JEE Main 2013 — Mathematics Coordinate Geometry

Let $l_1, m_1, n_1$l1​,m1​,n1​ and $l_2, m_2, n_2$l2​,m2​,n2​ be the d.c of line 1 and 2 respectively, then as given $l_1+m_1+n_1=0$l1​+m1​+n1​=0 and $l_2+m_2+n_2=0$l2​+m2​+n2​=0 and $l_1{ }^2+m_1{ }^2-n_1^2=0$l1​2+m1​2−n12​=0 and $l_2^2+m_2^2-n_2^2=0$l22​+m22​−n22​=0 $\left(\because l+m+n=0\right.$(∵l+m+n=0 and $\left.l^2+m^2-n^2=0\right)$l2+m2−n2=0) Angle between lines, $\theta$θ is $$\cos \theta=l_1 l_2+m_1 m_2+n_1 n_2$$cosθ=l1​l2​+m1​m2​+n1​n2​ As given $l^2+m^2=n^2$l2+m2=n2 and $l+m=-n$l+m=−n $\Rightarrow(-n)^2-2 l m=n^2 \Rightarrow 2 l m=0$⇒(−n)2−2lm=n2⇒2lm=0 or $l m=0$lm=0 So $l_1 m_1=0, l_2 m_2=0$l1​m1​=0,l2​m2​=0 If $l_1=0, m_1 \neq 0$l1​=0,m1​=0 then $l_1 m_2=0$l1​m2​=0 If $m_1=0, l_1 \neq 0$m1​=0,l1​=0 then $l_2 m_1=0$l2​m1​=0 If $l_2=0, m_2 \neq 0$l2​=0,m2​=0 then $l_2 m_1=0$l2​m1​=0 If $m_2=0, l_2 \neq 0$m2​=0,l2​=0 then $l_1 m_2=0$l1​m2​=0 Also $l_1 l_2=0$l1​l2​=0 and $m_1 m_2=0$m1​m2​=0 $$\begin{aligned} & l^2+m^2-n^2=l^2+m^2+n^2-2 n^2=0 \\ & \Rightarrow 1-2 n^2=0 \Rightarrow n=\pm \frac{1}{\sqrt{2}} \\ & \therefore n_1=\pm \frac{1}{\sqrt{2}}, n_2=\pm \frac{1}{\sqrt{2}} \\ & \therefore \cos \theta=\frac{1}{2} \theta=60^{\circ} \text { (acute angle) } \end{aligned}$$​l2+m2−n2=l2+m2+n2−2n2=0⇒1−2n2=0⇒n=±2​1​∴n1​=±2​1​,n2​=±2​1​∴cosθ=21​θ=60∘ (acute angle) ​