If the direction cosines $l, \mathrm{~m}, \mathrm{n}$ of two lines are connected by relations $l-5 m+3 n=0$…
If the direction cosines $l, \mathrm{~m}, \mathrm{n}$ of two lines are connected by relations $l-5 m+3 n=0$ and $7 l^2+5 \mathrm{~m}^2-3 \mathrm{n}^2=0$, then value of $l+\mathrm{m}+\mathrm{n}$ is
$\frac{2}{\sqrt{6}}$ or $\frac{6}{\sqrt{14}}$
$\frac{1}{\sqrt{6}}$ or $\frac{5}{\sqrt{14}}$
$\frac{2}{\sqrt{6}}$ or $\frac{5}{\sqrt{14}}$
$\frac{1}{\sqrt{6}}$ or $\frac{6}{\sqrt{14}}$
Solution
$\begin{aligned}
& l-5 \mathrm{~m}+3 \mathrm{n}=0 \text { and } 7 l^2+5 \mathrm{~m}^2-3 \mathrm{n}^2=0 \\
& \Rightarrow l=5 \mathrm{~m}-3 \mathrm{n} \text { and } 7 l^2=3 \mathrm{n}^2-5 \mathrm{~m}^2
\end{aligned}$
Putting $l=(5 \mathrm{~m}-3 \mathrm{n})$ in $7 l^2=3 \mathrm{n}^2-5 \mathrm{~m}^2$, we get
$\begin{aligned}
& 7(5 m-3 n)^2=3 n^2-5 m^2 \\
& \Rightarrow 7\left(25 m^2-30 m n+9 n^2\right)=3 n^2-5 m^2 \\
& \Rightarrow 180 m^2-210 m n+60 n^2=0 \\
& \Rightarrow 6 m^2-7 m n+2 n^2=0 \\
& \Rightarrow(3 m-2 h)(2 m-n)=0 \\
& \Rightarrow 3 m=2 n \text { or } 2 m=n
\end{aligned}$
If $3 \mathrm{~m}=2 \mathrm{n}$, then $l=\frac{\mathrm{n}}{3}$
$\therefore \quad \frac{\mathrm{m}}{2}=\frac{\mathrm{n}}{3}=\frac{l}{1}=\frac{1}{\sqrt{14}}$
$\therefore \quad l+\mathrm{m}+\mathrm{n}=\frac{6}{\sqrt{14}}$
If $2 \mathrm{~m}=\mathrm{n}$, then $l=\frac{-\mathrm{n}}{2}$
$\therefore \quad \frac{\mathrm{m}}{1}=\frac{\mathrm{n}}{2}=\frac{l}{-1}=\frac{1}{\sqrt{6}}$
$\therefore \quad l+\mathrm{m}+\mathrm{n}=\frac{2}{\sqrt{6}}$
$\therefore \quad$ The possible values of $l+\mathrm{m}+\mathrm{n}$ is $\frac{2}{\sqrt{6}}$ or $\frac{6}{\sqrt{14}}$