Mathematics - Straight Lines Question with Solution | TestHub

Let $l_1,m_1,n_1$ and $l_2,m_2,n_2$ be the d.c of line 1 and 2 respectively, then as given $l_1+m_1+n_1=0$ and $l_2+m_2+n_2=0$ and $l_1{}^2+m_1{}^2-n_1^2=0$ and $l_2^2+m_2^2-n_2^2=0$ $(\because l+m+n=0$ and $l^2+m^2-n^2=0)$ Angle between lines, $\theta$ is $$\cos \theta =l_1{l}_2+m_1{m}_2+n_1{n}_2$$ As given $l^2+m^2=n^2$ and $l+m=-n$ $\Rightarrow (-n{)}^2-2lm=n^2\Rightarrow 2lm=0$ or $lm=0$ So $l_1{m}_1=0,l_2{m}_2=0$ If $l_1=0,m_1\ne 0$ then $l_1{m}_2=0$ If $m_1=0,l_1\ne 0$ then $l_2{m}_1=0$ If $l_2=0,m_2\ne 0$ then $l_2{m}_1=0$ If $m_2=0,l_2\ne 0$ then $l_1{m}_2=0$ Also $l_1{l}_2=0$ and $m_1{m}_2=0$ $$\begin{array}{rl}& l^2+m^2-n^2=l^2+m^2+n^2-2n^2=0\\ & \Rightarrow 1-2n^2=0\Rightarrow n=\pm \frac{1}{\sqrt{2}}\\ & \therefore n_1=\pm \frac{1}{\sqrt{2}},n_2=\pm \frac{1}{\sqrt{2}}\\ & \therefore \cos \theta =\frac{1}{2}\theta =60^{\circ }\text{ (acute angle) }\end{array}$$