Question

Find the angle between the lines whose direction cosines satisfy the equations $\ell +m+n=0,$ ${\ell }^2+m^2-n^2=0$

Answer 1

Given

$l+m+n=\mathrm{0........................}\left(1\right)$

$and$

$l^2+m^2-n^2=\mathrm{0.....................}\left(2\right)$

By equation (1)

$l+m+n=0$

$l+m=-n$

$-(l+m)=n$

$n=-(l+m)$ put in equation (2)

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

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

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

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

$then$

$-2lm=0$

$lm=0$

Either $l=0$ or $m=0$

Let us put $l=0$ in equation (1), and we get

$l+m+n=0$

$0+m+n=0$

$m=-n$

Direction ratios $(l,m,n)=(0,1,-1)$

Let us put $m=0$ in equation (1) and we get

$l+m+n=0$

$l+0+n=0$

$l=-n$

Direction ratios $(l,m,n)=(1,0,-1)$

We know that angle between of two given line

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

First solve $b_1.b_2$

Then

$b_1.b_2=(1,0,-1)(0,1,-1)$

$b_1.b_2=0+0+1$

$b_1.b_2=1$

And $\left|b_1\right|=\sqrt{0^2+1^2+{(-1)}^2}=\sqrt{2}$

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

Now substituting the value in

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

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

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

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

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

Hence it is required solution