Find the direction cosines of two lines which are connected by the relation
$\text{ i) }\mathrm{\ell }-5\mathrm{m}+3\mathrm{n}=0\text{ and }7{\mathrm{\ell }}^2+5{\mathrm{m}}^2-3{\mathrm{n}}^2=0+$
ii) l + m + n = 0 and mn – 2nl – 2lm = 0

i) Given equations are
l - 5m + 3n = 0 .... (1)
7l² +5m² -3n² = 0 .... (2)
From (1), l = 5m – 3n
Substituting l = 5m – 3n in eq (2)
$\begin{array}{l}\Rightarrow 7(5\mathrm{m}-3\mathrm{n}{)}^2+5{\mathrm{m}}^2-3{\mathrm{n}}^2=0\\ \Rightarrow 7\left(25{\mathrm{m}}^2+9{\mathrm{n}}^2-30\mathrm{mn}\right)+5{\mathrm{m}}^2-3{\mathrm{n}}^2=0\\ \Rightarrow 175{\mathrm{m}}^2+63{\mathrm{n}}^2-210\mathrm{mn}+5{\mathrm{m}}^2-3{\mathrm{n}}^2=0\\ \Rightarrow 180{\mathrm{m}}^2-210\mathrm{mn}+60{\mathrm{n}}^2=0\\ \Rightarrow 30\left(6{\mathrm{m}}^2-7\mathrm{mn}+2{\mathrm{n}}^2\right)=0\\ \Rightarrow 6{\mathrm{m}}^2-7\mathrm{mn}+2{\mathrm{n}}^2=0\\ \Rightarrow 6{\mathrm{m}}^2-4\mathrm{mn}-3\mathrm{mn}+2{\mathrm{n}}^2=0\\ \Rightarrow 2\mathrm{m}(3\mathrm{m}-2\mathrm{n})-\mathrm{n}(3\mathrm{m}-2\mathrm{n})=0\\ \Rightarrow (3\mathrm{m}-2\mathrm{n})(2\mathrm{m}-\mathrm{n})=0\\ \Rightarrow 3\mathrm{m}-2\mathrm{n}=0\text{ (or) }2\mathrm{m}-\mathrm{n}=0\\ \Rightarrow 3\mathrm{m}=2\mathrm{n}\text{ (or) }2\mathrm{m}=\mathrm{n}\\ \mathrm{n}=\frac{3\mathrm{m}}{2}\end{array}$
$\text{ Case (i): If }\mathrm{n}=\frac{3\mathrm{m}}{2}$
$\text{ From eq (1), }\mathrm{l}=5\mathrm{m}-3\left(\frac{3\mathrm{m}}{2}\right)=\frac{\mathrm{m}}{2}$
$\mathrm{l}:\mathrm{m}:\mathrm{n}=\frac{\mathrm{m}}{2}:\mathrm{m}:\frac{3\mathrm{m}}{2}=1:2:3\text{; }$
dr’s of first line= (a, b, c) = (1, 2, 3)
dc’s of first line
$\begin{array}{l}=\left(\frac{\mathrm{a}}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2}},\frac{\mathrm{b}}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2}},\frac{\mathrm{c}}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2}}\right)\\ =\left(\frac{1}{\sqrt{1^2+2^2+3^2}},\frac{2}{\sqrt{1^2+2^2+3^2}},\frac{3}{\sqrt{1^2+2^2+3^2}}\right)\\ \left({\mathrm{l}}_1,{\mathrm{m}}_1,{\mathrm{n}}_1\right)=\left(\frac{1}{\sqrt{14}},\frac{2}{\sqrt{14}},\frac{3}{\sqrt{14}}\right)\end{array}$
Case (ii): if n = 2m ;
From equation (1),
l = 5m – 3n = 5m -3(2m) = -m
l : m : n = – m : m : 2m = –1 : 1 : 2
dr’s of 2nd line (a, b, c) = (–1, 1, 2)
dc’s of 2nd line
$=\left(\frac{\mathrm{a}}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2}},\frac{\mathrm{b}}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2}},\frac{\mathrm{c}}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2}}\right)$
$\left.\begin{array}{l}= (\frac{-1}{\sqrt{(-1{)}^2+1^2+2^2}},\frac{1}{\sqrt{(-1{)}^2+1^2+2^2}},\frac{2}{\sqrt{{(-1)}^2+1^2+2^2}} \\ \left({\mathrm{l}}_2,{\mathrm{m}}_2,{\mathrm{n}}_2\right)=\left(\frac{-1}{\sqrt{6}},\frac{1}{\sqrt{6}},\frac{2}{\sqrt{6}}\right)\end{array}\right)$
ii) Given equations are l + m + n = 0 .... (1)
mn – 2nl – 2lm = 0 .... (2)
from (1), l = –m – n
Substituting l = –m – n is eq.(2)
$\begin{array}{l}\Rightarrow \mathrm{mn}-2\mathrm{n}(-\mathrm{m}-\mathrm{n})-2\mathrm{m}(-\mathrm{m}-\mathrm{n})=0\\ \Rightarrow \mathrm{mn}+2\mathrm{mn}+2{\mathrm{n}}^2+2{\mathrm{m}}^2+2\mathrm{mn}=0\\ \Rightarrow 2{\mathrm{n}}^2+5\mathrm{mn}+2{\mathrm{m}}^2=0\\ \Rightarrow 2{\mathrm{n}}^2+4\mathrm{mn}+\mathrm{mn}+2{\mathrm{m}}^2=0\\ \Rightarrow 2\mathrm{n}(\mathrm{n}+2\mathrm{m})+\mathrm{m}(\mathrm{n}+2\mathrm{m})=0\\ \Rightarrow (\mathrm{n}+2\mathrm{m})(2\mathrm{n}+\mathrm{m})=0\\ \Rightarrow \mathrm{n}+2\mathrm{m}=0\text{ (or) }2\mathrm{n}+\mathrm{m}=0\\ \Rightarrow \mathrm{n}=-2\mathrm{m}\text{ (or) }\mathrm{n}=\frac{-\mathrm{m}}{2}\end{array}$
Case (i) : Put n = –2m in (1) $\Rightarrow$l = m
l : m : n = m : m : –2m = 1 : 1 : –2
dr’s = (a, b, c) = (1, 1, –2)
dc’s are (l₁, m₁, n₁) =
$\begin{array}{l}\left(\begin{array}{rl}\frac{1}{\sqrt{1^2+1^2+(-2{)}^2}},\frac{1}{\sqrt{1^2+1^2+(-2{)}^2}}\frac{-2}{\sqrt{1^2+1^2+(-2{)}^2}}& \\ & \end{array}\right)\\ =\left(\frac{1}{\sqrt{6}},\frac{1}{\sqrt{6}},\frac{-2}{\sqrt{6}}\right)\end{array}$
Case(ii) :
$\text{ Put }\mathrm{n}=\frac{-\mathrm{m}}{2}\text{ in (1) }\Rightarrow \mathrm{l}=\frac{-\mathrm{m}}{2}$
$\mathrm{l}:\mathrm{m}:\mathrm{n}=\frac{-\mathrm{m}}{2}:\mathrm{m}:\frac{-\mathrm{m}}{2}=1:-2:1$
d.r’s are (a, b, c) = (1, –2, 1)
d.c’s are (l₂, m₂, n₂) =
$$\begin{gathered} \left(\begin{array}{c}\frac{1}{\sqrt{1^2+(-2{)}^2+1^2}},\frac{-2}{\sqrt{1^2+(-2{)}^2+1^2}},\frac{1}{\sqrt{1^2+(-2{)}^2+1^2}}\\ \end{array}\right) \\ =\left(\frac{1}{\sqrt{6}},\frac{-2}{\sqrt{6}},\frac{1}{\sqrt{6}}\right) \end{gathered}$$