Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

Given relations are $l+m+n=0$ and $lm=0$.

From $lm=0$, we have two cases: $l=0$ or $m=0$.

Case 1: $l=0$
Substituting $l=0$ in $l+m+n=0$, we get $m=-n$.
The direction ratios of the first line are proportional to $(0,-n,n)$, which simplifies to $(0,-1,1)$.

Case 2: $m=0$
Substituting $m=0$ in $l+m+n=0$, we get $l=-n$.
The direction ratios of the second line are proportional to $(-n,0,n)$, which simplifies to $(-1,0,1)$.

The angle $\theta$ between the two lines is given by:
$\cos \theta =\frac{|a_1{a}_2+b_1{b}_2+c_1{c}_2|}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}$

Substituting the values:
$\cos \theta =\frac{|(0)(-1)+(-1)(0)+(1)(1)|}{\sqrt{0^2+(-1{)}^2+1^2}\sqrt{(-1{)}^2+0^2+1^2}}$

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

$\theta =60^{\circ }$

Answer: $60^{\circ }$