Question

If $P,Q,R$ are respectively $(2,3,5),(-1,3,2)$ and $(3,5,-2)$, find the direction cosines of the sides of the triangle $PQR$

Answer 1

Given sides of triangle are

$P=(2,3,5)$ $Q=(-1,3,2)$

Direction ratio's of PQ

$\begin{array}{c}=x_2-x_1,y_2-y_1,z_2-z_1\\ =-1-2,3-3,2-5\\ =-3,0,-3\\ PQ=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2+{\left(z_2-z_1\right)}^2},\\ =\sqrt{{\left(-3\right)}^2+{\left(0\right)}^2+{\left(-3\right)}^2},\\ =\sqrt{9+9}=\sqrt{18}=3\sqrt{2},\end{array}$

Now directions cosines are

$\begin{array}{c}\frac{-3}{3\sqrt{2}},\frac{0}{3\sqrt{2}},\frac{-3}{3\sqrt{2}}\\ \frac{-1}{\sqrt{2}},0,\frac{-1}{\sqrt{2}}\end{array}$

Direction ratio's of QR

$\begin{array}{c}=x_3-x_2,y_3-y_2,z_3-z_2\\ =3-\left(-1\right),5-3,-2-2\\ =3+1,2,-4\\ =4,2,-4\\ QR=\sqrt{{\left(4\right)}^2+{\left(2\right)}^2+{\left(-4\right)}^2}\\ =\sqrt{16+4+16}=\sqrt{36}=6\end{array}$

Now directions cosines are

$\begin{array}{c}\frac{4}{6},\frac{2}{6},\frac{-4}{6}\\ \frac{2}{3},\frac{1}{3},\frac{-2}{3}\end{array}$

Direction ratio's of RP

$\begin{array}{c}=x_3-x_1,y_3-y_1,z_3-z_1\\ =3-2,5-3,-2-5\\ =1,2,-7\\ RP=\sqrt{{\left(1\right)}^2+{\left(2\right)}^2+{\left(-7\right)}^2}\\ =\sqrt{1+4+49}=\sqrt{54}=3\sqrt{6}\end{array}$

Now direction cosines are

$\frac{1}{3\sqrt{6}},\frac{2}{3\sqrt{6}},\frac{-7}{7\sqrt{6}}$