Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

Given equations are $\ell +m+n=0$ and ${\ell }^2-m^2-n^2=0$.

From the first equation, $\ell =-(m+n)$.

Substituting this into the second equation:

$(-(m+n){)}^2-m^2-n^2=0$

$m^2+n^2+2mn-m^2-n^2=0$

$2mn=0\Rightarrow mn=0$

This gives two cases: $m=0$ or $n=0$.

Case 1: When $m=0$, $\ell =-n$.

The direction ratios of the first line are proportional to $(-n,0,n)$, which can be written as $(-1,0,1)$.

Case 2: When $n=0$, $\ell =-m$.

The direction ratios of the second line are proportional to $(-m,m,0)$, which can be written as $(-1,1,0)$.

The angle $\theta$ between the two lines is given by:

$\cos \theta =\left|\frac{a_1{a}_2+b_1{b}_2+c_1{c}_2}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}\right|$

$\cos \theta =\left|\frac{(-1)(-1)+(0)(1)+(1)(0)}{\sqrt{(-1{)}^2+0^2+1^2}\sqrt{(-1{)}^2+1^2+0^2}}\right|$

$\cos \theta =\frac{1}{\sqrt{2}\sqrt{2}}=\frac{1}{2}$

$\theta =\frac{\pi }{3}$

Answer: $\frac{\pi }{3}$