{17}},\frac{3}{\sqrt{17}}$

$[\therefore$ If a,b,c are direction ratios, then direction cosines are

$\frac{a}{\sqrt{a^2+b^2+c^2}},\frac{b}{\sqrt{a^2+b^2+c^2}},\frac{c}{\sqrt{a^2+b^2+c^2}}]$

Similary, the direction ratios of side BC are (-5-(-1)), (-5-1) and (-2-2) ie., (-4,-6,-4).

Magnitude of BC, $|BC|=\sqrt{(-4{)}^2+(-6{)}^2+(-4{)}^2}=\sqrt{68}=2\sqrt{17}$

Therefore, the direction cosines of BC are

$\frac{-4}{2\sqrt{17}},\frac{-6}{2\sqrt{17}},\frac{-4}{2\sqrt{17}}\Rightarrow \frac{-2}{\sqrt{17}},\frac{-3}{\sqrt{17}},\frac{-2}{\sqrt{17}}$

The direction ratios of side CA are (-5-3), (-5 -5) and (-2-(-4)) i.e., (-8,-10,2) or (8,10,-2).

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

Therefore, the direction cosines of AC are

$\frac{8}{2\sqrt{42}},\frac{10}{2\sqrt{42}},\frac{-2}{2\sqrt{42}}\Rightarrow \frac{4}{\sqrt{42}},\frac{5}{\sqrt{42}},\frac{-1}{\sqrt{42}}$