Find the angle between the lines whose direction cosines satisfy the equations
$\text{ i) }\mathrm{l}+\mathrm{m}+\mathrm{n}=0,{\mathrm{l}}^2+{\mathrm{m}}^2-{\mathrm{n}}^2=0$
$\text{ ii) }3\mathrm{l}+\mathrm{m}+5\mathrm{n}=0\text{ and }6\mathrm{mn}-2\mathrm{nl}+5\mathrm{lm}=0\text{. }$

i) Given equations are l + m + n = 0 ..... (1)
${\mathrm{l}}^2+{\mathrm{m}}^2-{\mathrm{n}}^2=0 ...\left(2\right)$
From eq (1), l = – m – n
Substituting in equation (2)
$\Rightarrow (-\mathrm{m}-\mathrm{n}{)}^2+{\mathrm{m}}^2-{\mathrm{n}}^2=0$
$\begin{array}{l}\Rightarrow {\mathrm{m}}^2+{\mathrm{n}}^2+2\mathrm{mn}+{\mathrm{m}}^2-{\mathrm{n}}^2=0\\ \Rightarrow 2{\mathrm{m}}^2+2\mathrm{mn}=0\Rightarrow 2\mathrm{m}(\mathrm{m}+\mathrm{n})=0\\ \Rightarrow 2\mathrm{m}=0\text{ or }\mathrm{m}+\mathrm{n}=0\Rightarrow \mathrm{m}=0\text{ or }\mathrm{m}=-\mathrm{n}\end{array}$
Case (i): put m = 0 in eq (1)
$\begin{array}{l}\Rightarrow \mathrm{l}=-\mathrm{n}\\ \mathrm{l}:\mathrm{m}:\mathrm{n}=-\mathrm{n}:0:\mathrm{n}=-1:0:1\end{array}$
dr’s of first line  are (– 1, 0, 1)
dc’s  of first line are (l₁, m₁, n₁) =
$\begin{array}{l}\left(\frac{-1}{\sqrt{(-1{)}^2+0^2+1^2}},\frac{0}{\sqrt{(-1{)}^2+0^2+1^2}}\right.,\frac{1}{\sqrt{(-1{)}^2+0^2+1^2}})\\ =\left(-\frac{1}{\sqrt{2}},0,\frac{1}{\sqrt{2}}\right)\end{array}$
Case (ii): put m = –n in (1)
$$\begin{gathered} \Rightarrow \mathrm{l}=0 \\ \mathrm{l}:\mathrm{m}:\mathrm{n}=0:-\mathrm{n}:\mathrm{n}=0:-1:1 \end{gathered}$$
dr’s  of second line are= (a, b, c) = (0, –1, 1)
dc’s  of second line are (l₂, m₂, n₂) =
$\begin{array}{l}\left(\frac{0}{\sqrt{0^2+(-1{)}^2+1^2}},\frac{-1}{\sqrt{0^2+(-1{)}^2+1^2}}, \frac{1}{\sqrt{0^2+{(-1)}^2+1^2}} )\right.\\ =\left(0,-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)\end{array}$
Let$\mathrm{\theta }$ be angle between the lines then
$\begin{array}{l}\cos \mathrm{\theta }=\left|{\mathrm{l}}_1{\mathrm{l}}_2+{\mathrm{m}}_1{\mathrm{m}}_2+{\mathrm{n}}_1{\mathrm{n}}_2\right|\\ =\left|-\frac{1}{\sqrt{2}}\left(0\right)+0\left(\frac{-1}{\sqrt{2}}\right)+\frac{1}{\sqrt{2}}\left(\frac{1}{\sqrt{2}}\right)\right|=\frac{1}{2}\\ \therefore \mathrm{\theta }={60}^{\circ }\end{array}$
ii) Given equations are
3l + m + 5n = 0 ..... (1)
6mn - 2nl + 5lm = 0 ...... (2)
From eq (1), m = –3l – 5n
Substituting in equation (2)
$\begin{array}{l}\Rightarrow 6(-3\mathrm{l}-5\mathrm{n})\mathrm{n}-2\mathrm{nl}+5\mathrm{l}(-3\mathrm{l}-5\mathrm{n})=0\\ \Rightarrow -18\ln -30{\mathrm{n}}^2-2\mathrm{nl}-15{\mathrm{l}}^2-25\mathrm{nl}=0\\ \Rightarrow -30{\mathrm{n}}^2-45\mathrm{nl}-15{\mathrm{l}}^2=0\\ \Rightarrow -15\left[2{\mathrm{n}}^2+3\mathrm{nl}+{\mathrm{l}}^2\right]=0\\ \Rightarrow 2{\mathrm{n}}^2+3\mathrm{nl}+{\mathrm{l}}^2=0\Rightarrow 2{\mathrm{n}}^2+2\mathrm{nl}+\mathrm{nl}+{\mathrm{l}}^2=0\end{array}$
$\begin{array}{l}\Rightarrow 2\mathrm{n}(\mathrm{n}+\mathrm{l})+\mathrm{l}(\mathrm{n}+\mathrm{l})=0\Rightarrow (\mathrm{n}+\mathrm{l})(2\mathrm{n}+\mathrm{l})=0\\ \Rightarrow \mathrm{n}+\mathrm{l}=0\text{ (or) }2\mathrm{n}+\mathrm{l}=0\end{array}$
Case (i): $\mathrm{l}+\mathrm{n}=0\Rightarrow \mathrm{l}=-\mathrm{n}$
From eq (1), $-3\mathrm{n}+\mathrm{m}+5\mathrm{n}=0\Rightarrow \mathrm{m}=-2\mathrm{n}$
$\mathrm{l}:\mathrm{m}:\mathrm{n}=-\mathrm{n}:-2\mathrm{n}:\mathrm{n}=-1:-2:1$
dr’s of first line are= (a, b, c) = (–1, –2, 1)
dc’s of second line are =
$\left(\frac{\mathrm{a}}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2}},\frac{\mathrm{b}}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2}},\frac{\mathrm{c}}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2}}\right)$
$=\left(\begin{array}{r}\frac{-1}{\sqrt{(-1{)}^2+(-2{)}^2+1^2}}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em},\frac{-2}{\sqrt{(-1{)}^2+(-2{)}^2+1^2}},\frac{1}{\sqrt{(-1{)}^2+(-2{)}^2+1^2}}\\ \end{array}\right)$
$=\left(\frac{-1}{\sqrt{6}},\frac{-2}{\sqrt{6}},\frac{1}{\sqrt{6}}\right)=\left({\mathrm{l}}_1,{\mathrm{m}}_1,{\mathrm{n}}_1\right)$
Case (ii): If $2\mathrm{n}+\mathrm{l}=0\Rightarrow \mathrm{l}=-2\mathrm{n}$
From eq (i), $-6\mathrm{n}+\mathrm{m}+5\mathrm{n}=0\Rightarrow \mathrm{m}=\mathrm{n}$
$\mathrm{l}:\mathrm{m}:\mathrm{n}=-2\mathrm{n}:\mathrm{n}:\mathrm{n}=-2:1:1$
dr’s = (a, b, c) = (–2, 1, 1)
dc’s = (l₂, m₂, n₂) =
$\left.\begin{array}{l}\left(\frac{-2}{\sqrt{(-2{)}^2+1^2+1^2}},\frac{1}{\sqrt{(-2{)}^2+1^2+1^2}},\right.\frac{1}{\sqrt{(-2{)}^2+1^2+1^2}}\\ =\left(\frac{-2}{\sqrt{6}},\frac{1}{\sqrt{6}},\frac{1}{\sqrt{6}}\right)\end{array}\right)$
$\text{ Let }\mathrm{\theta }\text{ is the angle between the lines }$
$\begin{array}{l}\cos \mathrm{\theta }=\left|{\mathrm{l}}_1{\mathrm{l}}_2+{\mathrm{m}}_1{\mathrm{m}}_2+{\mathrm{n}}_1{\mathrm{n}}_2\right|\\ =\left|\frac{-1}{\sqrt{6}}\left(\frac{-2}{\sqrt{6}}\right)+\left(\frac{-2}{\sqrt{6}}\right)\left(\frac{1}{\sqrt{6}}\right)+\frac{1}{\sqrt{6}}\left(\frac{1}{\sqrt{6}}\right)\right|\\ =\left|\frac{2}{6}-\frac{2}{6}+\frac{1}{6}\right|=\frac{1}{6}\\ \therefore \mathrm{\theta }={\cos }^{-1}\left(\frac{1}{6}\right)\end{array}$