Question

Find the direction cosine $l,m,n$ of line which are connected by the relations $l+m+n=0.$,$2mn+2ml-nl=0$

Answer 1

Given: $l+m+n=0----\left(1\right)$

$2mn+2ml+nl=0$

Now we have from $\left(1\right),n=-(u+m)$

Putting $m=-(l+m)$ in equation $\left(2\right)$, we get

$-2m(l+m)+2ml+(l+m)l=0$

$\Rightarrow -2ml-2m^2+2ml+l^2+ml$

$\Rightarrow l^2+ml-2m^2=0$

${\left(\frac{l}{m}\right)}^2+\left(\frac{l}{m}\right)-2=0$

Now g$\frac{l}{m}=\frac{l}{m}=\frac{1\pm \sqrt{9}}{2}=1,2$

Now when $\frac{l}{m}=1\Rightarrow l=m$

From $\left(1\right),2l+n=0\Rightarrow n=-2l$

$\therefore l:m:n=1:1:-2$

$\therefore$ Direction ratio of the line are $1,1,-2$

Direction cosiner are

$I=\frac{1}{\sqrt{1^2+1^2+(-2{)}^2}},\frac{1}{\sqrt{1^2+1^2+(-2{)}^2}},\frac{-2}{\sqrt{1^2+1^2+(-2{)}^2}}$

$\Rightarrow \frac{+1}{\sqrt{6}},\frac{1}{\sqrt{6}},\frac{-2}{\sqrt{6}}$ or $\frac{-1}{\sqrt{6}},\frac{-1}{\sqrt{6}},\frac{2}{\sqrt{6}}$

Case $II$ when $\frac{l}{m}=-2\Rightarrow l=-2m$

From $\left(1\right),-2m+m+n=0\Rightarrow n=m$

$\therefore l:m:n=-2m:m:m$

$=-2:1:1$

$\therefore$ Directio ratio of the line are $-2,1,1$

Direction codiner are

$\frac{-2}{\sqrt{-2^2+1^2+1^2}},\frac{-1}{\sqrt{(-2{)}^2+1^2+1^2}},\frac{-1}{\sqrt{(-2{)}^2+1^2+1^2}}$

$\Rightarrow \frac{-2}{\sqrt{6}},\frac{+1}{\sqrt{6}},\frac{1}{\sqrt{6}}$ or $\frac{2}{\sqrt{6}},\frac{-1}{\sqrt{6}},\frac{-1}{\sqrt{6}}$