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Q.
If the direction cosines of two lines are such that $l+m+n=0, l^{2}+m^{2}-n^{2}=0$, then the angle between them is
TS EAMCET 2018
Solution:
Given that,
$l+m+n=0\,\,\,\,\,\,\,\dots(i)$
$\Rightarrow \, l+m=-n$
$\Rightarrow \, -(l+m)=n$
and $( l^{2}+m^{2}-n^{2}=0\,\,\,\,\,\,\dots(ii)$
Let us substitute per 'n' in Eq. (ii), we get
$\Rightarrow \, l^{2}+m^{2}-l^{2}-m^{2}-2 m l=0$
$ \Rightarrow \,2ml=0$
i.e. either $l=0$ or $m=0$
Put $m=0$ in Eq. (i).
If $m=0$, then $l=-n$
Direction ratios $(l, m, n)=(1,0,-1)$
Put $l=0$ in Eq. (i), we get $m=-n$
Direction ratios $(l, m, n)=(0,1,-1)$
Here, $b_{1} \cdot b_{2}=(1,0,-1) \cdot(0,1,-1)=0+0+1 $
$\therefore \, \left(b_{1}\right)=\sqrt{0^{2}+1^{2}+1^{2}}=\sqrt{2}$
and $\left(b_{2}\right)=\sqrt{0^{2}+1^{2}+1^{2}}=\sqrt{2}$
$\therefore \, \cos \theta=\frac{ b _{1} \cdot b _{2}}{\left( b _{1}\right)\left( b _{2}\right)}$
$ \Rightarrow \, \frac{1}{\sqrt{2} \cdot \sqrt{2}}=\frac{1}{2}$
$\Rightarrow \, \theta=\frac{\pi}{3}$