Question

Find the direction cosines $l,m,n$ of a line which are connected by the relation $l+m-n=0$ and $2ml-2mn+nl=0$

• A. $\frac{-2}{\sqrt{6}},\frac{1}{\sqrt{6}},\frac{-1}{\sqrt{6}}$
• B. $\frac{2}{\sqrt{6}},\frac{-1}{\sqrt{6}},\frac{1}{\sqrt{6}}$
• C. $\frac{-2}{\sqrt{6}},\frac{-1}{\sqrt{6}},\frac{-1}{\sqrt{6}}$
• D. $\frac{2}{\sqrt{6}},\frac{1}{\sqrt{6}},\frac{1}{\sqrt{6}}$

Answer 1

Answer: (A), (B)

The correct options are
A $\frac{-2}{\sqrt{6}},\frac{1}{\sqrt{6}},\frac{-1}{\sqrt{6}}$
B $\frac{2}{\sqrt{6}},\frac{-1}{\sqrt{6}},\frac{1}{\sqrt{6}}$

Given $l+m-n=0$ ...$\left(1\right)$
$2ml-2mn+ln=0$ ...$\left(2\right)$
From $\left(1\right)$, $n=m+l$
Putting this value of $n$ in equation $\left(2\right)$, we get $2ml-2m(m+l)+l(m+l)=0$
$\Rightarrow 2ml-{2m}^2-2ml+ml+l^2=0\Rightarrow l^2+ml-{2m}^2=0$
$\Rightarrow (l-m)(l+2m)=0$
Either $l=m$ or $l=-2m$

Case I: $l=m$
$\Rightarrow n=m+l=2m$
$\therefore$ DR's are $<1,1,2>.$
$\therefore$ DC's are $\pm \frac{1}{\sqrt{1^2+1^2+2^2}},\pm \frac{1}{\sqrt{1^2+1^2+2^2}},\pm \frac{2}{\sqrt{1^2+1^2+2^2}}$
$\therefore$ DC's are $<\frac{1}{\sqrt{6}},\frac{1}{\sqrt{6}},\frac{2}{\sqrt{6}}>$ or $<\frac{-1}{\sqrt{6}},\frac{-1}{\sqrt{6}},\frac{-2}{\sqrt{6}}>$

Case II: $l=-2m\Rightarrow n=l+m=-2m+m=-m$
$\therefore$ DR's are $-2,1,-1$
DC's are $\Rightarrow \pm \frac{-2}{\sqrt{{(-2)}^2+1^2+{(-1)}^2}},\pm \frac{1}{\sqrt{{(-2)}^2+1^2+{(-1)}^2}},\pm \frac{-1}{\sqrt{{(-2)}^2+1^2+{(-1)}^2}}$
$\Rightarrow$ DC's are $<\frac{-2}{\sqrt{6}},\frac{1}{\sqrt{6}},\frac{-1}{\sqrt{6}}>$ or $<\frac{2}{\sqrt{6}},\frac{-1}{\sqrt{6}},\frac{1}{\sqrt{6}}>$

Hence, options $A$ and $B$ are correct.