Mathematics - 3D-Coordinate Geometry Question with Solution | TestHub

Given equations of direction cosies

$2\ell +2m-n=0 ...............\left(i\right)$
$mn+n\ell +\ell m=0..............\left(ii\right)$
$\ell m+n(\ell +m)=0$

From equation $\left(i\right)$

$n=2\left(\ell +m\right)$
$\ell m+2(\ell +m{)}^2=0$
$2{\ell }^2+2m^2+5\ell m=0$

Dividing by $m^2$ on both sides
$2{\left(\frac{\ell }{m}\right)}^2+2+5\left(\frac{\ell }{m}\right)=0$
Let $\frac{\ell }{m}=t$
$2t^2+5t+2=0$

$2t^2+4t+t+2=0$

$\left(t+2\right)\left(2t+1\right)=0$
$t=-2,-\frac{1}{2}$

Case 1

$\begin{array}{r}\frac{\ell }{m}=-\frac{1}{2}\end{array}$

$\begin{array}{r}m=-2\ell \end{array}$, $n=-2\ell$

$(\ell ,-2\ell ,-2\ell )\Rightarrow (1,-2,-2)$

Case 2

$\frac{\ell }{m}=-2$

$\ell =-2m, n=-2m$

$\left(-2m,m,-2m\right)$$\Rightarrow \left(-2,1,-2\right)$

$\cos \theta =\frac{1\times \left(-2\right)+\left(-2\right)\times 1+\left(-2\right)\times \left(-2\right)}{\sqrt{1^2+{\left(-2\right)}^2+{\left(-2\right)}^2}\sqrt{{\left(-2\right)}^2+1^2+{\left(-2\right)}^2}}$

$\cos \theta =\frac{-2-2+4}{9}=0$

$\theta =\frac{\pi }{2}$