Question

Find the angle between the line whose direction cosines are given by $l+m+n=0$ and $l^2+m^2=n^2$

Answer 1

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

$\Downarrow$

$l+m=-n$

$-(l+m)=n$

put $m=0$ in (1)

So $l=-n$

$(l,m,n)d{r}^{\prime }s=(1,0,-1)=b_1$

put $l=0$ in (1)

so $m=-n$

$(l,m,n)d{r}^{\prime }s=(0,1,-1)=b_2$

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

$\left|b_2\right|=\sqrt{0^2+1^2+(-1{)}^2}=\sqrt{2}$

$l^2+m^2=n^2\to l^2+m^2-n^2=0---\left(2\right)$

substitute '$n$'

$l^2+m^2-[-(l+m){]}^2=0$

$l^2+m^2-l^2-m^2-2lm=0$

$2lm=0$

either $l=0\left(or\right)m=0$

i.e;

$\cos \theta =\frac{b_1.b_2}{\left|b_1\right|\left|b_2\right|}$

$=\frac{1}{\sqrt{2}.\sqrt{2}}$

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

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